ST 311
Evening Problem Session Solutions
Week 10
1. p. 576, Question 26 (Module 9.1) [Learning Objectives H2, H3, H6-H9, H11, H12]
During an angiogram, heart problems can be examined via a small tube (a catheter)
threaded into the heart from a vein in the patient’s leg. It’s important that the company
that manufactures the catheter maintain a diameter of 2.00 mm. The company regularly
tests to see if they are meeting the standard of 2.00 mm in order to ensure a certain level of
quality of product. For the most recent test, the company randomly sampled 36 catheters
and found an average diameter of 2.03 mm with a standard deviation of 0.15 mm. Answer the following questions about the hypothesis test you would use to report back to the
company’s quality control manager:
a) Is this question addressing means or proportions?
b) Are the necessary assumptions to make inference satisfied?
c) Should you use a one or two-sided hypothesis test? Write the hypotheses.
d) Calculate the appropriate test statistic for the hypothesis test and find the p-value.
e) State your conclusion both in terms of your hypotheses and how you would explain the
results to the company.
f) The company decides to validate your findings with another random sample of 121
catheters and again finds an average diameter of 2.03 mm with a standard deviation of
0.15 mm. Repeat parts a) through e) for these new values.
g) Do the two samples reach the same conclusion?
h) Calculate a 95% confidence interval for each of the samples. Do these confidence intervals make sense given the results from your hypothesis test?
Solutions:
a) This question addresses means
b) Yes, we are randomly selecting more than 30, so the conditions are satisfied.
c) Here we want to use a two-sided hypothesis test with the following hypotheses:
H0 ∶ µ = 2.00
HA ∶ µ ≠ 2.00
d) The appropriate test statistic for means is a t statistic.
t=
2.03 − 2.00
0.15
√
36
= 1.2
We need to know the number of degrees of freedom for this problem. We sampled 36
catheters, so we have 36 - 1 = 35 degrees of freedom. Looking at the t-table on the row
for 35 degrees of freedom, we find that t = 1.2 with 35 degrees of freedom indicates that
the p-value is greater than 0.2.
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e) Based upon the fact that p > 0.2 we fail to reject H0 . In this context of this problem,
there is not significant evidence to suggest that the catheter diameter is different from
0.2 mm
f) When we increase our sample size, the t-statistic will change. The new t-statistic will
be
t=
2.03 − 2.00
0.15
√
121
= 2.2
Additionally our degrees of freedom increases to 120. Looking at the t-table on the row
for 120 degrees of freedom, we find that t = 2.2 with 120 degrees of freedom indicates
that 0.05 > p − value > 0.02. Therefore based upon the fact that p < .05 we should reject
H0 in favor of HA . Thus, in the context of this problem, there is significant evidence to
suggest that the catheter diameters are different from 0.2 mm.
g) No, we do not arrive at the same conclusions for each of these samples.
h) There is a very strong relationship between a 95% confidence and a two-sided hypothesis
test where we compare our p-value to 0.05. If we construct a 95% confidence interval
and it contains the null value from our hypothesis test, then we will fail to reject the
null hypothesis. If we construct a 95% confidence interval and it does not contain the
null value from our hypothesis test, then we will reject the null hypothesis. For this
problem, the two 95% confidence intervals are:
= (1.979, 2.081)
i) n = 36. Confidence Interval: 2.03 ± 2.030 √.15
36
ii) n = 121. Confidence Interval: 2.03 ± 1.980 √.15
= (2.003, 2.057)
121
Yes, in the first test, we failed to reject the null hypothesis, and the 95% confidence interval includes the null value of 2.00. In the second test, we rejected the null hypothesis,
and the 95% confidence interval does not include the null value of 2.00.
