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252y0551s 10/28/05 (Open in ‘Print Layout’ format)
ECO252 QBA2 Name _Combined Key______
FIRST HOUR EXAM Hour of class registered _____
October 17-18 2005 Class attended if different ____
This key is extremely long and shows most of what I did to put together the exam.
Show your work! Make Diagrams! Exam is normed on 50 points. Answers without reasons are not
usually acceptable.
Version 1
I. (8 points) Do all the following. If you do not use the standard Normal table, explain!
x ~ N 2.5, 6 Make diagrams. For x draw a Normal curve with a vertical line in the center at 2.5. For z
draw a Normal curve with a vertical line in the center at zero.
19.5 2.5 3 2.5
1. P19.5 x 3 P z P3.67 z 0.08
6 6
P3.67 z 0 P0 z 0.08 .4999 .0319 .5318
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0 2.5 2.5 2.5
2. P0 x 2.50 P z P0.42 z 0 .1628
6 6
10.22 2.5
3. Px 10.22 P z Pz 1.29 Pz 0 P0 z 1.29 .5 .4015 .0985
6
2
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4. x.06 There is no excuse for not being able to do this at this point, since you should have done it 3 times
in the take-home.
To find z.06 make a Normal diagram for z showing a mean at 0 and 50% above 0, divided into 6% above
z.06 and 44% below z.04 . So P0 z z.06 .4400 The closest we can come is P0 z 1.55 .4394
or P0 z 1.56 .4406 . So use z .06 1.555 . x.06 z.06 2.5 1.5556 11.83
Version 2
I. (8 points) Do all the following. If you do not use the standard Normal table, explain!
x ~ N 4.5, 6 Make diagrams. For x draw a Normal curve with a vertical line in the center at 4.5. For z
draw a Normal curve with a vertical line in the center at zero.
17.5 4.5 3 4.5
1. P17.5 x 3 P z P3.67 z 0.25
6 6
P3.67 z 0 P0.25 z 0 .4999 .0987 .4012
3
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4.5 4.5 7 4.5
2. P4.5 x 7.00 P z P0 z 0.42 .1628
6 6
10.22 4.5
3. Px 10.22 P z Pz 0.95 Pz 0 P0 z 0.95 .5 .3289 .1711
6
4
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
4. x.06 There is no excuse for not being able to do this at this point, since you should have done it 3 times
in the take-home.
To find z.06 make a Normal diagram for z showing a mean at 0 and 50% above 0, divided into 6% above
z.06 and 44% below z.04 . So P0 z z.06 .4400 The closest we can come is P0 z 1.55 .4394
or P0 z 1.56 .4406 . So use z .06 1.555 . x.06 z.06 4.5 1.5556 13.83
Generation of solutions and graphs for Normal distribution using Minitab
————— 10/18/2005 7:35:09 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > WOpen "C:\Documents and Settings\rbove\My Documents\Minitab\252x0505-
4.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0505-4.MTW'
Worksheet was saved on Tue Oct 18 2005
Results for: 252x0505-4.MTW
MTB > let c6 = c3-c1 #computing values of z.
MTB > let c7 = c4-c1
MTB > let c6 = c6/c2
MTB > let c7 = c7/c2
MTB > print c1-c5 #original data
Data Display
Row mn sd ll ul C5
1 2.5 6 -19.50 3.0 0
2 2.5 6 0.00 2.5 0
3 2.5 6 10.22 100.0 2
4 4.5 6 -17.50 3.0 0
5 4.5 6 4.50 7.0 0
6 4.5 6 10.22 100.0 2
MTB > print c3 c4 c6 c7 #values of x and z
Data Display
Row ll ul zl zu
1 -19.50 3.0 -3.66667 0.0833
2 0.00 2.5 -0.41667 0.0000
3 10.22 100.0 1.28667 16.2500
4 -17.50 3.0 -3.66667 -0.2500
5 4.50 7.0 0.00000 0.4167
6 10.22 100.0 0.95333 15.9167
MTB > let c6 = 100*c6 #rounding z to 2 places to right of decimal point
MTB > round c6 c6
MTB > let c7 = 100*c7
MTB > round c7 c7
MTB > let c6 = c6/100
MTB > let c7 = c7/100
MTB > print c3 c4 c6 c7 #value of x and z
Data Display
Row ll ul zl zu
1 -19.50 3.0 -3.67 0.08
2 0.00 2.5 -0.42 0.00
3 10.22 100.0 1.29 16.25
4 -17.50 3.0 -3.67 -0.25
5 4.50 7.0 0.00 0.42
5
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6 10.22 100.0 0.95 15.92
MTB > stack c3 c6 c3 #putting values of z in c3 and c4
MTB > stack c4 c7 c4
MTB > stack c5 c5 c5
MTB > erase c6 c7
MTB > print c1-c5 #Setup for normarea6c
Data Display
Row mn sd ll ul C5
1 2.5 6 -19.50 3.00 0
2 2.5 6 0.00 2.50 0
3 2.5 6 10.22 100.00 2
4 4.5 6 -17.50 3.00 0
5 4.5 6 4.50 7.00 0
6 4.5 6 10.22 100.00 2
7 0.0 1 -3.67 0.08 0
8 0.0 1 -0.42 0.00 0
9 0.0 1 1.29 16.25 2
10 0.0 1 -3.67 -0.25 0
11 0.0 1 0.00 0.42 0
12 0.0 1 0.95 15.92 2
MTB > %normarea6c #call to macro
Executing from file: normarea6c.MAC
Executing from file: NormArea6.MAC
...working... #Graph stored in jpg format as graph 20505-401 and copied
#from storage into document.
Normal Curve Area
...working... #402
Normal Curve Area
6
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
...working... #403
Normal Curve Area
...working... #404
Normal Curve Area
...working... #405
Normal Curve Area
7
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
...working... #406
Normal Curve Area
...working... #407
Normal Curve Area
...working... #408
Normal Curve Area
8
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
...working... #409
Normal Curve Area
...working... #410
Normal Curve Area
...working... #411
Normal Curve Area
9
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...working... #412
Normal Curve Area
MTB >
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II. (5 points-2 point penalty for not trying part a.) A dealer wishes to test the manufacture’s claim that the
Toyota Caramba gets 28 miles per gallon. The data below is found. (Recomputing what I’ve done for you
is a great way to waste time.)
a. Compute the sample standard deviation, s , of the miles per gallon. Show your work! (2)
b. Is the population mean significantly different (using a 95% confidence level) from 28 mpg?
Show your work! (3)
c. (Extra Credit) Find an approximate p value for your null hypothesis. (2)
Version 1
Row x x2 Solution: a. Compute the sample
1 28.82 830.592 standard deviation, s , of the miles per
2 23.71 562.164 gallon. Show your work!(2)
3
4
29.25
25.16
855.563
633.026 x
x 264.02
26.402
5 27.60 761.760 n 10
6
7
27.41
27.23
751.308
741.473 s2
x 2 nx 2 7018.253 1026.4022
8 26.42 698.016 n 1 9
9 21.75 473.063 47.59696
5.28855
10 26.67 711.289 9
sum 264.02 7018.253
s 5.28855 2.299685.
Note that excessive rounding can throw this answer way off. Using x 26 , I got
7018 10(26) 2 7018 6760 258
s2 28.667 and s 5.3541.
9 9 9
b. Is the population mean significantly different (using a 95% confidence level) from 28 mpg?
Show your work! (3)
9
Given: x 26.402 s 2.2997, n 7 and .05. DF n 1 9 and t .025 2.262 .
We are testing H 0 : 28 . Use only one of the following!
s 2.2997 5.28855
Confidence Interval: s x 0.52855 0.7272.
n 10 10
x t 2 s x 26.402 2.262 0.7272 26.402 1.6449 or 24.757 to 28.047.
Since 28 is on the confidence interval, it is not significantly different from the sample mean.
s 2.2997 5.28855
Critical Value: s x 0.52855 0.7272.
n 10 10
x cv t 2 s x 28 2.262 0.7272 28 1.6449 or 26.36 to 29.65.
Since 26.402 is between the critical values, it is not significantly different from the population
mean.
s 2.2997 5.28855
Test Ratio: s x 0.52855 0.7272.
n 10 10
x 0 26.402 28
t 2.197 . The ‘accept’ zone is between -2.262 and 2.262. Since -2.197
sx 0.7272
is between these values, do not reject the null hypothesis.
c. (Extra Credit) Find an approximate p value for your null hypothesis. (2)
9
2.197 is between t .025 2.262 and t .05 1.833 . Since pval 2Pt 2.197 , it is between .05
9
and .10.
————— 10/12/2005 6:39:53 PM ————————————————————
Welcome to Minitab, press F1 for help.
11
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
MTB > WOpen "C:\Documents and Settings\rbove\My Documents\Minitab\252x0505-2.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0505-2.MTW'
Worksheet was saved on Wed Oct 12 2005
Version 1
Results for: 252x0505-2.MTW
MTB > sum x1
Sum of x1
Sum of x1 = 264.02
MTB > ssq x1
Sum of Squares of x1
Sum of squares (uncorrected) of x1 = 7018.253
MTB > print c1 c3
Data Display
Row x1 x1sq
1 28.82 830.592
2 23.71 562.164
3 29.25 855.563
4 25.16 633.026
5 27.60 761.760
6 27.41 751.308
7 27.23 741.473
8 26.42 698.016
9 21.75 473.063
10 26.67 711.289
264.02 7018.253
MTB > describe c1
Descriptive Statistics: x1
Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum
x1 10 0 26.402 0.727 2.300 21.750 24.798 26.950 27.905 29.250
MTB > Onet c1.
