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```					Practice Test 6 (Chapters 18-21)

1. For each of the following, write the null and alternative hypotheses.
a) An airline claims that 90% of lost luggage is recovered and returned within 24 hours. A consumer group
surveyed 122 people who had lost luggage on that airline and found that 103 of them were reunited with the
missing items by the next day. Does this cast doubt on the airline’s claim?
Ho:__p = 0.90___                Ha:__p < 0.90______
b) Commercial fishermen working certain parts of the Atlantic Ocean sometimes have trouble with the presence
of whales. They would like to scare away the whales without frightening the fish. In the past, sonar researchers
have determined that 40% of all whales in an area do leave on their own, probably to get away from the noise of
the fishing boat. The fishermen plan to try a new technique to increase that figure. The technique involves
transmitting the sounds of a killer whale underwater.

State the appropriate null and alternative hypotheses for testing whether or not the new technique works in
terms of the percent of whales that leave the area.

H0:     p = 0.40                               HA:    p > 0.40

c) Is a randomly chosen newborn baby equally likely to be a girl or a boy? Suppose that a researcher decides to
investigate this question by checking the records of births for the past 10 years at a large metropolitan hospital.
The null hypothesis can be stated as “The proportion of newborns that are girls is 0.50”, or Ho: p = 0.50.

Give the appropriate alternative hypothesis in words:
The proportion of newborns that are girls is not equal to 0.50.

Give the appropriate alternative hypothesis in symbols:

HA:     p 0.50

d). A researcher thinks that if expectant mothers use vitamin pills, the birth weight of the babies will increase.
The average of the birth weights of the population is 8.6 pounds.
State the appropriate null and alternative hypotheses.

H0:     mu = 8.6 pounds                        HA:    mu > 8.6 pounds
5. Suppose that the population mean adult weight is 180 pounds with a standard deviation of 28 pounds. An
elevator in our building has a weight limit of 10 persons or 2000 pounds. What’s the probability that the 10
people who get on the elevator will overload its weight limit? This is the same as asking what the probability is
of having a sample mean weight of more than 200 pounds, or P(xbar>200).
a) What assumptions must you make to solve this problem?

1) Random Sample
2) Independent values in sample
3) sample size is < 10% of population

b) What are the mean and standard deviation of the sampling distribution of the mean weight?

Mu(xbar) = 180 lbs (same as the population mean)

        28.
 ( xbar)                           8.854pounds
n        10
         28.
c) What is the probability of overloading the weight limit of the elevator?       ( xbar)                          8.85
n         10
200  180
z                        2.2587
28
P(xbar>200) = ??               Calculate z-score:               10

Go to z-table or Zarea tool, find area under the curve: P(xbar<200) = 0.9880
But we want P(xbar > 200) = 1-0.9880 = 0.0120.

21. In a marketing survey, a random sample of 730 U.S. women shoppers revealed that 628 remained loyal to
their favorite supermarket during the past year (ie, did not switch stores).

a) What is the population?

U.S. women shoppers
b) Give the numerical value of the
statistic p that estimates p.      phat = 628 / 730 = 0.860274

c) Find a 90% confidence interval for the true, unknown population proportion, p.

For 90% CI, z* = 1.645.

0.860274( 1  0.860274)
0.860274  1.645
`                                 730

0.839165 < p < 0.881383
We are 90% confident that the true proportion of US women shoppers remaining loyal to their favorite
supermarket is between 0.839165 and 0.881383
9. In May 2002, the Gallup Organization asked a random sample of 537 American adults this question: If you
could choose between the following two approaches, which do you think is the better penalty for murder, the
death penalty or life imprisonment, with absolutely no possibility of parole? Of those polled, 280 chose the
death penalty.
a) Is the sampling distribution model for the sample proportion likely to be Normal? [yes/ no]
Random: stated.
Independent values: most likely to be true in a random sample.
n < 10% population: 537 is definitely < 10% of all American adults
n*phat > 10, n* qhat > 10: we have more than 10 people for success and failure.
b) What is the population of interest?
c) Estimate the population proportion of American adults who would choose the death penalty, at a 99%
confidence level. Explain your confidence interval in context.

280
phat                  0.521415
537                                      For 99% confidence, z* = 2.576

0.521415 ( 1  0.521415 )
0.521415  2.576
537
0.465885< p < 0.576946       We are 99% confident, based on the sample, that the true population proportion of
American Adults that think the death penalty is better is between 0.465885 and 0.576946.
d) How many people must Gallup poll if they want the margin of error of their 99% confidence interval to be no
larger than 4%? Assume the sample proportion remains the same.

