# Problem solving tips for EXAM 1 - UAH Personal Web Pages

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```					Which statistical method do you have to use to solve a problem? (Chapter 6,7,8,9)

1. If the question asks you to find the probability (percentage) of a mean AND the keywords
“sample”, “sampling”, “sample size” does NOT exist, use the method in Chapter 6.
(Chapter 6 METHOD 1)

2. If the question asks you to find “the value of something” (population mean) given the
probability (percentage) of that “thing”, use the method in Chapter 6
(Chapter 6 METHOD 1)

Step 1. Draw the standard normal probability distribution curve
Step 2. Draw the area (you have to use the information provided by the problem to be able to
draw the area) based on the percentage/probability value given in the problem
Step 3. Compute the total area to the left of Z (Remember: The total area to the left of Z is
one minus the total area to the right of Z)
Step 4. Use the table to find the Z based on the area you found in Step 3
Step 5. Convert the Z to the X (X = μ + (σ)(Z))

3. If the question asks you to find the probability (percentage) of a sample mean AND the
keywords “sample”, “sampling”, or “sample size” exists, use the method in Chapter 7.
(Chapter 7 METHOD 1)

4. If the question asks you to find the probability (percentage) of a sample proportion, use the
method in Chapter 7.
(Chapter 7 METHOD 2)

5. If the question asks you to find the confidence interval, the interval estimate, or margin of error,
use the method in Chapter 8.
σ known =      Chapter 8 METHOD 1
s known =      Chapter 8 METHOD 2
p=             Chapter 8 METHOD 3

6. If the question asks you to test a hypothesis, use the method in Chapter 9.
σ known =      Chapter 9 METHOD 1
s known =      Chapter 9 METHOD 2
p=             Chapter 9 METHOD 3
How to determine if a standard deviation is a sample standard deviation or a population standard
deviation?

 Some problems will tell you explicitly if the standard deviation is population (σ) or sample (s)
 Some problems will not. In those cases, look at the following keywords for population standard
deviation:
o Based on prior studies, the standard deviation is...
o Based on historical studies, the standard deviation is...
o Based on past studies, the standard deviation is...

Chapter 8: Computing the confidence interval or margin of error

1. If the problem gives you the population standard deviation σ, use the Z value method (and the Z
table)

2. If the problem gives you the sample standard deviation s, use the t value method (and the t
table)

Chapter 9: Testing hypothesis about population mean

1. If the problem gives you the population standard deviation σ, use the Z value method (and the Z
table)

2. If the problem gives you the sample standard deviation s, use the t value method (and the list of
p-values for different t values and degrees of freedom that will be given in the exam)

Chapter 9: How to identify the hypothesis in the problem description?

 Find the sentence in the problem that says that the average/mean the proportion is
o greater than (exceed, has increased),
o less than (fewer than, has been reduced),
o not equal to (is not, different from, has changed).

Proportion = percentage
Confidence interval = interval estimate
Mean = average
Example of the p-value from the t-value table (Chapter 9 - Hypothesis test for mean if s is known) in the Exam

t-value         d.f.        Upper-tail p-value          Lower-tail p-value           Two-tail p-value

0.5           15                 0.035                      0.965                       0.070

0.5           16                 0.030                      0.970                       0.060

1.0           99                 0.025                      0.975                       0.050

1.0           100                0.020                      0.980                       0.040

2.0           35                 0.027                      0.973                       0.054

2.0           36                 0.023                      0.977                       0.046

2.5           120               0.0069                      0.9931                      0.0138

2.5           121               0.0062                      0.9938                      0.0124

2.5           99                 0.015                      0.985                       0.030

2.5           100                0.010                      0.990                       0.020
EXTRA: IF you have to find range probability of z that is less than -3 or greater than +3

• IF a ≥ +3 (If a is any value greater than 3),
P( z < a ) = 1
P( z < 100) = 1

• IF a ≥ +3 (If a is any value greater than 3),
P( z > a ) = 0
P( z > 100 ) = 0

• IF a ≤ -3 (If a is any value less than -3),
P( z < a ) = 0
P( z < -500 ) = 0

• IF a ≤ -3 (If a is any value less than -3),
P( z > a ) = 1
P( z > -500 ) = 1
For the hypothesis test problems, the key provided at the end of each problem is NOT the
final answer. It is only the value of the Z or the t.
You will then have to find the P-value by using this Z or t, and compare the P-value to the α
to determine whether you should accept the result.
Problem 1.