One-Sample T: x1
Variable N Mean StDev SE Mean 95% CI
x1 10 26.4020 2.2997 0.7272 (24.7569, 28.0471)
Version 2
Row x x2
1 23.56 555.074
2 26.09 680.688
3 27.55 759.003
4 26.92 724.686
5 29.20 852.640
6 29.02 842.160
7 22.48 505.350
8 27.45 753.503
9 25.90 670.810
10 25.81 666.156
sum 263.98 7010.070
a. Compute the sample standard deviation, s , of the miles per gallon. Show your work! (2)
12
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
x
x 263.98 26.398 s2
x 2
nx 2
7010.07 1026.3982
n 10 n 1 9
41.52596
4.61400 s 4.61400 2.148021 Note that excessive rounding can throw this
9
7010 10(26) 2 7010 6760 250
answer way off. Using x 26 , I got s 2 27.778 and
9 9 9
s 5.3541.
b. Is the population mean significantly different (using a 95% confidence level) from 28 mpg? Show your
work! (3)
9
Given: x 26.398 s 2.14802, n 7 and .05. DF n 1 9 and t .025 2.262 .
We are testing H 0 : 28 . Use only one of the following!
s 2.14802 4.61400
Confidence Interval: s x 0.461400 0.6793.
n 10 10
x t 2 s x 26.398 2.262 0.6793 26.398 1.5366 or 21.861 to 27.935.
Since 28 is not on the confidence interval, it is significantly different from the sample mean.
s 2.14802 4.61400
Critical Value: s x 0.461400 0.6793.
n 10 10
x cv t 2 s x 28 2.262 0.6793 28 1.5366 or 26.46 to 29.53.
Since 26.398 is not between the critical values, it is significantly different from the population
mean.
s 2.14802 4.61400
Test Ratio: s x 0.461400 0.6793.
n 10 10
x 0 26.398 28
t 2.358 . The ‘accept’ zone is between -2.262 and 2.262. Since -2.197
sx 0.6793
is not between these values, reject the null hypothesis.
c. (Extra Credit) Find an approximate p value for your null hypothesis. (2)
9
2.358 is between t .025 2.262 and t .01 2.821 . Since pval 2Pt 2.358 , it is between .02
9
and .05.
MTB > sum c2
Sum of x2
Sum of x2 = 263.98
MTB > ssq c2
Sum of Squares of x2
Sum of squares (uncorrected) of x2 = 7010.07
MTB > describe c2
Descriptive Statistics: x2
Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum
x2 10 0 26.398 0.679 2.148 22.480 25.248 26.505 27.918 29.200
MTB > Onet c2.
One-Sample T: x2
Variable N Mean StDev SE Mean 95% CI
x2 10 26.3980 2.1480 0.6793 (24.8614, 27.9346)
13
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
III. Do all of the following problems (2 each unless marked otherwise adding to 18+ points). Show your
work except in multiple choice questions. (Actually – it doesn’t hurt there either.) If the answer is ‘None
of the above,’ put in the correct answer.
Version 1
1. For sample sizes less than 10, the sampling distribution of the mean will be approximately
normally distributed
a) Regardless of the shape of the population.
b) *Only if the shape of the population is symmetrical.
c) Only if the standard deviation of the samples are known.
d) Only if the sample is normally distributed.
2. The width of a confidence interval estimate for a proportion will be
a) Narrower for 99% confidence than for 95% confidence.
b) Wider for a sample size of 100 than for a sample size of 50.
c) *Narrower for 90% confidence than for 95% confidence.
d) Narrower when the sample proportion is 0.50 than when the sample proportion is 0.20
3. Which of the following would be an appropriate null hypothesis?
a) *The mean of a population is equal to 55.
b) The mean of a sample is equal to 55.
c) The mean of a population is greater than 55.
d) Only (a) and (c) are true.
4. A Type II error is committed when
a) We reject a null hypothesis that is true.
b) We don't reject a null hypothesis that is true.
c) We reject a null hypothesis that is false.
d) *We don't reject a null hypothesis that is false.
5. If we are performing a two-tailed test of whether = 100, the power of the test in detecting a shift
of the mean to 105 will be ________ its power detecting a shift of the mean to 94.
a) *Less than 94 is further from 100 than 105.
b) Greater than
c) Equal to
e) Not comparable to [10]
Version 2
1. For sample sizes greater than 100, the sampling distribution of the mean will be approximately
normally distributed
a) *Regardless of the shape of the population.
b) Only if the shape of the population is symmetrical.
c) Only if the standard deviation of the samples are known.
d) Only if the population is normally distributed.
2. When determining the sample size for a proportion for a given level of confidence and sampling
error, the closer to 0.50 that p is estimated to be the __________ the sample size required.
a) Smaller
b) *Larger
c) Sample size is not affected.
d) The effect cannot be determined from the information given .
3. Which of the following would be an appropriate null hypothesis?
a) The population proportion is less than 0.65.
14
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
b) The sample proportion is less than 0.65.
c) *The population proportion is at most 0.65.
d) The sample proportion is at most 0.65.
4. A Type I error is committed when
a) *We reject a null hypothesis that is true.
b) We don't reject a null hypothesis that is true.
c) We reject a null hypothesis that is false.
d) We don't reject a null hypothesis that is false
5. If we are performing a two-tailed test of whether = 100, the power of the test in detecting a shift
of the mean to 105 will be ________ its power detecting a shift of the mean to 96.
a) *Less than 96 is closer than 105 to 100.
b) Greater than
c) Equal to
d) Not comparable to
Version 1
6. From a sample of 100 students we find that the mean expenditure for books is $316.40 with a
standard deviation of $43.20. You are asked to test whether the (population) mean expenditure is
above $302 using a 10% significance level.
a) What are the null and alternative hypotheses? (2)
b) What is the ‘rejection zone’ (in terms of x )? (2)
c) What is your conclusion? (2)
Solution: This is basically the revised version of Exercise 9.54 [9.52 in 9th] (9.50 in 8th edition)
that I warned you about.
a) Since the problem asks if the mean is above $302 302 , and this does not contain an equality,
it must be an alternate hypothesis. Our hypotheses are H 0 : 302 , (The average cost of
textbooks per semester at a large university is no more than $302.) and H1 : 302 (The
average cost of textbooks per semester at a large university is more than $302.) This is a
right-sided test.
b) Given: 0 302.00, s 43.20, n 100, df n 1 99, x 316.40 and .10. So
4.320. Note that tn 1 t ..99 1.290. We need a critical value for x
s 43.20
sx 10
n 100
above $302. Common sense says that if the sample mean is too far above $302, we will not
believe H 0 : 302 . The formula for a critical value for the sample mean is
x t n1 s , but we want a single value above 302, so use x t n 1 s
cv 0 x cv 0 x
2
302.00 1.2904.320 307.57 . Make a diagram showing an almost Normal curve with
a mean at 302 and a shaded 'reject' zone above 305.57.
c) Since x 316.40 is in the ‘reject’ zone, we reject the null hypothesis and conclude that the mean is
above $302.
Version 2
6. From a sample of 100 students we find that the mean expenditure for books is $316.40 with a
sample standard deviation of $43.20. You are asked to test whether the (population) mean
expenditure is above $314 using a 10% significance level.
a) What are the null and alternative hypotheses? (2)
b) What is the ‘rejection zone’ (in terms of x ) (2)
c) What is your conclusion? (2) [16]
Solution: This is basically the revised version of Exercise 9.54 [9.52 in 9th] (9.50 in 8th edition)
that I warned you about.
15
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
a) Since the problem asks if the mean is above $314 314 , and this does not contain an equality,
it must be an alternate hypothesis. Our hypotheses are H 0 : 314 , (The average cost of
textbooks per semester at a large university is no more than $314.) and H1 : 314 (The
average cost of textbooks per semester at a large university is more than $314.) This is a
right-sided test.
b) Given: 0 314.00, s 43.20, n 100, df n 1 99, x 316.40 and .10. So
4.320. Note that tn 1 t ..99 1.290. We need a critical value for x
s 43.20
sx 10
n 100
above $314. Common sense says that if the sample mean is too far above $314, we will not
believe H 0 : 314 . The formula for a critical value for the sample mean is
x t n1 s , but we want a single value above 314, so use x t n 1 s
cv 0 x cv 0 x
2
314.00 1.2904.320 319.57 . Make a diagram showing an almost Normal curve with
a mean at 314 and a shaded 'reject' zone above 319.57.
c) Since x 316.40 is not in the ‘reject’ zone, we do not reject the null hypothesis and cannot
conclude that the mean is above $314.
Version 1
7. From a sample of 12 students we find that the mean expenditure for books is $316.40 with a
sample standard deviation of $43.20. You are asked to test whether the (population) mean
expenditure is below some number. You compute a t ratio and find that it is 1.645. The p-value is
a) Exactly .05
b) Between 1.60 and 1.80
c) Between .80 and .90
d) Between .10 and .20
e) Between .05 and .10
f) * None of the above – provide a more suitable answer. [18]
Explanation: For this see the writeup that I announced two weeks ago. If our alternate hypothesis
is that the mean expenditure is below some number, we want Pt 1.645 . There are
n 1 11 degrees of freedom. The table says that t .11 1.363 and t .11 1.796 . Thus we can see
10 05
that .05 Pt 1.645 .10 , and we can conclude that .90 Pt 1.645 .95 .