2. p. 577, Question 30, v. 1 (Module 9.1) [Learning Objectives H3, H6-H9]
A company with a large fleet of cars hopes to keep gasoline costs down and sets a goal
of attaining a fleet average of at least 26 miles per gallon. The company’s fleet manager
does not believe that they have met this goal, and would like to test his hypothesis. So,
they check the gasoline usage for 46 company trips chosen at random, finding a mean of
25.02 mpg and a standard deviation of 4.83 mpg. Answer the following questions about
the hypothesis test you would use to report back to the company’s fleet manager:
a) Is this question addressing means or proportions?
b) Are the necessary assumptions to make inference satisfied?
c) Should you use a one or two-sided hypothesis test? Write the hypotheses.
d) Calculate the appropriate test statistic for the hypothesis test and find the p-value.
e) State your conclusion both in terms of your hypotheses and how you would explain the
results to the company.
Solutions:
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Evening Problem Session Solutions
Week 10
a) This question is addressing means
b) The necessary assumptions are satisfied. We are taking a random sample. We do not
know if the trip mileages come from a normal distribution; however, in this case n ≥ 30,
so we are ok.
c) We should use a one-sided hypothesis test:
H0 ∶ µ = 26
HA ∶ µ < 26
d) Because we are researching means here, and we do not know the population standard
deviation, the appropriate statistic for this question is a t statistic:
t=
25.02 − 26
4.83
√
46
= −1.38
We are interested in finding P (t < −1.38). We need to know the number of degrees
of freedom for this problem. We sampled 46 trips, so we have 46 - 1 = 45 degrees of
freedom. One thing you might notice is that our t-table only includes positive values,
but the statistic we calculated is negative. This is actually ok because the t-distribution
is symmetric and centered around 0. Therefore P (t < −1.38) = P (t > 1.38). So, looking
at the t-table on the row for 45 degrees of freedom, and we find that t = 1.38 with 45
degrees of freedom indicates that the p-value is between 0.1 and 0.05.
e) Because our p-value is greater than 0.05, we fail to reject the null hypothesis. The
sentence we write to the company’s fleet manager should be worded as follows. There is
not significant evidence to suggest that the fleet has not met the fuel efficiency standards
for the company.
3. p. 577, Question 30, v. 2 (Modules 8.1-8.4) [Learning Objectives H3, H6-H9]
Another company with a large fleet of cars has decided that their fleet is fuel efficient if
three-quarters of the cars average at least 24 miles per gallon. The company’s fleet manager
believes that they have met this goal, so he would like to test his hypothesis. They check the
gasoline usage for 90 company cars chosen at random, finding that 13 cars have a mileage
of less than 24 miles per gallon. Answer the following questions about the hypothesis test
you would use to report back to the company’s fleet manager:
a) Is this question addressing means or proportions?
b) Are the necessary assumptions to make inference satisfied?
c) Should you use a one or two-sided hypothesis test? Write the hypotheses.
d) Calculate the appropriate test statistic for the hypothesis test and find the p-value.
e) State your conclusion both in terms of your hypotheses and how you would explain the
results to the company.
Solutions:
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Evening Problem Session Solutions
Week 10
a) This question is addressing proportions
b) Yes the necessary assumptions are satisfied. We are taking a random sample. We need
to check to see if n ⋅ p0 ≥ 10 and n ⋅ q0 ≥ 10. In this example, the two values we obtain
are 67.5 and 22.5 respectively.
c) We should use a one-sided hypothesis test:
H0 ∶ p = 0.75
HA ∶ p > 0.75
d) Since this question is about proportions, the appropriate test statistic is a z-statistic.
p̂ =
0.86 − 0.75
77
= 2.41
= 0.86 z = √
90
(0.75)(0.25)
90
We are interested in finding P (z > 2.41). So we need to look at the z-table. From the
z-table we see that P (z < 2.41) = 0.9920, so
P (z > 2.41) = 1 − P (z < 2.41) = 1 − 0.9920 = 0.008
e) Because our p-value is less than 0.05, we reject the null hypothesis. The sentence we
write to the company’s fleet manager should be worded as follows. There is significant
evidence to suggest that the fleet has met the fuel efficiency standards for the company.