2                                      2
z phat qhat                      2.576 ( 0.521415 ) ( 1  0.521415 )
n                                                                                      1034.938011
2                                             2
ME                                           0.04
They must sample at least 1035 people.
e) Without doing any calculations, indicate whether the answer to (c) would be larger, smaller, or the same if
larger

f) How will the margin of error’s size change if the confidence level is decreased?
It will get smaller.
10. The article “Credit Cards and College Students: Who Pays, Who Benefits?” (Journal of College Student
Development, 1998, 50-56) described a study of credit card payment practices of college students. According to
the authors of the article, the credit card industry asserts that at most 50% of college students carry a credit card
balance from month to month. However, the authors of the article report that, in a random sample of 325
college students, 175 carried a balance each month. Do the data support the credit card industry’s claim?
a) State the null and alternative hypotheses.
Ho:___ p = 0.50__________

Ha:__p > 0.50___________
b) Show if the conditions for hypothesis testing are satisfied.
Random: given.
Independent values: given
N< 10% pop: yes       325*0.5 = 162.5 >10
Np>10, nq> 10: yes 325*0.5 = 162.5 >10
c) Sketch the appropriate model and calculate the test statistic.

0.53846  0.50
z                                1.38675
175                                            0.5( 0.5)
phat
325                                              325

d) Determine the P-value.
P-value = 1-0.9177 = 0.0823
(0.9177 is from the z-table, but we want the right side, so must subtract).
e) At a significance level of 5%, what is your conclusion (IN CONTEXT)?
We fail to reject the Null: the evidence suggests that the true proportion of students with credit card
balances is 50%.
f) If your conclusion in (e) is incorrect, what type of error have you committed?
Type II error
2. Five years ago the proportion of student’s passing a Statistic teacher’s class was 77%. A sample of this year’s
class showed 117 of 145 students passing the class. Has the proportion of students passing changed?

a) State the Null and Alternative Hypothesis.

Ho: p = 0.77            Ha: p 0.77

b) Sketch the appropriate model and calculate the test statistic.

0.8069  0.77
z                               1.0557
117                                  ( 0.77) ( .23)
phat                 0.8069
145                                       145

c) Determine the P-value.

From the z-table, P(z>1.0557) = 1-0.8554 = 0.1446
Since it is two-tailed, we must multiply this by 2 to get a P-value of 0.2892.

d) At a significance level of 10%, what is your conclusion (IN CONTEXT)?
Fail to reject the null: At an alpha level of 0.10, with a p-value of 0.2892, we fail to reject the null. The
evidence is strong that the passing rate has not changed from 77%.

8. The U.S. Census Bureau reports that 26% of all U.S. businesses are owned by women. A
Colorado consulting firm surveys a random sample of 520 businesses in the Denver area and
finds that 120 of them have women owners. Should the firm conclude that its area has a lower
a) Which test procedure will you use? Why?

z-test for proportions, since we are testing a proportion.

b) State the null and alternative hypotheses.
Ho: p = 0.26 (or 26%)             Ha: p < 0.26 (or 26%)
c) Calculate the test statistic.

Phat = 120/520 = 0.230769

0.230769  0.26
 1.519647
0.26( 1  0.26)
520

d) Determine the P-value.

Using Table A, we look up z = -1.52 to find the P-value.
P-value = 0.0643

e) At a significance level of 5%, what is your conclusion (reject or fail to reject Ho) and explain
it in context of the problem.

The P-value is above the alpha level of 5%. We will fail to reject the null. The are does not have
a lower percentage of women-owned businesses.

2. Investors remember 1987 as the year stocks lost 20% of their value in a single day. For 1987
as a whole, though, the mean return of all common stocks on the New York Stock Exchange
was μ = -3.5%. (that is, these stocks lost an average of 3.5% of their value in 1987.) The
standard deviation of the returns was about σ = 26%. The distribution of annual returns for
stocks is roughly normal.

a) What percent of stocks lost money in 1987? (this is the same as the probability that a single
stock chosen at random has a return less than 0)

P(x<0) = ??          Find z  z = (0- (-0.035))/0.26 = 0.135
From Zarea Tool, P(z<0.135) = 0.55354
55.35% of stocks lost money in 1987.

b) Suppose that you held a portfolio of five stocks chosen at random from the NYSE stocks.
What are the mean and standard deviation of the returns of randomly chosen portfolios of 5
stocks (ie, the mean return of the 5 stocks)?