8. An employer states that the median wage in a factory is $45000. You believe that it is lower. From
a sample of 300 employees you find that 125 have wages above $45000. The median for the
sample is $40000, the mean is $46273 and the sample standard deviation is $4001. The test should
reflect you beliefs and, if you use p in a hypothesis, you must define it. Your hypotheses are (3)
H 0 : 45000
a)
H 1 : 45000
H 0 : 45000
b)
H 1 : 45000
H 0 : 45000
c)
H 1 : 45000
H 0 : 45000 H 0 : p .5
d) and
H 1 : 45000 H 1 : p .5
H 0 : 45000 H 0 : p .5
e) and
H 1 : 45000 H 1 : p .5
16
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
H 0 : 45000 H 0 : p .5
f) and
H 1 : 45000 H 1 : p .5
H 0 : 45000 H 0 : p .5
g) and
H 1 : 45000 H 1 : p .5
H 0 : 45000 H 0 : p .5
h) and
H 1 : 45000 H 1 : p .5
H 0 : 45000 H 0 : p .5
i) and
H 1 : 45000 H 1 : p .5 [21]
From the outline.
Hypotheses about Hypotheses about a proportion
a median If p is the proportion If p is the proportion
above 0 below 0
H 0 : 0 H 0 : p .5 H 0 : p .5
H 1 : 0 H 1 : p .5 H 1 : p .5
H 0 : 0 H 0 : p .5 H 0 : p .5
H 1 : 0 H 1 : p .5 H 1 : p .5
H 0 : 0 H 0 : p .5 H 0 : p .5
H 1 : 0 H 1 : p .5 H 1 : p .5
9. (Extra credit) An employer states that the median wage in a factory is $45000. You believe that it
is lower. From a sample of 300 employees you find that 125 have wages above $45000. The median
for the sample is $40000, the mean is $46273 and the sample standard deviation is $4001. The test
should reflect you beliefs and, if you use p in a hypothesis, you must define it. Finish the problem. (3)
Version 2:
7. From a sample of 10 students we find that the mean expenditure for books is $316.40 with a
sample standard deviation of $43.20. You are asked to test whether the (population) mean
expenditure is below some number. You compute a t ratio and find that it is 1.960. The p-value is
a) Between .025 and .05
b) Between .05 and .10
c) Between .95 and .975
d) Between 1.90 and 1.95
e) Exactly .025
f) None of the above – provide a more suitable answer. [18]
Explanation: For this see the writeup that I announced two weeks ago. If our alternate hypothesis
is that the mean expenditure is below some number, we want Pt 1.960 . There are
n 1 9 degrees of freedom. The table says that t .9 2.262 and t .9 1.833 . Thus we can see
025 05
that .025 Pt 1.960 .05 , and we can conclude that .95 Pt 1.645 .975 .
8. An employer states that the median wage in a factory is $45000. You believe that it is lower. From
a sample of 300 employees you find that 170 have wages below $45000. The median for the
sample is $41000, the mean is $47273 and the sample standard deviation is $4051. The test should
reflect you beliefs and, if you use p in a hypothesis, you must define it. Your hypotheses are (3)
H 0 : 45000
a)
H 1 : 45000
17
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
H 0 : 45000
b)
H 1 : 45000
H 0 : 45000
c)
H 1 : 45000
H 0 : 45000 H 0 : p .5
d) and
H 1 : 45000 H 1 : p .5
H 0 : 45000 H 0 : p .5
e) and
H 1 : 45000 H 1 : p .5
H 0 : 45000 H 0 : p .5
f) and
H 1 : 45000 H 1 : p .5
H 0 : 45000 H 0 : p .5
g) and
H 1 : 45000 H 1 : p .5
H 0 : 45000 H 0 : p .5
h) and
H 1 : 45000 H 1 : p .5
H 0 : 45000 H 0 : p .5
i) and
H 1 : 45000 H 1 : p .5 [21]
From the outline.
Hypotheses about Hypotheses about a proportion
a median If p is the proportion If p is the proportion
above 0 below 0
H 0 : 0 H 0 : p .5 H 0 : p .5
H 1 : 0 H 1 : p .5 H 1 : p .5
H 0 : 0 H 0 : p .5 H 0 : p .5
H 1 : 0 H 1 : p .5 H 1 : p .5
H 0 : 0 H 0 : p .5 H 0 : p .5
H 1 : 0 H 1 : p .5 H 1 : p .5
9. (Extra credit) An employer states that the median wage in a factory is $45000. You believe that it
is lower. From a sample of 300 employees you find that 170 have wages below $45000. The median
for the sample is $41000, the mean is $47273 and the sample standard deviation is $4051. The test
should reflect your beliefs and, if you use p in a hypothesis, you must define it. Finish the problem.
(3)
Version 1
10. (Extra Credit – Nasty but not that hard – problem due to Prem S. Mann.) Use .01. A quality
control inspector is checking machines that are supposed to produce packages of cookies with a mean
weight of 32 oz and a standard deviation of .015 oz. She takes a sample of 25 packages and finds a
sample standard deviation of .029 oz. Do a one-sided test on the variance to see if the machine is out
of control. Do the test by finding a critical value for the sample variance (3) (You can do it in a more
familiar way, but you won’t get as much credit!) [27]
Solution: H 0 : .015 H 1: .015 , n 25 , s .029 and .01. Table 3 says
.2 .5 2 0
5
2
n 1 n 1
s cv
2
, where the 2 , for a 2-sided test would be 2 or 2 1 . Since this is a
n 1 2 2
right sided test we want a critical value above .015, we take the higher value of the two and change it
18
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n1 24
to 2 2 .01 42.9798 . So s cv
2 42.9798.0152 0.0004029. This means that
24
s cv 0.0004029 .02007 . This is a right-side test, so that the ‘reject’ region is the area under the
curve to the right of 0.02007. Since s .029 is above this value, reject the null hypothesis and fix the
machine.
Version 2
10. (Extra Credit – Nasty but not that hard – problem due to Prem S. Mann.) Use .01. A quality
control inspector is checking machines that are supposed to produce packages of cookies with a mean
weight of 32 oz and a standard deviation of .015 oz. She takes a sample of 40 packages and finds a
sample standard deviation of .029 oz. Do a one-sided test on the variance to see if the machine is out
of control. Do the test by finding a critical value for the sample variance (3) (You can do it in a more
familiar way, but you won’t get as much credit!) [27]
Solution: H 0 : .015 H 1: .015 , n 40 , s .029 and .01. Table 3 says
2DF . If we are doing a right sided test we want a critical value above .015, we use
s cv
z 2 2DF
.0158.84176
2 DF 239 78 8.84176 s cv
0.13222
z z.01 2.372 . .0203.
2.327 8.84176 6.5148
This is a right-side test, so that the ‘reject’ region is the area under the curve to the right of 0.0203.
Since s .029 is above this value, reject the null hypothesis and fix the machine.
19
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ECO252 QBA2
FIRST EXAM
October 17-18 2005
TAKE HOME SECTION
-
Name: _________________________
Student Number and class: _________________________
IV. Do sections adding to at least 20 points - Anything extra you do helps, and grades wrap around) .
Show your work! State H 0 and H 1 where appropriate. You have not done a hypothesis test unless
you have stated your hypotheses, run the numbers and stated your conclusion. (Use a 95% confidence
level unless another level is specified.) Answers without reasons are not usually acceptable. Neatness
counts!
1. (Prem S. Mann - modified)
Exhibit T1: According to a 1992 survey, 45% of the American population would support higher taxes to
pay for health insurance. A state government is considering offering a health insurance plan and took a
survey of 400 residents and found that 50% would support higher taxes. Use this result to test whether
the results of the 1992 survey apply in the state. Use a 99% confidence level.
Personalize these results as follows. Change 45% by replacing 5 by the last digit of your student number.
We will call your result the ‘proportion of interest.’
(Example: Seymour Butz’s student number is 976512, so he changes 45% to 42%; 42% is his proportion of
interest.)
a. State the null and alternative hypotheses in each case and find a critical value for each case. What is the
‘reject’ region? Compute a test ratio and find a p-value for the hypothesis in each case. (12)
(i) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is below the proportion of interest
(ii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is above the proportion of interest.
(iii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is equal to the proportion of interest. [12]
b. (Extra credit) Find the power of the test if the true proportion is 50% and:
(i) The alternate hypothesis is that the fraction of people is above the proportion of interest. (2)
(ii) The alternate hypothesis is that the fraction of people does not equal the proportion of interest. (2)
c. Assuming that the proportion of interest is correct, is the sample size given above adequate to find the
true proportion within .005 (1/2 of 1%)? (Don’t say yes or no without calculating the size that you actually
need!) (2)
d. If the alternate hypothesis is that the fraction of people is above the proportion of interest, create an
appropriate confidence interval for the hypothesis test. (2) [14]
Solution: a) From the formula table we have:
Interval for Confidence Hypotheses Test Ratio Critical Value
Interval
Proportion p p z 2 s p H 0 : p p0 p p0 p cv p 0 z 2 p
z
pq H1 : p p0 p p0 q0
sp p
n n
q 1 p q0 1 p0
20
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p0 q0 .40.60
Version 0 a) p 0 .40 p .0006 .024495
n 400
(i) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is below the proportion of interest
H 0 : p .40, H 1 : p .40, n 400, p .50 and .01. z.01 2.327 .