4. p. 523, Question 28 (Modules 8.1-8.4) [Learning Objectives H3, H6-H10]
A company is willing to renew its advertising contract with a local radio station only if
the station can prove that more than 20% of the residents of the city have heard the ad
and recognize the company’s product. However, the company is skeptical that the radio
station has actually met this goal. The radio station conducts a random phone survey of
400 people, and only 62 people remember the ad. Answer the following questions about
the hypothesis test you would use to report back to the company’s ad rep.
a) Is this question addressing means or proportions?
b) Are the necessary assumptions to make inference satisfied?
c) Should you use a one or two-sided hypothesis test? Write the hypotheses.
d) Calculate the appropriate test statistic for the hypothesis test and find the p-value.
e) State your conclusion both in terms of your hypotheses and how you would explain the
results to the company.
f) Would your conclusion change if you used a 10% significance level? How about a 1%
level.
Solutions:
a) This question is addressing proportions
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b) Yes the necessary assumptions are satisfied. We are taking a random sample. We need
to check to see if n ⋅ p0 ≥ 10 and n ⋅ q0 ≥ 10. In this example, the two values we obtain
are 80 and 320 respectively.
c) We should use a one-sided hypothesis test:
H0 ∶ p = 0.20
HA ∶ p < 0.20
d) Since this question is about proportions, the appropriate test statistic is a z-statistic.
p̂ =
62
0.16 − 0.20
= 0.16 z = √
= −2.00
400
(0.20)(0.80)
400
We are interested in finding P (z < −2.00). So we need to look at the z-table. From the
z-table we see that
P (z < −2.00) = 0.0228
e) Because our p-value is less than 0.05, we reject the null hypothesis. The sentence we
write to the company’s ad rep should be worded as follows. There is significant evidence
to suggest that less than 20% of residents of the city have heard the ad and recognize
the company’s product.
f) We would still reject the null hypothesis at the 10% level since the p-value is less than
0.10. However, at the 1% significance level, we would fail to reject since the p-value is
greater than 0.01.
5. Additional Question 1 (Module 9.1) [Learning Objectives H3, H6-H9, H11]
A potato chip manufacturer lists the net weight of one of its packages as 28.3 grams. It
is believed that the population of these net weights follows a normal distribution. On an
annual basis the company tests its machines that fill the bags to see if they need to be
recalibrated. From a random sample of 20 bags from one of the machines, they find average
weight of 30 grams with a standard deviation of 4.0 grams.
a) Is this significant evidence that this machine should be recalibrated? Make sure you
identify each of the necessary parts of a hypothesis test in answer this question.
b) Calculate the 95% confidence interval for this statistic. Does this confidence interval
make sense given the results from your hypothesis test?
Solutions: The solution to the problem below will follow the steps listed for each of the
problems above.
a) In this problem, we are dealing with means. Yes, we are randomly selecting from a normal distribution, so the conditions are satisfied. This should be a two-sided hypothesis,
with
H0 ∶ µ = 28.3
HA ∶ µ ≠ 28.3
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Evening Problem Session Solutions
Week 10
Because we are testing for means, and we do not know the population standard deviation, we need to calculate a t-statistic:
t=
30 − 28.3
√4
20
= 1.90
We need to know the number of degrees of freedom for this problem. We sampled 20
bags, so we have 20 - 1 = 19 degrees of freedom. Looking at the t-table on the row
for 19 degrees of freedom, we find that t = 1.9 with 19 degrees of freedom indicates
0.10 > p − value > 0.05. Then since the p-value is greater than 0.05, we fail to reject the
null hypothesis. Therefore, there is not significant evidence to suggest that this machine
needs to be recalibrated.
b) The 95% confidence interval here is 30±2.093 √420 = (28.13, 31.87). This is not surprising
because in our hypothesis test, we failed to reject the null hypothesis, and the 95%
confidence interval includes the null value of 28.3.
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