A bank has kept records of the checking balances of its customers and determined that the average daily balance of its
customers is \$300 with a standard deviation of \$48. A random sample of 144 checking accounts is selected.
a.     What is the probability that the sample mean will be more than \$306.60?
306 .60  300
Ch. 7, proc. 1: P( x >306.60) = P(Z>                   ) = P(Z>1.65) = 1 - P(Z<1.65) = 1-0.9505 = 0.0495
48
144
b.      What is the probability that the sample mean will be less than \$308?
308  300
Ch. 7, proc. 1: P( x <308) = P(Z<               ) = P(Z<2) = 0.9772
48
144
c.      What is the probability that the sample mean will be between \$302 and \$308?
302  300     308  300
Ch. 7, proc. 1: P(302< x <308) = P(               <Z<           ) = P(0.5<Z<2) = P(Z<2) - P(Z<0.5) = 0.9772 –
48            48
144              144
0.6915 = 0.2857

a.      0.0495
b.      0.9772
c.      0.2857

Problem 2.

A student believes that MORE THAN 20% of the students who finish a statistics course get an A. A random sample of
100 students was taken. Twenty-four percent of the students in the sample received A's. Test the student belief.
Ch. 9, proc. 3:
Step 1. Ha: P> 0.20
Step 2. p =24% = 0.24. Assuming that sample proportion is an estimate for population proportion, Ha is
supported by the sample.
0.24  0.20        0.04
Step 3. Z =                     =      =1
0.20 (1  0.20 )   0.04
100
Step 4. P-value = P(Z>1) = 1-P(Z<1) = 1-0.8413 = 0.1587
Step 5. α = 0.01 or 0.05 or 0.10.
Step 6. We reject the result because the P-value is greater than the α. Conclusion: Our sample does not
provide support on Ha (more than 20% of the students who finish a statistics course get an A).

1

Problem 3.
A random sample of 87 airline pilots had an average yearly income of \$99,400 with a standard deviation of \$12,000.
Develop a 95% confidence interval for the average yearly income of all pilots.
Ch. 8, proc. 2.
s                    12000
x ±tα/2           = 99400±1.988           = 99400 - 2557.63 to 99400 + 2557.63 = 96842.37 to 101957.60
n                      87

\$96,842.37 to \$101,957.60

Problem 4.

An official of a large national union claims that the fraction of women in the union is significantly different from one-half.
Test this statement at 95% confidence level.

sample size            400
women                  168
men                    232
Ch. 9, proc. 3:
Step 1. Ha: P ≠ 0.50
Step 2. p =168/400 = 0.42. Assuming that sample proportion is an estimate for population proportion, Ha
is supported by the sample.
0.42  0.50        0.08
Step 3. Z =                    =        = -3.2
0.50 (1  0.50 )   0.025
400
Step 4. P-value = 2 x P(Z<-3.2) = 2 x 0 = 0
Step 5. α = 1 – c = 1 – 0.95 = 0.05
Step 6. We accept the result because the P-value is less than the α. Conclusion: Our sample provides
support on Ha (the fraction of women in the union is significantly different from one half).

-3.2

Problem 5.

The average starting salary for this year's graduates at a large university (LU) is \$20,000 with a standard deviation of
\$8,000. Furthermore, it is known that the starting salaries are normally distributed.

a.      What is the probability that a randomly selected LU graduate will have a starting salary of at least \$30,400?
30400  20000
Ch. 6, proc. 1: P( x ≥ 30400) = P(Z ≥                     ) = P(Z≥1.30) = 1 - P(Z≤1.30) = 1-0.9032 = 0.0968
8000
b.      Individuals with starting salaries of less than \$15,600 receive a low income tax break. What percentage of the
15600  20000
Ch. 6, proc. 1: P( x < 15600) = P(Z <                     ) = P(Z<-0.55) = 0.2912 = 29.12%
8000

a.      0.0968
b.      29.12
Problem 6.