The critical value must be below .40. p cv p 0 z p .40 2.327.024495 = .3430. The ‘reject’ zone is
.50 .40
below .3430. pval P p .50 P z Pz 4.08 1
.024495
(ii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is above the proportion of interest.
H 0 : p .40, H 1 : p .40, n 400, p .50 and .01. z.01 2.327 .
The critical value must be above .40. p cv p 0 z p .40 2.327.024495 = .4570 . The ‘reject’ zone
.50 .40
is above .4570. pval P p .50 P z Pz 4.08 0
.024495
(iii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is equal to the proportion of interest. [12]
H 0 : p .40, H 1 : p .40, n 400, p .50 and .01. z.005 2.576 .
The critical value must be on either side of .40. p cv p 0 z 2 p .40 2.576.024495 .40 .0631 .
The ‘reject’ zone is below .3369 and above .4631.
.50 .40
pval 2 P p .50 2 P z 2 Pz 4.08 0
.024495
.5.5 .5
b) (i) H 1 : p .40, P p .4570 p1 .50 p
pq
.025.
n 400 20
.4570 .5
P z Pz 1.72 .5 .4573 .0427 Power 1 .0427 .9573
.025
.3369 .5 .4631 .5
(ii) H 1 : p .40, P.3369 p .4631 p1 .50 P z
.025 .025
P6.52 z 1.48 .5 .4306 .0694 Power 1 .0694 .9306
pqz 2 .40.602.5762
c) p 0 .40 n 63703.42 This is above 400, so the sample size is inadequate.
e2 .0052
.5.5 .5
p p z s p .5 2.327.025 .4418 If the null hypothesis is
pq
d) s p .025.
n 400 20
H 0 : p .40, we must reject it.
p0 q0 .41.59
Version 1 a) p 0 .41 p .0006048 .0245917
n 400
(i) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is below the proportion of interest
H 0 : p .41, H 1 : p .41, n 400, p .50 and .01. z.01 2.327 .
The critical value must be below .40. p cv p 0 z p .41 2.327.0245917 = .3528. The ‘reject’ zone is
.50 .41
below .3528. pval P p .50 P z Pz 3.65 .5 .4999 .9999
.0245917
(ii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is above the proportion of interest.
21
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H 0 : p .41, H 1 : p .41, n 400, p .50 and .01. z.01 2.327 .
The critical value must be above .41. p cv p 0 z p .41 2.327.0245917 = .4672 . The ‘reject’
.50 .41
zone is above .467270. pval P p .50 P z Pz 3.66 .5 .4999 .0001
.0245917
(iii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is equal to the proportion of interest. [12]
H 0 : p .41, H 1 : p .41, n 400, p .50 and .01. z.005 2.576 .
The critical value must be on either side of .40. p cv p 0 z 2 p .41 2.576.0245917 .41 .0633 .
The ‘reject’ zone is below .3467 and above .4733.
.50 .41
pval 2 P p .50 2 P z 2 Pz 3.66 2.5 .4999 .0002
.0245917
.5.5 .5
b) (i) H 1 : p .41, P p .4672 p1 .50 p
pq
.025.
n 400 20
.4672 .5
Pz Pz 1.31 .5 .4049 .0951 Power 1 .0951 .9049
.025
(ii) H 1 : p .41,
.3467 .5 .4733 .5
P.3467 p .4733 p1 .50 P z P 6.13 z 1.07
.025 .025
.5 .3577 .1423 Power 1 .1423 .8577
pqz 2 .41.592.5762
c) p 0 .40 n 64207.8 This is above 400, so the sample size is inadequate.
e2 .0052
.5.5 .5
p p z s p .5 2.327.025 .4418 If the null hypothesis is
pq
d) s p .025.
n 400 20
H 0 : p .41, we must reject it.
p0 q0 .42.58
Version 2 a) p 0 .42 p .0006090 .0246779
n 400
(i) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is below the proportion of interest
H 0 : p .42, H 1 : p .42, n 400, p .50 and .01. z.01 2.327 .
The critical value must be below .42. p cv p 0 z p .42 2.327.0246779 = .3626. The ‘reject’ zone
.50 .42
is below .3626. pval P p .50 P z Pz 3.24 .5 .4994 .9994
.0246779
(ii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is above the proportion of interest.
H 0 : p .42, H 1 : p .42, n 400, p .50 and .01. z.01 2.327 .
The critical value must be above .42. p cv p 0 z p .42 2.327.02446779 = .4774 . The ‘reject’
.50 .42
zone is above .4774. pval P p .50 P z Pz 3.24 .5 .4994 .0006
.0246779
(iii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is equal to the proportion of interest. [12]
H 0 : p .40, H 1 : p .40, n 400, p .50 and .01. z.005 2.576 .
22
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The critical value must be on either side of .42. p cv p 0 z 2 p .42 2.576.0246779 .40 .0636 .
The ‘reject’ zone is below .3564 and above .4836.
.50 .40
pval 2 P p .50 2 P z 2 Pz 3.24 .0012
.0246779
.5.5 .5
b) (i) H 1 : p .42, P p .4774 p1 .50 p
pq
.025.
n 400 20
.4774 .5
P z Pz 0.90 .5 .3159 .1841 Power 1 .1841 .8159
.025
(ii) H 1 : p .42,
.3564 .5 .4836 .5
P.3564 p .4836 p1 .50 P z P 5.74 z 0.66
.025 .025
.5 .2454 .2546 Power 1 .2546 .7454
pqz 2 .42.582.5762
c) p 0 .40 n 64659.0 This is above 400, so the sample size is inadequate.
e2 .0052
.5.5 .5
p p z s p .5 2.327.025 .4418. If the null hypothesis is
pq
d) s p .025.
n 400 20
H 0 : p .42, we must reject it.
p0 q0 .43.57
Version 3 a) p 0 .43 p .0006128 .0247538
n 400
(i) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is below the proportion of interest
H 0 : p .43, H 1 : p .43, n 400, p .50 and .01. z.01 2.327 .
The critical value must be below .43. p cv p 0 z p .43 2.327.0247538 = .3724. The ‘reject’ zone is
.50 .43
below .3724. pval P p .50 P z Pz 2.83 .5 .4977 .9977
.0247538
(ii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is above the proportion of interest.
H 0 : p .43, H 1 : p .43, n 400, p .50 and .01. z.01 2.327 .
The critical value must be above .43. p cv p 0 z p .43 2.327.027538 = .4876. The ‘reject’ zone is
.50 .43
above .4876. pval P p .50 P z Pz 2.83 .5 .4977 .0023
.0247538
(iii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is equal to the proportion of interest. [12]
H 0 : p .43, H 1 : p .43, n 400, p .50 and .01. z.005 2.576 .
The critical value must be on either side of .40. p cv p 0 z 2 p .43 2.576.0247538 .43 .0638 .
The ‘reject’ zone is below .3662 and above .4938.
.50 .43
pval 2 P p .50 2 P z 2 Pz 2.83 2.0023 .0046
.0247538
.5.5 .5
b) (i) H 1 : p .43, P p .4876 p1 .50 p
pq
.025.
n 400 20
.4876 .5
P z Pz 0.50 .5 .1915 .3085 Power 1 .3085 .6915
.025
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(ii) H 1 : p .43,
.3662 .5 .4938 .5
P.3662 p .4938 p1 .50 P z P 5.35 z 0.24
.025 .025
.5 .0948 .4052 Power 1 .4052 .5948
pqz 2 .43.572.5762
c) p 0 .43 n 65057.1 This is above 400, so the sample size is inadequate.
e2 .005
.5.5 .5
p p z s p .5 2.327.025 .4418 If the null hypothesis is
pq
d) s p .025.
n 400 20
H 0 : p .43, we must reject it.
p0 q0 .44.56
Version 4 a) p 0 .44 p .0006160 .0248193
n 400
(i) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is below the proportion of interest
H 0 : p .44, H 1 : p .44, n 400, p .50 and .01. z.01 2.327 .
The critical value must be below .44. p cv p 0 z p .44 2.327.0248193 = .3822. The ‘reject’ zone is
.50 .44
below .3822. pval P p .50 P z Pz 2.42 .5 .4922 .9922
.0248193
(ii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is above the proportion of interest.
H 0 : p .44, H 1 : p .44, n 400, p .50 and .01. z.01 2.327 .
The critical value must be above .44. p cv p 0 z p .44 2.327.0248193 = .4978. The ‘reject’ zone
.50 .44
is above .4978. pval P p .50 P z Pz 2.42 .5 .4922 = .0078
.0248193
(iii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is equal to the proportion of interest. [12]
H 0 : p .44, H 1 : p .44, n 400, p .50 and .01. z.005 2.576 .
The critical value must be on either side of .44.
p cv p 0 z 2 p .44 2.576.0248193 .44 .0639 .
The ‘reject’ zone is below .3761 and above .5039.
.50 .40
pval 2 P p .50 2 P z 2 Pz 2.42 2.0078 .0156
.028193
.5.5 .5
b) (i) H 1 : p .44, P p .4977 p1 .50 p
pq
.025.
n 400 20
.4977 .5
P z Pz 0.90 .5 .3159 .1841 Power 1 .1841 .8159
.025
.3369 .5 .4631 .5
(ii) H 1 : p .40, P.3369 p .4631 p1 .50 P z
.025 .025
P6.52 z 1.48 .5 .4306 .0694 Power 1 .0694 .9306
pqz 2 .44.602.5762
c) p 0 .44 n 65402.2 This is above 400, so the sample size is inadequate.
e2 .0052
.5.5 .5
p p z s p .5 2.327.025 .4418 If the null hypothesis is
pq
d) s p .025.
n 400 20
H 0 : p .44, we must reject it.
