Math 1342
Test #3 Review
Chapters 9 and 10
Concepts to know in Chapter 9:
Compute a point estimate
Construct a Confidence Interval
Interpret a Confidence Interval
Margin of Error
Sample Size
Concepts to know in Chapter 10:
1. State the Claim in English
2. Determine the null and alternative Hypotheses
3. State the Level of significance
4. Find the Critical Value
5. Draw the Picture
6. Find the Test Statistic
7. Find the P-value
8. Make a Decision
9. State the Conclusion
Summary of Major Concepts for Chapters 9 and 10
p̂ 
x
n
 z
Sample Size Estimate for Proportions: n  p̂q̂ 2
 E

Confidence Interval for µ:
Confidence Interval for p:
T-Interval
1-PropZInt
Confidence Interval for  2 :
Lower bound =




2
 z
if a p̂ is known and n  0.25 2
 E

(n  1)s 2
 2
2
Test Statistics:
Means:
T-Test
Proportions: 1-PropZTest
(n  1)s 2
SD/Variance:  2 
2
Upper bound =
(n  1)s 2
2 
1
2
2

 otherwise


Math 1342
Test #3 Review
Chapters 9 & 10
1. A simple random sample of size n = 25 is drawn from a population that is normally distributed.
The sample variance is found to be s2 = 3.97. Construct and interpret a 95% confidence interval for
the population standard deviation.
2. A simple random sample of size n = 20 is drawn from a population that is normally distributed.
The sample variance is found to be 49.3. Determine whether the population variance is less than 95
at the =0.1 level of significance.
3. A simple random sample of size n = 320 adults was asked their favorite ice cream flavor. Of the
320 individuals surveyed, 85 responded that they preferred mint chocolate chip. Do less than 25% of
adults prefer mint chocolate chip ice cream? Use the =0.01 level of significance.
4. A simple random sample of size n = 14 is drawn from a population that is normally distributed.
The sample mean is found to be 60 with a sample standard deviation of 20. Construct and interpret
a 90% confidence interval for the population mean.
5. In October 1945, the Gallup organization asked 1,487 randomly sampled Americans, “Do you
think we can develop a way to protect ourselves from atomic bombs in case other countries tried to
use them against us?” with 788 responding yes. Did a majority of Americans feel the United States
could develop a way to protect itself from atomic bombs in 1945? Use the =0.05 level of
significance.
6. Fifty rounds of a new type of ammunition were fired from a test weapon, and the muzzle velocity
of the projectile was measured.
The sample had a mean muzzle velocity of 863 meters per second and a standard deviation of 2.7
meters per second. Construct and interpret a 99% confidence interval for the mean muzzle velocity.
7. According to the Insurance Information Institute, the mean expenditure for auto insurance in the
United States was $774 in 2002. An insurance sales person believes that mean expenditure for auto
insurance is different now. He obtains a random sample of 35 auto insurance policies and
determines the mean expenditure to be $735 with a standard deviation of $48.31. Is there enough
evidence to conclude that the mean expenditure for auto insurance is different from the 2002 amount
at the =0.01 level of significance?
8. A pharmaceutical company manufactures a 200-milligram (mg) pain reliever. Company
specifications require that the standard deviation of the amount of the active ingredient must not
exceed 5 mg. The quality-control manager selects a random sample of 30 tablets from a certain
batch and finds that the sample standard deviation is 5.9 mg. Assume that the amount of the active
ingredient is normally distributed. Determine whether the standard deviation of the amount of the
active ingredient is greater than 5 mg at the =0.05 level of significance.
9. In a February 2007 Harris Poll, 881 of 1,013 randomly selected adults said that they always wear
seat belts. Construct and interpret a 98% confidence interval for the proportion of adults who always
wear seat belts. Express the confidence interval in the form of p̂ ± E.
10. With a previous contractor, the mean time to replace a streetlight was 3.2 days. A city
councilwoman thinks that the new contractor is not getting the streetlights replaced as quickly. She
selects a random sample of 12 streetlight service calls and obtains the following times to replacement
(in days). Is there enough evidence to support the councilwoman’s belief at the =0.05 level of
significance?
6.2
7.1
5.4
5.5
7.5
2.6
4.3
2.9
3.7
0.7
5.6
1.7
11. A city councilwoman selects a random sample of 12 streetlight service calls and obtains the
following times to replacement (in days). Construct and interpret a 98% confidence interval for the
mean amount of time for replacement of streetlights.
6.2
4.3
7.1
2.9
5.4
3.7
5.5
0.7
7.5
5.6
2.6
1.7
12. A simple random sample of size n = 320 adults was asked their favorite ice cream flavor. Of
the 320 individuals surveyed, 85 responded that they preferred mint chocolate chip. State p̂ .
Construct and interpret a 95% confidence interval for the true percentage of all adults that have high
blood pressure.