It is said that more males register to vote in a national election than females. A research organization selected a random
sample of 300 registered voters and reported that 165 of the registered voters were male. Test the claim at 95% confidence
level.

1.73

Problem 7.

Ten percent of the items produced by a machine are defective. A random sample of 100 items is selected and checked for
defects.
a.       What is the probability that the sample will contain more than 2.5% defective units?
b.       What is the probability that the sample will contain more than 13% defective units?

a.      0.9938
b.      0.1587

Problem 8.

A student believes that the average grade on the statistics final examination was NOT 87. A sample of 36 final
examinations was taken. The average grade in the sample was 83.96 with a standard deviation of 12. Test the student
belief at 95% confidence level.

-1.52

Problem 9.

The life expectancy in the United States is 75 with a standard deviation of 7 years. A random sample of 49 individuals is
selected.
a.       What is the probability that the sample mean will be larger than 77 years?
b.       What is the probability that the sample mean will be less than 72.7 years?
c.       What is the probability that the sample mean will be between 73.5 and 76 years?

a.      0.0228
b.      0.0107
c.      0.7745

Problem 10.

A university planner is interested in determining the percentage of spring semester students who will attend summer
school. She takes a pilot sample of 160 spring semester students discovering that 56 will return to summer school.
a.      Construct a 95% confidence interval estimate for the percentage of spring semester students who will return to
summer school.

a.      0.276 to 0.424

Problem 11.
A professor at a local community college noted that the grades of his students were normally distributed with a mean of
74 and a standard deviation of 10. The professor has informed us that 6.3 percent of his students received A's while only
2.5 percent of his students failed the course and received F's.
a.    What is the minimum score needed to make an A?
Ch. 6, proc. 2: P( x ≥ XGradeA) = 6.3% ⇔ P( z ≥ ZGradeA) = 6.3% ⇔ P( z ≤ ZGradeA) = 1 - 6.3% ⇔ P( z ≤
ZGradeA) = 93.7%⇔ P( z ≤ 1.53) = 0.9370 ⇔ ZGradeA = 1.53 ⇔ XGradeA = μ + (σ)(ZGradeA) = 74 + (10)(1.53) =
89.3
b.     What is the maximum score among those who received an F?
Ch. 6, proc. 2: P( x ≤ XGradeF) = 2.5% ⇔ P( z ≤ ZGradeF) = 2.5% ⇔ P( z ≤ ZGradeF) = 0.0250 ⇔ P( z ≤ -1.96) =

a.      89.3
b.      54.4

Problem 12.

A producer of various kinds of batteries has been producing "D" size batteries with a life expectancy of 87 hours. Due to
an improved production process, management believes that there has been an increase in the life expectancy of their "D"
size batteries. A sample of 36 batteries showed an average life of 88.5 hours. Assume from past information that it is
known that the standard deviation of the population is 9 hours.
a.       At 99% confidence, test management's belief.

1

Problem 13.

In a large university, 20% of the students are business majors. A random sample of 100 students is selected, and their
majors are recorded.
a.       What is the probability that the sample contains at least 12 business majors?
b.       What is the probability that the sample contains less than 15 business majors?
c.       What is the probability that the sample contains between 12 and 14 business majors?

a.      0.9772
b.      0.1056
c.      0.044

Problem 14.

The average monthly electric bill of a random sample of 256 residents of a city is \$90 with a standard deviation of \$24.
a.     Construct a 90% confidence interval for the mean monthly electric bills of all residents.

a.      87.5325 to 92.4675

Problem 15.

In a random sample of 400 registered voters, 120 indicated they plan to vote for Candidate A. Determine a 95%
confidence interval for the proportion of all the registered voters who will vote for Candidate A.