24
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p0 q0 .45.55
Version 5 a) p 0 .45 p .0006188 .0248747
n 400
(i) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is below the proportion of interest
H 0 : p .45, H 1 : p .45, n 400, p .50 and .01. z.01 2.327 .
The critical value must be below .45. p cv p 0 z p .45 2.327.0248747 = .3921. The ‘reject’ zone
.50 .45
is below .3921. pval P p .50 P z Pz 2.01 .5 .4778 .9778
.0248747
(ii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is above the proportion of interest.
H 0 : p .45, H 1 : p .45, n 400, p .50 and .01. z.01 2.327 .
The critical value must be above .40. p cv p 0 z p .45 2.327.0248747 = .5079 . The ‘reject’
.50 .45
zone is above .5029. pval P p .50 P z Pz 2.01 .5 .4778 .0452
.0248747
(iii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is equal to the proportion of interest. [12]
H 0 : p .45, H 1 : p .45, n 400, p .50 and .01. z.005 2.576 .
The critical value must be on either side of .45. p cv p 0 z 2 p .45 2.576.0248747 .45 .0641 .
The ‘reject’ zone is below .3859 and above .5141.
.50 .45
pval 2 P p .50 2 P z 2 Pz 2.01 2.0452 .0904
.0248747
.5.5 .5
b) (i) H 1 : p .45, P p .5079 p1 .50 p
pq
.025.
n 400 20
.5079 .5
P z Pz 0.32 .5 .1255 .6255 power 1 .6255 .3745
.025
(ii) H 1 : p .45,
.3859 .5 .5141 .5
P.3859 p .5141 p1 .50 P z P 4.56 z 0.56
.025 .025
.5 .2123 .2877 Power 1 .2877 .7123
pqz 2 .45.552.5762
c) p 0 .45 n 65694.2 This is above 400, so the sample size is inadequate.
e2 .0052
.5.5 .5
.025. p p z s p .5 2.327.025 .4418. If the null hypothesis is
pq
d) s p
n 400 20
H 0 : p .45, p could be between .4418 and .45 so we must not reject the null hypothesis.
p0 q0 .46.54
Version 6 a) p 0 .46 p .0006210 .0249199
n 400
(i) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is below the proportion of interest
H 0 : p .46, H 1 : p .46, n 400, p .50 and .01. z.01 2.327 .
The critical value must be below .46. p cv p 0 z p .46 2.327.0249199 = .4020. The ‘reject’ zone
.50 .46
is below .4020. pval P p .50 P z Pz 1.60 .5 .4452 .9452
.0249199
25
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(ii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is above the proportion of interest.
H 0 : p .46, H 1 : p .46, n 400, p .50 and .01. z.01 2.327 .
The critical value must be above .46. p cv p 0 z p .46 2.327.0249199 = .5180 . The ‘reject’
.50 .46
zone is above .5180. pval P p .50 P z Pz 1.60 .5 .4452 .0548
.0249199
(iii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is equal to the proportion of interest. [12]
H 0 : p .46, H 1 : p .46, n 400, p .50 and .01. z.005 2.576 .
The critical value must be on either side of .46. p cv p 0 z 2 p .46 2.576.0249199 .46 .0642 .
The ‘reject’ zone is below .3958 and above .5242.
.50 .46
pval 2 P p .50 2 P z 2 Pz 1.60 2.0548 .1096
.0249199
.5.5 .5
b) (i) H 1 : p .46, P p .5180 p1 .50 p
pq
.025.
n 400 20
.5180 .5
P z Pz 0.72 .5 .2642 .7642 Power 1 .7642 .2358
.025
(ii) H 1 : p .46,
.3958 .5 .5242 .5
P.3958 p .5242 p1 .50 P z P 4.17 z 0.97
.025 .025
.5 .3340 .8340 Power 1 .8340 .1660
pqz 2 .46.542.5762
c) p 0 .46 n 2 65933.1 This is above 400, so the sample size is inadequate.
e .0052
.5.5 .5
.025. p p z s p .5 2.327.025 .4418 If the null hypothesis is
pq
d) s p
n 400 20
H 0 : p .46, p could be between .4418 and .46, so we must not reject the null hypothesis.
p0 q0 .47.53
Version 7 a) p 0 .47 p .0006228 .0249550
n 400
(i) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is below the proportion of interest
H 0 : p .47, H 1 : p .47, n 400, p .50 and .01. z.01 2.327 .
The critical value must be below .47. p cv p 0 z p .47 2.327.0249550 = .4119. The ‘reject’ zone
.50 .47
is below .4119. pval P p .50 P z Pz 1.20 .5 .3849 .8849
.0249550
(ii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is above the proportion of interest.
H 0 : p .47, H 1 : p .47, n 400, p .50 and .01. z.01 2.327 .
The critical value must be above .40. p cv p 0 z p .47 2.327.0249550 = .5281 . The ‘reject’
.50 .46
zone is above .5281. pval P p .50 P z Pz 1.20 .5 .3849 = .1151
.0249550
(iii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is equal to the proportion of interest. [12]
H 0 : p .47, H 1 : p .47, n 400, p .50 and .01. z.005 2.576 .
26
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The critical value must be on either side of .47. p cv p 0 z 2 p .47 2.576.0249550 .40 .0643 .
The ‘reject’ zone is below .4057 and above .5343.
.50 .47
pval 2 P p .50 2 P z 2 Pz 1.20 2.1151 .2302
.024495
.5.5 .5
b) (i) H 1 : p .47, P p .5281 p1 .50 p
pq
.025.
n 400 20
.5281 .5
P z Pz 1.12 .5 .3708 .8708 Power 1 .8708 .1292
.025
(ii) H 1 : p .47,
.4057 .5 .5343 .5
P.4057 p .5343 p1 .50 P z P 3.77 z 1.37
.025 .025
.4999 .4147 .9146 Power 1 .9146 .0854
pqz 2 .47.532.5762
c) p 0 .47 n 66118.9 This is above 400, so the sample size is inadequate.
e2 .0052
.5.5 .5
.025. p p z s p .5 2.327.025 .4418 If the null hypothesis is
pq
d) s p
n 400 20
H 0 : p .47, p could be between .4418 and .47, so we must not reject the null hypothesis.
p0 q0 .48.52
Version 8 a) p 0 .48 p .0006240 .0249800
n 400
(i) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is below the proportion of interest
H 0 : p .48, H 1 : p .48, n 400, p .50 and .01. z.01 2.327 .
The critical value must be below .40. p cv p 0 z p .48 2.327.0249800 = .4219. The ‘reject’ zone is
.50 .48
below .4219. pval P p .50 P z Pz 0.80 .5 .2881 .7881
.0249800
(ii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is above the proportion of interest.
H 0 : p .48, H 1 : p .48, n 400, p .50 and .01. z.01 2.327 .
The critical value must be above .40. p cv p 0 z p .48 2.327.0249800 = .5381 . The ‘reject’
.50 .48
zone is above .5381. pval P p .50 P z Pz 0.80 .5 .2119 .2881
.0249800
(iii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is equal to the proportion of interest. [12]
H 0 : p .48, H 1 : p .48, n 400, p .50 and .01. z.005 2.576 .
The critical value must be on either side of .48. p cv p 0 z 2 p .48 2.576.0249800 .48 .0643 .
The ‘reject’ zone is below .4157 and above .5443.
.50 .48
pval 2 P p .50 2 P z 2 Pz 0.80 .5762
.0249800
.5.5 .5
b) (i) H 1 : p .48, P p .5381 p1 .50 p
pq
.025.
n 400 20
.5381 .5
P z Pz 1.53 .5 .4370 .9370 Power 1 .9370 .6300
.025
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(ii) H 1 : p .48,
P.4156 p .5443 p1 .50
.4156 .5 .5443 .5
P z P 3.37 z 1.77 .4996 .4616 .9612
.025 .025
P6.52 z 1.48 .5 .4306 .0694 Power 1 .9612 .0388
pqz 2 .48.522.5762
c) p 0 .48 n 66251.6 This is above 400, so the sample size is inadequate.
e2 .0052
.5.5 .5
.025. p p z s p .5 2.327.025 .4418. If the null hypothesis is
pq
d) s p
n 400 20
H 0 : p .48, p could be between .4418 and .48, so we must not reject the null hypothesis.
p0 q0 .49.51
Version 9 a) p 0 .49 p .0006248 .0249950
n 400
(i) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is below the proportion of interest
H 0 : p .49, H 1 : p .49, n 400, p .50 and .01. z.01 2.327 .
The critical value must be below .49. p cv p 0 z p .49 2.327.0249950 = .4318. The ‘reject’ zone
.50 .49
is below .4318. pval P p .50 P z Pz 0.40 .5 .1554 .6554
.0249950
(ii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is above the proportion of interest.
H 0 : p .49, H 1 : p .49, n 400, p .50 and .01. z.01 2.327 .
The critical value must be above .49. p cv p 0 z p .49 2.327.0249950 = .5481 . The ‘reject’
.50 .49
zone is above .5481. pval P p .50 P z Pz 0.40 .5 .1554 .3446
.0249950
(iii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is equal to the proportion of interest. [12]
H 0 : p .49, H 1 : p .49, n 400, p .50 and .01. z.005 2.576 .
The critical value must be on either side of .49. p p 0 z 2 p .49 2.576.0249950 .49 .0643 .