13. A school administrator wishes to estimate the proportion of high school seniors who are
planning to go to college after graduation. How many students must the administrator sample to
be within 2.5 percentage points of the true proportion with 96% confidence?
14. In a previous study, it was reported that 68% of a local town’s high school seniors planned to go
to college after graduation. A school administrator needs an updated estimate of the proportion of
high school seniors who are planning to go to college after graduation. How many students must
the administrator sample to be within 2.5 percentage points of the true proportion with 96%
confidence?
Math 1342
Answers Test #3 Review
1. s2 = 3.97 n = 25 df = 24
,
95% CI
Chapters 9 & 10
= .05 X2 = 12.401 and 39.364 (from chi-square table)
(1.556, 2.772)
We are 95% confident that the population standard deviation is between 1.556 and 2.772.
2. Claim: The population variance is less than 95.
Hypothesis: H0:
H1:
2
= 95
< 95
2
_________________
Level of significance:
ts
=0.1
11.651
Critical Value: X2 = 11.651 (from chi-square table, df = 19)
Test Statistic: 9.86
ts = 19(49.3)
95
Decision: Reject H0
Conclusion: There is sufficient evidence to claim that the population variance is less than 95.
3. Claim: Less than 25% of adults prefer mint chocolate chip ice cream.
Hypothesis: H0: = .25
H1: < .25
Level of significance:
=0.01
Critical Value: z = -2.33 Calculator: z = invNorm(.01)
Test Statistic: 0.645
Calculator: 1-PropZTest x = 85 n = 320
P-value: .7407
Calculator: 1-PropZTest
Decision: Do Not Reject H0
_______ ts______
-2.33
ts is NOT in the rejection region; p >
Conclusion: There is NOT sufficient evidence to conclude that less than 25% of adults prefer mint
chocolate chip ice cream.
4.(50.534, 69.466)
Calculator: TInterval
We are 90% confident that the population mean is between 50.534 and 69.466.
5. n = 1487, x = 788, p̂ = 788/1487 = 0.53
Claim: A majority of Americans feel the US could develop a way to protect itself.
Hypothesis: H0: = .5
H1: > .5
Level of significance:
=0.05
______________
Critical Value: z = 1.645
Calculator: z = invNorm(.05)
Test Statistic: 2.31
Calculator: 1-PropZTest
P-value: .01
Calculator: 1-PropZTest
Decision: Reject H0
ts is in the rejection region; p <
1.645 ts = 2.31
Conclusion: There is sufficient evidence to support the claim that the majority of Americans feel that
the US could develop a way to protect itself.
6. (861.98, 864.02) Calculator: TInterval
Interpretation: If many different samples of size 50 were selected and, based on each sample, a
confidence interval were constructed, in the long run 99% of the confidence intervals would contain
the true value of µ.
7. Claim: The mean expenditure for auto insurance is different than $774.
Hypothesis: H0: = 774
H1: ≠ 774
Level of significance:
=0.01
Critical Value: t = ±2.728
(from t-table, df = 34)
Test Statistic: -4.776
Calculator: T-Test
P-value: .00003
Calculator: T-Test
Decision: Reject H0
ts -2.728
2.728
ts is in the rejection region [ts < cv]; p <
Conclusion: There is sufficient evidence to support the claim that the mean expenditure for auto
insurance is different than $774.
8. Claim: The standard deviation of the amount of the active ingredient is greater than 5 mg.
Hypothesis: H0: = 5
H1: > 5
Level of significance:
Critical Value:
X2
=0.05
ts
42.557
= 42.557 (from chi-square table, df = 29)
ts = 29(5.92)
52
Decision: Do Not Reject H0
Test Statistic: 40.38
No p-value
ts is NOT in the rejection region
Conclusion: There is Not sufficient evidence to support the claim that the standard deviation of the
amount of the active ingredient is greater than 5 mg.
8. (.845, .894)
p̂ ± E:
Calculator: 1-PropZInt
0.894  0.845
±
2
0.894  0.845
= 0.8695 ± 0.0245
2
Interpretation: If many different samples of size 1,013 were selected and, based on each sample, a
confidence interval were constructed, in the long run 98% of the confidence intervals would contain
the true value of p.
10. Claim: Streetlights are not getting replaced as quickly as 3.2 days.
Hypothesis: H0: = 3.2
H1: > 3.2
Level of significance:
=0.05
Critical Value: t = 1.796 (from the t-table, df = 11)
_____________
1.796
Test Statistic: 1.989
Calculator: T-Test
P-value: .0361
Calculator: T-Test
Decision: Reject H0
ts is in the rejection region [ts > cv]; p <
ts = 1.989
Conclusion: There is sufficient evidence to support the claim. Based on the evidence, it can be
concluded that streetlights are not getting replaced as quickly as 3.2 days.
11. (2.748, 6.1186)
Calculator: TInterval
We are 98% confident that the true mean amount of time for replacement of streetlights is between
2.748 and 6.1186 days.
12. (0.2172, 0.3140)
21.72% < p < 31.4%
Calculator: 1-PropZInt
We are 95% confident that the true percentage of all adults that have high blood pressure
is between 21.72% and 31.4%.
13.  = 0.04

2
= 0.02
z  / 2 = 2.05
E = 2.5% = 0.025
2
 2.05 
n = .25 
 = 1681
 0.025 
14.  = 0.04

= 0.02 z  / 2 = 2.05
E = 2.5% = 0.025
2
2
 2.05 
n = 0.68∙0.32 
 = 1463.1424 round up
 0.025 
n = 1464
p̂ = 0.68
q̂ = 1 – 0.68 = 0.32