0.255 to 0.345
Problem 16.

MNM Corporation gives each of its employees an aptitude test. The scores on the test are normally distributed with a
mean of 75 and a standard deviation of 15. A simple random sample of 25 is taken from a population of 500.
a.     What is the probability that the average aptitude test in the sample will be between 70.14 and 82.14?
b.     What is the probability that the average aptitude test in the sample will be greater than 82.68?

a.      0.9387
b.      0.0052

Problem 17.

In a sample of 200 individuals, 120 indicated they are Democrats. Develop a 95% confidence interval for the proportion
of people in the population who are Democrats.

.5321 to .6679

Problem 18.

Scores on a recent national statistics exam were normally distributed with a mean of 80 and a standard deviation of 6.
a.      What is the probability that a randomly selected exam will have a score of at least 71?
b.      What percentage of exams will have scores between 89 and 92?

a.      .9332
b.      .04

Problem 19.

Some people who bought X-Game gaming systems complained about having received defective systems. The industry
standard for such systems has been ninety-eight percent non-defective systems. In a sample of 120 units sold, 6 units were
defective. At 95% confidence using the critical value approach, test to see if the percentage of defective systems produced
by X-Game has exceeded the industry standard.

Ch. 9, proc. 3:
Step 1. Ha: P > 2%
Step 2. p =6/120 = 0.05 = 5%. Assuming that sample proportion is an estimate for population proportion,
Ha is supported by the sample.
0.05  0.02         0.03
Step 3. Z =                    =         = 2.35
0.02 (1  0.02 )   0.01278
120
Step 4. P-value = P(Z>2.35) = 1 - P(Z<2.35) = 1 – 0.9906 = 0.0094
Step 5. α = 1 – c = 1 – 0.95 = 0.05
Step 6. We accept the result because the P-value is less than the α. Conclusion: Our sample provides
support on Ha (the percentage of defective systems produced by X-Game has exceeded the industry
standard).

2.35

Problem 20.
A random sample of 81 credit sales in a department store showed an average sale of \$68.00. From past data, it is known
that the standard deviation is \$27.00.
With a 0.95 probability, what can be said about the size of the margin of error?

5.88

Problem 21

"DRUGS R US" is a large manufacturer of various kinds of liquid vitamins. The quality control department has noted that
the bottles of vitamins marked 6 ounces vary in content with a standard deviation of 0.3 ounces. Assume the contents of
the bottles are normally distributed.
a.       What percentage of all bottles produced contains more than 6.51 ounces of vitamins?
b.       What percentage of all bottles produced contains less than 5.415 ounces?
c.       What percentage of bottles produced contains between 5.46 to 6.495 ounces?

a.      4.46%
b.      2.56%
c.      91.46%

Problem 22

Choo Choo Paper Company makes various types of paper products. One of their products is a 30 mils thick paper. In
order to ensure that the thickness of the paper meets the 30 mils specification, random cuts of paper are selected and the
thickness of each cut is measured. A sample of 256 cuts had a mean thickness of 30.3 mils with a standard deviation of 4
mils.
At 95% confidence, test to see if the mean thickness is significantly more than 30 mils.

1.2

Problem 23

An automotive repair shop has determined that the average service time on an automobile is 2 hours with a standard
deviation of 32 minutes. A random sample of 64 services is selected.
a.      What is the probability that the sample of 64 will have a mean service time greater than 114 minutes?

a.      0.9332

Problem 24

A major department store has determined that its customers charge an average of \$500 per month, with a standard
deviation of \$80. Assume the amounts of charges are normally distributed.
a.      What percentage of customers charges more than \$380 per month?
b.      What percentage of customers charges less than \$340 per month?
c.      What percentage of customers charges between \$644 and \$700 per month?

a.      93.32%
b.      2.28%
c.      2.97%
Problem 25

In a local university, 10% of the students live in the dormitories. A random sample of 100 students is selected for a
particular study.
a.       What is the probability that the sample proportion (the proportion living in the dormitories) is between 0.172 and
0.178?
b.       What is the probability that the sample proportion (the proportion living in the dormitories) is greater than 0.025?