From the Central Limit Theorem: The mean of the sampling distribution of
portfolios of 5 stocks is the same as the mean of the population of stocks, -3.5%.
The standard deviation s_x = sigma/sqrt(n) = 0.26/sqrt(5) = 0.11628
c) What percent of such portfolios lost money? Why is this percentage different from the result
in (a)?

P(average return of 5 stocks < 0) = ??
Z = ( 0 –(-0.035))/0.11628 = 0.301
From Zarea, P(z<0.301) = 0.61829
61.829% of portfolios lost money.

This is different because there is less variability in the mean of samples of size 5 than there is
for the return of a single stock. Therefore, a value that is 3.5% below the mean is further away
from the mean in terms of standard deviation in the sampling distribution.

d) What happens to the mean of the sampling distribution as the sample size is increased? What
happens to the standard deviation of the sampling distribution as the sample size is increased?

As n increases, SD of the sampling distribution of xbar gets smaller and smaller (approaches 0).
The mean does not change- it stays at –3.5%. The shape of the distribution becomes more and
more normal.

3. A local newspaper polls 400 voters to attempt to predict whether a school budget will pass.
Suppose the budget actually has the support of 52% of the voters.
What is the probability of the newspaper predicting the defeat of the budget? (ie, P(phat
<50%)).
Sampling Distribution problem: Calculate z-score.
0.50  0.52
 0.800641
0.52( 0.48)
400
P(z<-0.800) = 0.2119

So the probability of predicting defeat based on the sample proportion is 21.19%.
12. The McClatchy News Service reported on a sample of prime-time television hours. The
following table summarizes the information reported for four networks.

Mean number of violet
Network acts per hour
15.6
ABC
CBS                  11.9
FOX                  11.7
NBC                  11.0
Suppose that each of these sample means was computed on the basis of viewing n =50
randomly selected prime-time hours and that the sample standard deviation for each of the four
networks is s = 5.
a) What is an appropriate model for the sampling distribution of the mean hours for
ABC?
N(15.6, 0.707) since the SE(xbar) = s/sqrt(n) = 5/sqrt(50)
b) Would you expect a 95% confidence interval to be wider or narrower than the 90%
confidence interval you calculated above? Explain your reasoning.
Wider- as confidence level increases, z* increases, which increases ME.
c) A similar study was done for n = 100 randomly selected prime-time hours for each
network, and a 90% confidence interval was constructed. Which of the 90%
confidence intervals (for n=50 or n=100) would have the smaller margin of error?
Why?
N = 100 will have the smaller margin of error. ME of error decreases as sqrt(n), so the larger n, the smaller

ME.

14. A national health organization warns that 30% of middle school students nationwide have
been drunk. Concerned, a local health agency randomly surveyed 110 of the 1212 middle
school students in its city. Only 21 of them reported having been drunk.
a) What is the population?                          b) Give the numerical value of the
statistic p that estimates p.
The city’s 1212
middle school students                                               phat       = 21/110 =
0.190909
c) Calculate a 90% confidence interval for the true, unknown population parameter.

Use CI formula with z* = 1.645

0.1909( 1  0.1909)
0.1909  1.645
130

with 90% confidence, 0.134198 < p < 0.247602

d) Does the true (population) proportion lie in the confidence interval that you obtained above?
Why or why not?

You don’t know. You are 90% confident that the true proportion
lies in the interval, but not 100%, though. Therefore, you
cannot say whether the proportion definitely lies in the
confidence interval, or not.

19. The amount of potato chips that a packaging machine puts into bags is believed to have a
near-Normal distribution with a mean of 10.2 ounces and a standard deviation
of 0.12 ounces.
a) The bags are labeled as containing 10.0 ounces. What is the probability that a randomly
selected bag will be underweight (have less than 10.0 ounces)?

10.0  10.2
z
0.12 z = -1.6666                           Look this up in z Table, or use
the Zarea tool.
From Table, P(x<10.0) = 0.0475

b) The potato chips are also sold in “mega-bargain multi-packs” of 8 bags. What is the mean of
the sampling distribution of xbar (the sample mean of the multi-packs)?

The mean of the sampling dist of xbar is the same as the population: 10.2
ounces.

c) What is the standard deviation of the sampling distribution of xbar?

            .12
x                            0.042426
n                8
d) What is the probability that the mean weight of a multi-pack will be underweight?

10.0  10.2
z
0.12
8  z = -4.7140      This is off the table: use the
closest z, or Zarea.
Using Zarea, P(xbar < 10.0) =0.000001214

e) How would the sampling distribution of the sample mean change if the sample size (the
multi-pack size) were increased? Ie, how would the shape, mean AND the standard deviation
change? BE SPECIFIC.