The ‘reject’ zone is below .4256 and above .5544.
.50 .49
pval 2 P p .50 2 P z 2 Pz 0.40 2.3446 .6892
.0249950
.5.5 .5
b) (i) H 1 : p .49, P p .5481 p1 .50 p
pq
.025.
n 400 20
..5481 .5
P z Pz 1.93 .5 .4732 .9732 Power 1 .9732 .0268
.025
(ii) H 1 : p .40,
.4256 .5 .5544 .5
P.4256 p .5544 p1 .50 P z P 2.98 z 2.18
.025 .025
.4986 .4854 .9840 Power 1 .9840 .0160
pqz 2 .49.512.5762
c) p 0 .49 n 2 66331.2 . This is above 400, so the sample size is inadequate.
e .0052
28
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
.5.5 .5
.025. p p z s p .5 2.327.025 .4418 If the null hypothesis is
pq
d) s p
n 400 20
H 0 : p .49, p could be between .4418 and .49, so we must not reject the null hypothesis.
————— 10/17/2005 7:01:49 PM ————————————————————
Welcome to Minitab, press F1 for help.
Results for: 252x0505-3.MTW
MTB > WSave "C:\Documents and Settings\rbove\My Documents\Minitab\252x0505-
3.MTW";
SUBC> Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0505-3.MTW'
MTB > let c2=1-c1
MTB > let c3 c1*c2
MTB > let c3=c1*c2
MTB > let c3=c3/400
MTB > let c4 = sqrt(c3)
MTB > let c5 = c1*c2
MTB > let c5=6.635776 * c5
MTB > let c5 = c5/.005
MTB > let c5 = c5/.005
MTB > let c6 = c1-2.327*c3
MTB > let c6 = c1 - 2.327 * c4
MTB > let c7 = c1 + 2.327 * c4
MTB > let c8 = c1 + 2.576* c4
MTB > let c9 = c1 - 2.576 * c4
MTB > let c10= .5-c1
MTB > let c10 = c10/c4
MTB > let c11 = let c11 = c8
MTB > let c11 = c8
MTB > let c8 = c9
MTB > let c9 = c11
MTB > let c11 = c7 - .5
MTB > let c11 = c11/.025
MTB > let c12 = c8 - .5
MTB > let c12 = c12/.025
MTB > let c13 = c9 - .5
MTB > let c13 = c13/.025
MTB > print c1-c7
Data Display
p0 qo p0 q0 pqz 2
p 0 .40 q 1 p 2
p p n 2 p cv p 0 z p p cv p 0 z p
n n e
Row p q var sqrt n cv1 cv2
1 0.40 0.60 0.0006000 0.0244949 63703.4 0.343000 0.457000
2 0.41 0.59 0.0006048 0.0245917 64207.8 0.352775 0.467225
3 0.42 0.58 0.0006090 0.0246779 64659.0 0.362574 0.477426
4 0.43 0.57 0.0006128 0.0247538 65057.1 0.372398 0.487602
5 0.44 0.56 0.0006160 0.0248193 65402.2 0.382245 0.497755
6 0.45 0.55 0.0006188 0.0248747 65694.2 0.392117 0.507883
7 0.46 0.54 0.0006210 0.0249199 65933.1 0.402011 0.517989
8 0.47 0.53 0.0006228 0.0249550 66118.9 0.411930 0.528070
9 0.48 0.52 0.0006240 0.0249800 66251.6 0.421872 0.538128
10 0.49 0.51 0.0006248 0.0249950 66331.2 0.431837 0.548163
MTB > print c8-c13
Data Display
29
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
.50 p 0 cv2 .5 cv3 .5 cv4 .5
p cv p 0 z 2 p p cv p 0 z 2 p z z z z
p .025 .025 .025
Row cv3 cv4 tr1 tr2 tr3 tr4
1 0.336901 0.463099 4.08248 -1.72001 -6.52395 -1.47605
2 0.346652 0.473348 3.65978 -1.31101 -6.13393 -1.06607
3 0.356430 0.483570 3.24176 -0.90298 -5.74281 -0.65719
4 0.366234 0.493766 2.82785 -0.49592 -5.35063 -0.24937
5 0.376065 0.503935 2.41747 -0.08982 -4.95739 0.15739
6 0.385923 0.514077 2.01008 0.31534 -4.56309 0.56309
7 0.395806 0.524194 1.60514 0.71954 -4.16774 0.96774
8 0.405716 0.534284 1.20217 1.12281 -3.77136 1.37136
9 0.415652 0.544348 0.80064 1.52514 -3.37394 1.77394
10 0.425613 0.554387 0.40008 1.92653 -2.97548 2.17548
MTB > let c14=2.576*c4
MTB > print c1 c14
Data Display
p0 z.005 x
Row p er
1 0.40 0.0630989
2 0.41 0.0633481
3 0.42 0.0635703
4 0.43 0.0637658
5 0.44 0.0639346
6 0.45 0.0640772
7 0.46 0.0641936
8 0.47 0.0642840
9 0.48 0.0643485
10 0.49 0.0643871
MTB > print c7 c11
Data Display
Row cv2 tr2
cv2 .5
p cv p 0 z p z
.025
1 0.457000 -1.72001
2 0.467225 -1.31101
3 0.477426 -0.90298
4 0.487602 -0.49592
5 0.497755 -0.08982
6 0.507883 0.31534
7 0.517989 0.71954
8 0.528070 1.12281
9 0.538128 1.52514
10 0.548163 1.92653
MTB > print c8 c9 c12 c13
Data Display
cv3 .5 cv4 .5
p cv p 0 z 2 p p cv p 0 z 2 p z z
.025 .025
Row cv3 cv4 tr3 tr4
1 0.336901 0.463099 -6.52395 -1.47605
2 0.346652 0.473348 -6.13393 -1.06607
30
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
3 0.356430 0.483570 -5.74281 -0.65719
4 0.366234 0.493766 -5.35063 -0.24937
5 0.376065 0.503935 -4.95739 0.15739
6 0.385923 0.514077 -4.56309 0.56309
7 0.395806 0.524194 -4.16774 0.96774
8 0.405716 0.534284 -3.77136 1.37136
9 0.415652 0.544348 -3.37394 1.77394
10 0.425613 0.554387 -2.97548 2.17548
MTB > Save "C:\Documents and Settings\rbove\My Documents\Minitab\252x0505-
3.MTW";
SUBC> Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0505-3.MTW'
Existing file replaced.
31
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
2. Customers at a bank complain about long lines and a survey shows a median waiting time of 8 minutes.
Personalize the data as follows: Use the second-to-last digit of your student number – subtract it from the
last digit of every single number. (Example: Seymour Butz’s student number is 976512, so he changes {6.7
6.3 5.7 ….} to {6.6 6.1 5.6 …}) You may drop any number in the data set that you use for this problem
that is exactly equal to 8. Use .05 . Do not assume that the population is Normal!
Exhibit T2: Changes are made and a new survey of 32 customers is taken. The times are as follows:
6.7 6.3 5.7 5.5 4.9 7.0 7.1 7.2 7.3 7.3 7.4
7.6 7.6 7.7 7.7 8.0 8.0 8.1 8.2 8.4 8.4 8.4
8.5 8.9 9.0 9.1 9.3 9.5 9.6 9.7 9.8 10.3
a. Test that the median is below 8. State your null and alternate hypotheses clearly. (2) [16]
b. (Extra credit) Find a 95% (or slightly more) two-sided confidence interval for the median. (2)
Solution: The data sets for this problem are below. Let n be the count of the sample and x be the number
of numbers below 8.
Row x1 x2 x3 x4 x5 x6 x7 x8 x9 x10
1 4.8 4.7 4.6 4.5 4.4 4.3 4.2 4.1 4.0 4.9
2 5.4 5.3 5.2 5.1 5.0 4.9 4.8 4.7 4.6 5.5
3 5.6 5.5 5.4 5.3 5.2 5.1 5.0 4.9 4.8 5.7
4 6.2 6.1 6.0 5.9 5.8 5.7 5.6 5.5 5.4 6.3
5 6.6 6.5 6.4 6.3 6.2 6.1 6.0 5.9 5.8 6.7
6 6.9 6.8 6.7 6.6 6.5 6.4 6.3 6.2 6.1 7.0
7 7.0 6.9 6.8 6.7 6.6 6.5 6.4 6.3 6.2 7.1
8 7.1 7.0 6.9 6.8 6.7 6.6 6.5 6.4 6.3 7.2
9 7.2 7.1 7.0 6.9 6.8 6.7 6.6 6.5 6.4 7.3
10 7.2 7.1 7.0 6.9 6.8 6.7 6.6 6.5 6.4 7.3
11 7.3 7.2 7.1 7.0 6.9 6.8 6.7 6.6 6.5 7.4
12 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 6.7 7.6
13 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 6.7 7.6
14 7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 7.7
15 7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 7.7
16 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 8.1
17 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 8.2
18 8.1 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 8.4
19 8.3 8.2 7.9 7.8 7.7 7.6 7.5 7.4 7.3 8.4
20 8.3 8.2 8.1 8.1 7.9 7.8 7.7 7.6 7.5 8.4
21 8.3 8.2 8.1 8.5 7.9 7.8 7.7 7.6 7.5 8.5
22 8.4 8.3 8.1 8.6 7.9 7.8 7.7 7.6 7.5 8.9
23 8.8 8.7 8.2 8.7 8.4 7.9 7.8 7.7 7.6 9.0
24 8.9 8.8 8.6 8.9 8.5 8.3 8.2 8.1 8.1 9.1
25 9.0 8.9 8.7 9.1 8.6 8.4 8.3 8.2 8.2 9.3
26 9.2 9.1 8.8 9.2 8.8 8.5 8.4 8.3 8.4 9.5
27 9.4 9.3 9.0 9.3 9.0 8.7 8.6 8.5 8.6 9.6
28 9.5 9.4 9.2 9.4 9.1 8.9 8.8 8.7 8.7 9.7
29 9.6 9.5 9.3 9.9 9.2 9.0 8.9 8.8 8.8 9.8
30 9.7 9.6 9.4 9.3 9.1 9.0 8.9 8.9 10.3
31 10.2 10.1 9.5 9.8 9.2 9.1 9.0 9.4
32 10.0 9.7 9.6 9.5
n 31 31 32 29 31 32 32 3132 30
x 17 18 19 19 22 23 23 2323 15
a) Test that the median is below 8. State your null and alternate hypotheses clearly. (2)
Hypotheses about Hypotheses about a proportion
a median If p is the proportion If p is the proportion
above 0 below 0
H 0 : 0 H 0 : p .5 H 0 : p .5
H 1 : 0 H 1 : p .5 H 1 : p .5
32
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
It seems easiest to let p be the proportion below 8. If you defined p as the proportion above .8, the p-
H 0 : p .5
values should be identical. Our Hypotheses are . According to the outline, in the absence of a
H 1 : p .5
x
binomial table we must use the normal approximation to the binomial distribution. If p is our
n
p p0
observed proportion, we use z . But for the sign test, p .5 and q 1 .5 .5 . So
p0 q0
n
x
n .5
x
n .5 x .5n n x .5n 2x n
z
.5
2 .