a.      0.0035
b.      0.9938

Problem 26

The life expectancy of Timely brand watches is normally distributed with a mean of four years and a standard deviation of
eight months.
a.       What is the probability that a randomly selected watch will be in working condition for more than five years?
b.       The company has a three-year warranty period on their watches. What percentage of their watches will be in
operating condition after the warranty period?

a.      0.0668
b.      93.32%

Problem 27

A lathe is set to cut bars of steel into lengths of 6 centimeters. The lathe is considered to be in perfect adjustment if the
average length of the bars it cuts is 6 centimeters. A sample of 121 bars is selected randomly and measured. It is
determined that the average length of the bars in the sample is 6.08 centimeters with a standard deviation of 0.44
centimeters. Determine whether or not the lathe is in perfect adjustment.
Ch. 9, proc. 2:
Step 1. Ha: μ ≠ 6 (The lathe is NOT in perfect adjustment)
Step 2. x = 6.08. Assuming that sample proportion is an estimate for population proportion, Ha is
supported by the sample.
6.08  6 0.08
Step 3. t =          =        =2
0.44     0.04
121
Step 4. P-value for t = 2, d.f. = n – 1 = 120, two-tail P-value = 0.0478
Step 5. If the problem does not give the α, use the standard α = 0.05
Step 6. We accept the result because the P-value is less than the α. Conclusion: Our sample provides
support on Ha (the lathe is NOT in perfect adjustment).

2

Problem 28

The life expectancy in the United States is 75 with a standard deviation of 7 years. A random sample of 49 individuals is
selected.
a.       What is the probability that the sample mean will be larger than 77 years?
b.       What is the probability that the sample mean will be less than 72.7 years?
c.       What is the probability that the sample mean will be between 73.5 and 76 years?
a.       0.0228
b.       0.0107
c.       0.7745

Problem 29

Ahmadi, Inc. has been manufacturing small automobiles that have averaged 50 miles per gallon of gasoline in highway
driving. The company has developed a more efficient engine for its small cars and now advertises that its new small cars
average more than 50 miles per gallon in highway driving. An independent testing service road-tested 64 of the
automobiles. The sample showed an average of 51.5 miles per gallon with a standard deviation of 4 miles per gallon.
Determine whether or not the manufacturer's advertising campaign is legitimate.

3

Problem 30

A new brand of chocolate bar is being market tested. Four hundred of the new chocolate bars were given to consumers to
try. The consumers were asked whether they liked or disliked the chocolate bar. You are given their responses below.

Response          Frequency
Liked               300
Disliked             100
400
Construct a 95% confidence interval for the true proportion of people who liked the chocolate bar.

0.71 to 0.79

Problem 31

In order to estimate the average electric usage per month, a sample of 196 houses was selected, and their electric usage
determined. Assume a population standard deviation of 350-kilowatt hours. With a 0.95 probability, determine the margin
of error.

49

Problem 32

The Bureau of Labor Statistics reported that the average yearly income of dentists in the year 2005 was \$110,000. A
sample of 81 dentists, which was taken in 2006, showed an average yearly income of \$120,000. Assume the standard
deviation of the population of dentists in 2006 is \$36,000. We want to test to determine if there has been a significant
increase in the average yearly income of dentists.

0.0062

Problem 33

A sample of 81 account balances of a credit company showed an average balance of \$1,200 with a standard deviation of
\$126.
Determine whether the mean of all account balances is significantly different from \$1,150.
3.57

Problem 34

Automobiles manufactured by the Efficiency Company have been averaging 42 miles per gallon of gasoline in highway
driving. It is believed that its new automobiles average more than 42 miles per gallon. An independent testing service
road-tested 36 of the automobiles. The sample showed an average of 42.8 miles per gallon with a standard deviation of 1.2
miles per gallon. Test the belief at 95% confidence level.

4.0

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