From our knowledge of sampling distributions: The mean will not
change. The standard deviation will decrease (at a rate of
sqrt(n)). The shape will get closer and closer to NORMAL as n
increases.

23. The US Department of Transportation, National Highway Traffic Safety
Administration, reported that 77% of all fatally injured automobile drivers were intoxicated. A
random sample of 27 records of automobile driver fatalities in Kit Carson County, Colorado,
showed that 15 involved an intoxicated driver.
Do these data indicate that the population proportion of driver fatalities related to alcohol
is less than 77% in Kit Carson County?

a) State the appropriate test procedure and why.

z-test for proportions, since we are testing a proportion.

b) State the null and alternative hypotheses.

Ho: p = 77%                     Ha: p < 77%

c) Calculate the test statistic.

Phat = 15/27 = 0.5555

0.5555  0.77
 2.6485
0.77( 1  0.77)
27
d) Determine the P-value.

From the z-table, the Pvalue is 0.0040.

e) At a significance level of 1%, what is your conclusion (reject or fail to reject Ho) and explain
it in context of the problem.

We reject the null hypothesis. The data suggests that the true proportion of driver
fatalities related to alcohol is less than 77% in Kit Carson County.

4. The National Center for Health Statistics (NCHS) reports that the mean systolic blood
pressure for U.S. males 35 to 44 years of age is 128. The medical director of a large company
looks at a random sample of the medical records of 27 executives in this age group and finds
that the mean systolic blood pressure for this sample is 132.07 with standard deviation of 9.2.

a. Identify each of the following: (If a value is not given, write NG.)
 = 128        = NG          y = 132.07     s = 9.2

b. Suppose that the medical director thinks that the mean systolic blood pressure of company
employees is significantly higher than that recorded by the NCHS. Give the null hypothesis
and the alternative hypothesis.

Ho: mu=128            Ha: mu>128

c. What test should you perform? Explain your choice.

t-test, because we don’t know the population SD, sigma.

d. Perform the hypothesis test at the  = .05 significance level. Give t statistic and P-value.

132.07  128
t                 2.2987
9.2
27

From the t-table, with df = 26, we find the P-value to be between 0.02 and 0.01.

e. Draw the standard normal curve and label it with your test statistic, and shade the area
represented by the P-value.

Normal curve, centered at 128, with xbar labeled at 132.07, and the area to the right of that shaded.
f. What is your conclusion (reject or fail to reject) and why?

We reject Ho at the 5% significance level, since P-value is below 0.05.

g. Explain your conclusion in terms of mean systolic blood pressure.

There is strong evidence that the true mean of the male company employees 35-44 is
higher than the national average.

10. As the statistician in charge of quality control at a local food packaging plant, you take a random sample of
25 cans that are on the way to be filled with spinach. To ensure that the lids fit securely inside the rim of the
can, you measure the diameter of the lid. For these 25 cans, the mean diameter was 3.13 inches and the standard
deviation was 0.12 inches.
a. Calculate a 95% confidence interval for the mean diameter of spinach cans.

Xbar +/- t*s/sqrt(n)
For df = 24, 95% CI, t = 2.064 (from T-table)
3.13 +- 2.064*0.12/sqrt(25)

3.13 +- 0.049536

3.0805 < mu < 3.1795

b) What is the margin of error for the interval above?

0.049536

c) In a sentence or two, explain what the interval above means?
We are 95% confident that the true mean of the spinach can diameters is within 0.049536 inches
of our sample mean of 3.13 inches.

12. A random sample of 12 large camping tents listed in Consumer Reports: Special Outdoor
Issue gave the following prices (in \$).
115     140    80      150     250    230     110     135    110     210    120     130

The mean of this sample is x = \$148.33, and the standard deviation is s = \$53.02.

a) Which procedure, z or t, will you use to calculate a confidence interval? Why?

t, because we do not know the population standard deviation and this is a problem dealing
with means.
b) Find an 80% confidence interval for the mean price of all large-sized tents.

Df = 11. so t* = 1.363

148.33+- 1.363 (53.02/sqrt(12))

We are 80% confident that
127.4685 < mu < 169.1915

c) What is the margin of error for the interval above?

\$20.86

d) In a sentence, explain the confidence interval above in context of the problem.

We are 80% confident that the true mean price of large camping tents is between \$127.47 and
\$169.19.

e) Would a 90% confidence interval be wider or narrower?

It would be wider.

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