.25 .5 n n n
n n
2x 1 n
(For relatively small values of n , a continuity correction is advisable, so try z , where the +
n
n n
applies if x , and the applies if x . ) Values of x and n are repeated below for all ten possible
2 2
2x n 2x 1 n n
cases. Both z1 and the more correct z 2 (except when x . ) are computed. Since
n n 2
the alternative hypothesis is p .5, the probabilities in the two right columns are p-values.
2x n 2x 1 n
x n z1 z2 Pz z1 Pz z 2
n n
1 17 31 0.53882 0.71842 .5-.2054=.2946 .5-.2642=.2358
2 18 31 0.89803 1.07763 .5-.3133=.1867 .5-.3599=.1401
3 19 32 1.06066 1.23744 .5-.3554=.1445 .5-.3925=.1075
4 19 29 1.67126 1.85695 .5-.4525=.0475* .5-.4686=.0314*
5 22 31 2.33487 2.51447 .5-.4901=.0099* .5-.4940=.0060*
6 23 32 2.47487 2.65165 .5-.4934=.0066* .5-.4960=.0040*
7 23 32 2.47487 2.65165 .5-.4934=.0066* .5-.4960=.0040*
8 23 32 2.47487 2.65165 .5-.4934=.0066* .5-.4960=.0040*
9 23 31 2.69408 2.87368 .5-.4964=.0036* .5-.4979=.0021*
10 15 30 0.00000 0.00000 .5 .5
Since .05 , and the starred items are below .05, these are the cases in which we reject the hypothesis
that the median is, at least 8. Note that z2 is wrong!!! See 252y0551h for correct values.
Computation of exact probabilities for this Problem
————— 10/31/2005 4:26:58 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > WOpen "C:\Documents and Settings\rbove\My Documents\Minitab\252x0505-
1c.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0505-1c.MTW'
Worksheet was saved on Mon Oct 31 2005
Results for: 252x0505-1c.MTW
MTB > let c3 = c1-1
MTB > cdf c3 c4; #Computes F x 1 for n 29
SUBC> binomial 29 .5.
MTB > cdf c3 c5;
33
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
SUBC> binomial 30 .5.
MTB > cdf c3 c6;
SUBC> binomial 31 .5.
MTB > cdf c3 c7;
SUBC> binomial 32 .5.
MTB > let c4=1-c4 #so if x is 17, we now have Px 17
MTB > let c5=1-c5
MTB > let c6 = 1-c6
MTB > let c7 = 1-c7
MTB > print c1-c7 #Note that correct answers are in ‘()’
Data Display
Row x n x-1 n29 n30 n31 n32
1 17 31 16 0.229129 0.292332 (0.360050) 0.430025
2 18 31 17 0.132465 0.180797 (0.236565) 0.298307
3 19 32 18 0.068023 0.100244 0.140521 (0.188543)
4 19 29 18 (0.068023) 0.100244 0.140521 0.188543
5 22 31 21 0.004065 0.008062 (0.014725) 0.025051
6 23 32 22 0.001158 0.002611 0.005337 (0.010031)
7 23 32 22 0.001158 0.002611 0.005337 (0.010031)
8 23 31 22 0.001158 0.002611 (0.005337) 0.010031
9 15 30 14 0.500000 (0.572232) 0.639950 0.701693
MTB >
b) (Extra credit) Find a 95% (or slightly more) two-sided confidence interval for the median. (2) I am going
to assume that you continued to use the data sets above. The outline says that an approximate value for the
n 1 z .2 n
index of the lower limit of the interval is k . In this formula z 3 z .025 1.96.
2
We use this formula with the following results.
Row n sqrtn k k * rounded down n k *1 interval
1 31 5.56776 10.5436 10 22 7.2 8.4
2 31 5.56776 10.5436 10 22 7.1 8.3
3 32 5.65685 10.9563 10 23 7.0 8.2
4 29 5.38516 9.7225 9 23 6.9 8.6
5 31 5.56776 10.5436 10 22 6.8 8.4
6 32 5.65685 10.9563 10 23 6.7 7.9
7 32 5.65685 10.9563 10 23 6.6 7.8
8 32 5.65685 10.9563 10 23 6.5 7.7
9 31 5.56776 10.5436 10 22 6.4 7.5
10 30 5.47723 10.1323 10 21 7.3 8.5
————— 10/17/2005 9:38:18 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > let c3 = 2*c1
MTB > let c3=c3-c2
MTB > let c5 = sqrt(c2)
MTB > let c4 = c3+1
MTB > let c3=c3/c5
MTB > let c4=c4/c6
MTB > let c4 = c4/c5
MTB > print c1-c5
34
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
Data Display
Row x n z1 z2 sqrtn
1 17 31 0.53882 0.71842 5.56776
2 18 31 0.89803 1.07763 5.56776
3 19 32 1.06066 1.23744 5.65685
4 19 29 1.67126 1.85695 5.38516
5 22 31 2.33487 2.51447 5.56776
6 23 32 2.47487 2.65165 5.65685
7 23 32 2.47487 2.65165 5.65685
8 23 32 2.47487 2.65165 5.65685
9 23 31 2.69408 2.87368 5.56776
10 15 30 0.00000 0.18257 5.47723
MTB > print c1-c4
Data Display
Row x n z1 z2
1 17 31 0.53882 0.71842
2 18 31 0.89803 1.07763
3 19 32 1.06066 1.23744
4 19 29 1.67126 1.85695
5 22 31 2.33487 2.51447
6 23 32 2.47487 2.65165
7 23 32 2.47487 2.65165
8 23 32 2.47487 2.65165
9 23 31 2.69408 2.87368
10 15 30 0.00000 0.18257
MTB > MTB > let c6 = c2+1-1.960*c5
MTB > let c6 = c6/2
MTB > print c2 c5 c6
Data Display
Row n sqrtn k
1 31 5.56776 10.5436
2 31 5.56776 10.5436
3 32 5.65685 10.9563
4 29 5.38516 9.7225
5 31 5.56776 10.5436
6 32 5.65685 10.9563
7 32 5.65685 10.9563
8 32 5.65685 10.9563
9 31 5.56776 10.5436
10 30 5.47723 10.1323
35
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
3. Use the personalized data from Problem 2 (but do not drop any numbers.) test that the mean is below 8.
Minitab found the following the original numbers, which were in C4:
Descriptive Statistics: C4
Variable N Mean SE Mean StDev Minimum Q1 Median Q3 Maximum
C4 32 7.944 0.228 1.291 4.900 7.225 8.000 8.975 10.300
Sum of squares (uncorrected) of C4 = 2070.94
I do not know the values for your numbers, but the following (copied from last year’s exam) should be
useful: x a x na, x a2 x 2 2a x na2 . Your value of a is negative or
zero. Can you say that the population mean is below 8? Use .05
a. State your null and alternative hypotheses (1)
b. Find a critical value for the sample mean and a ‘reject’ zone. Make a diagram! (1)
c. Do you reject the null hypothesis? Use the diagram to show why. (1)
d. Find an approximate p-value for the null hypothesis. Make sure that I know where you got your results.
e. Test to see if the population standard deviation is 2. (2)
f. (Extra credit) What would the critical values be for this test of the standard deviation? (1)
g. Assume that the population standard deviation is 2, restate your critical value for the mean and create a
power curve for your test. (5)
h. Assume that the population standard deviation is 2 and find a 94% (this does not say 95%!) two sided
confidence interval for the mean. (1)
i. Using a 96% confidence level and assuming that the population standard deviation is 2, test that the mean
is below 8. (1)
j. Using a 96% confidence interval an assuming that the population standard deviation is 2, how large a
sample do you need to have an error in the mean of .005 ? (1) [30]
Solution:
a. State your null and alternative hypotheses (1)
Since the problem asks if the mean is below 8 , and this does not contain an equality, it must
be an alternate hypothesis. Our hypotheses are H 0 : 8 , (The average wait is at least 8
minutes. ) and H 1 : 8 (The average wait is less than 8 minutes.) This is a left-sided test.
b. Find a critical value for the sample mean and a ‘reject’ zone. Make a diagram! (1)
Given: 0 8, s 1.291 n 32, df n 1 31, and .05. So
,
0.228. Note that tn 1 t .10 1.696. We need a critical value for x below 8.
s 1.291
sx 31
n 32
Common sense says that if the sample mean is too far below 8, we will not believe H 0 : 8 . The
formula for a critical value for the sample mean is x t n1 s , but we want a single value
cv 0 x
2
below 8, so use x cv
0 tn 1 s x 8 1.6960.228 7.613 . Make a diagram showing an
almost Normal curve with a mean at 8 and a shaded 'reject' zone below 7.613.
c. Do you reject the null hypothesis? Use the diagram to show why. (1)
x
7.844 Do not reject Not below cv
7.744 Do not reject Not below cv
7.644 Do not reject Not below cv
7.544 Reject Below cv
7.444 Reject Below cv
7.344 Reject Below cv
7.244 Reject Below cv
7.144 Reject Below cv
7.044 Reject Below cv
7.944 Do not reject Not below cv
36
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
d. Find an approximate p-value for the null hypothesis. Make sure that I know where you got your results.
n 32 means df 31. The closest values from the t table to the computed t are shown.
x 0 x 8
The alternative hypothesis, H 1 : 8 , means that the p-value is P t P t
sx 0.228
Version x t Nearest t Values Approx. p-value
1 7.844 -0.68421 t .25 0.680 t .20 0.853
31 31
p-value is between .20 and .25.
2 7.744 -1.12281 t .15 1.054 t .10 1.309
31 31
p-value is between .10 and .15.
3 7.644 -1.56140
31
t .05 1.309 t .025 2.040
31
p-value is between .025 and .05
4 7.544 -2.00000
31
t .05 1.309 t .025 2.040
31
p-value is between .025 and .05
5 7.444 -2.43860 t 31 2.040 t 31 2.453
.025 .01 p-value is between .01 and .025
6 7.344 -2.87719 t .01 2.453 t .01 2.453
31 31
p-value is between .005 and .01.
7 7.244 -3.31579
31 31
t .005 2.744 t .001 3.375 p-value is between .001 and .005.
8 7.144 -3.75439 t 31 3.375
.001 p-value is below .005
9 7.044 -4.19298
31
t .001 3.375 p-value is below .005
10 7.944 -0.24561 t .45 0.127 t .40 0.256
31 31
p-value is between .40 and .45.
e. Test to see if the population standard deviation is 2. s 1.291 df 31. , (2) The outline says
To test H 0 : 0 against H1 : 0
i. Test Ratio: 2
n 1s 2 or for large samples z 2 2 2DF 1
0
2
0
2 2
12 2 0
2
ii. Critical Value: 2
s cv 2
or or for large samples (from table 3) s cv 2DF .
n 1 n 1 z 2 2DF
iii. Confidence Interval:
n 1s 2 2
n 1s 2 or for large samples
22
12 2
s 2DF s 2DF
z 2 2DF z 2 2DF
If we use a test ratio, 2
n 1s 2
311.2912
12.9168. Since our degrees of freedom are too large
0
2
22
for the table, use z 2 2 2DF 1 212.9168 231 1 25.8336 61
5.0827 7.8102 2.7275 . If we are doing a 2-sided 5% test, we do not reject the null hypothesis if z
is between -1.960 and 1.960. It’s not, so we reject the null hypothesis.
f. (Extra credit) What would the critical values be for this test of the standard deviation? (1)
2 231 27.8740
Table 3 says s cv 2DF . The two critical values are
z 2DF 1.960 231 7.8740 1.960
2
27.8740 15.748 27.8740 15.748
1.6014 and 2.2663 . Since the standard deviation is
7.8740 1.960 9.834 7.8740 1.960 5.914
not between them, we reject the null hypothesis.
37
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
g. Assume that the population standard deviation is 2, restate your critical value for the mean and create a
power curve for your test. (5). Our hypotheses are H 0 : 8 , (The average wait is at least 8 minutes. )
and H 1 : 8 (The average wait is less than 8 minutes.) This is a left-sided test. If 2 , n 32 ,
2 2
x 0.35355 and .05 , our critical value is 1.645 8 0.3762 7.4184 . We use the
32 32
following points: 8, 7.709, 7.4184, 7.127 and 6.836. Our results are as follows.
7.4184 8
Px 7.4184 8 P z
1 8 Pz 1.645 = .95
.35355
Power 1 .95 .05
7.4184 7.709
Px 7.4184 7.709 P z
1 7.812 Pz 0.82 =.5 + 0.2939 =.7939
.35355
Power 1 .7939 .2061
7.4184 7.4184
Px 7.4184 7.4184 P z
1 7.6238 Pz 0 .5
.35355
Power 1 .5 .5
7.4184 7.127
Px 7.4184 7.127 P z
1 7.436 Pz 0.82 = .5-.2939 = .2061
.35355
Power 1 .2061 .7939
7.4184 6.836
Px 7.4184 6.836 P z
1 7.248. Pz 1.647 = .5 - .4500 = .0500
.35355
Power 1 .05 .95
h. Assume that the population standard deviation is 2 and find a 94% (this does not say 95%!) two sided
2
confidence interval for the mean. (1) .06 . We know x z 2 x . x 0.35355 . To find
32
z .03 make a Normal diagram for z showing a mean at 0 and 50% above 0, divided into 3% above z .03 and
47% below z .03 . So P0 z z.03 .4700 The closest we can come is P0 z 1.88 .4699 . So
z.03 1.88 x z 2 x x 1.88.35355 x 0.665 . Substitute your value of the sample mean.
i. Using a 96% confidence level and assuming that the population standard deviation is 2, test that the mean
is below 8. (1) Our hypotheses are H 0 : 8 , (The average wait is at least 8 minutes. ) and H 1 : 8
(The average wait is less than 8 minutes.) This is a left-sided test. We can do this by a critical value, a test
ratio or a confidence interval.
(i) .04 To find z.04 make a Normal diagram for z showing a mean at 0 and 50% above 0,
divided into 4% above z.04 and 46% below z.04 . So P0 z z.04 .4600 The closest we can
come is P0 z 1.75 .4579 . So xcv 0 z x 8 1.750.35355 7.381 If your value of
the sample mean is below 7.381, reject the null hypothesis.
x 0 x 8 x 8
(ii) z . To get a p-value find P z . So, for example, if
x 0.35355 0.35355
7.944 8
x 7.944 , pval P z Pz 0.16 .5 .0636 .4364 . If the p-value is below
0.35355
.04, reject the null hypothesis.
(iii) The one-sided confidence interval is x z.04 x x 1.75.35355 x 0.6187 . If this
value is below 8, reject the null hypothesis.
38
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
j. Using a 96% confidence level and assuming that the population standard deviation is 2, how large a
sample do you need to have an error in the mean of .005 ? (1)
.04 To find z.02 make a Normal diagram for z showing a mean at 0 and 50% above 0, divided into
2% above z.02 and 48% below z.04 . So P0 z z.02 .4800 The closest we can come is
z 2 2 2.08 2 2 2
2
P0 z 2.08 .4798 From the outline n 3461.12. Use 3462.
e2 .005
————— 10/17/2005 11:10:20 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > WOpen "C:\Documents and Settings\rbove\My Documents\Minitab\252x0505-
1.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0505-1.MTW'
Worksheet was saved on Tue Oct 11 2005
Results for: 252x0505-1.MTW
MTB > describe c1-c10
Descriptive Statistics: C1, C2, C3, C4, C5, C6, C7, C8, C9, C10
Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum
C1 32 0 7.844 0.228 1.291 4.800 7.125 7.900 8.875 10.200
C2 32 0 7.744 0.228 1.291 4.700 7.025 7.800 8.775 10.100
C3 32 0 7.644 0.228 1.291 4.600 6.925 7.700 8.675 10.000
C4 32 0 7.544 0.228 1.291 4.500 6.825 7.600 8.575 9.900
C5 32 0 7.444 0.228 1.291 4.400 6.725 7.500 8.475 9.800
C6 32 0 7.344 0.228 1.291 4.300 6.625 7.400 8.375 9.700
C7 32 0 7.244 0.228 1.291 4.200 6.525 7.300 8.275 9.600
C8 32 0 7.144 0.228 1.291 4.100 6.425 7.200 8.175 9.500
C9 32 0 7.044 0.228 1.291 4.000 6.325 7.100 8.075 9.400
C10 32 0 7.944 0.228 1.291 4.900 7.225 8.000 8.975 10.300
MTB > let c13=c12-8
MTB > let c13=c13/0.288
MTB > let c14=c12-8
MTB > let c14=c14/.3536
MTB > print c12-c14
Data Display
x 8 x 8
Row x t z
sx x
1 7.844 -0.54167 -0.44118
2 7.744 -0.88889 -0.72398
3 7.644 -1.23611 -1.00679
4 7.544 -1.58333 -1.28959
5 7.444 -1.93056 -1.57240
6 7.344 -2.27778 -1.85520
7 7.244 -2.62500 -2.13801
8 7.144 -2.97222 -2.42081
9 7.044 -3.31944 -2.70362
10 7.944 -0.19444 -0.15837
MTB >
39
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