1)
– Side effects of Lipitor
The drug Lipitor is meant to reduce total cholesterol and LDL-cholesterol. In clinical trials, 19 out of 863 patients taking 10 mg of Lipitor daily complained of flu-like symptoms. Suppose that it is known that 1.9% of patients taking competing drugs complain of flu-like symptoms.
Part I - Is there significant evidence to support the claim that more than 1.9% of Lipitor users experience flu-like symptoms as a side effect at t he α = 0.01 level of significance?
Part II
– Construct a 98% confidence interval estimate for the proportion of all patients who experience flu-like symptoms when taking 10 mg Lipitor daily. (Go to page 8 to do this)
Part I - Is there significant evidence to support the claim that more than 1.9% of Lipitor users experience flulike symptoms as a side effect at the α = 0.01 level of significance? a) Describe in words the population and success attribute
People taking 10 mg of Lipitor daily
Experience flu-like symptoms b) List all relevant statistics x = 19, n = 863, From this we get p-hat = 19/863 = .022
α = 0.01, p = .019 (proportion of people who take other drug and have flu like symptoms)
Note: the proportion experiencing flu-like symptoms is higher in the sample of Lipitor users; is that due to chance or could it be because of the drug? c) Verify assumptions
We are going to assume that for patients taking 10 mg of Lipitor daily the proportion experiencing flu like symptoms is the same as the one for the competing drug; that is 1.9%
Using p = 0.019, we check the assumptions:
Np = 863 (.019) = 16.4, and n(1-p) = 863(.981) = 846.6
Since both are greater than 10, the distribution of p-hat (for samples of size 863) is approximately normal
The mean is p = 0.019 and the standard deviation of p-hat is SQRT( 0.019*0.981/863) = 0.004647 d) Set both hypothesis
H0: p = .019 right tailed test Ha: p > .019 e) Sketch graph, shade rejection region, label, and indicate possible locations of the point estimate in the graph.
You do this. The point estimate is p-hat = .022
****You should be wondering: Is p-hat =_ 0.022
__ higher than p = _ .019
_ by chance, or is it significantly higher? The p-value found below will help you in answering this. f) Run the test: 1-prop-ZTest: and obtain prop > 0.019, test statistic z = 0.65, p-value = P(p-hat > 0.022) = 0.258 > 0.01
If the true proportion is 0.019, it is very likely to observe a sample proportion p-hat of 0.022 or a more extreme one in samples of size 863. P-hat of 0.022 is higher than 0.019 by chance.
At the 1% significance level, the sample data do not provide enough evidence to support the claim that more than 1.9% of Lipitor users experience flu-like symptoms as a side effect.
Notice that the conclusion will be the same at any significance level because the p-value is very high.

g) Use formulas to find the test-statistic z
  
.65
(use the z-table to find the area to the left of z = 0.65 which is 0.7422) pq .019*.981
n 863 h) Find the p-value showing all steps
P(p-hat > .022) = P(z > .65) = 1 - .7422 = .2578 > significance level
Part II – Construct a 98% confidence interval estimate for the proportion of all patients who experience flu-like symptoms when taking 10 mg Lipitor daily. a) Verify assumptions
For confidence intervals the assumptions are that np and n(1-p) are both greater than 15. This holds on true in this case
Np = 863 (.019) = 16.4, and n(1-p) = 863(.981) = 846.6 b) Construct the interval with the calculator:
Use 1-prop-ZInterval and get (.0104 , .03364)
We are 98% confident that the proportion of Lipitor users experiencing flu-like symptoms is somewhere between 0.0104 and 0.03364. Since the interval contains .019, it’s possible for p to be
0.019. With 98% confidence we can say that we don’t have enough evidence to support the claim that more than 1.9% of Lipitor users experience flu-like symptoms.
This conclusion agrees with the hypothesis testing conclusion. c) Now with the formulas: p
 z c
  n 863
.0104 < p < .0336 d) Complete the following:
(notice the margin of error = 0.0116 ~ 0.012 = 1.2%)
 We are __ 98 ___% confident that the percentage of patients taking 10 mg of Lipitor daily complaining of flu-like symptoms is between ___ 1.0
% and __ 3.4
%
 With __ 98 % confidence we can say that ___ 2.2
% of patients who take 10 mg of Lipitor daily complain of flu like symptoms, with a margin of error of ____ 1.2
%
 The statement “98% confident” means that, if 100 samples of size __ 863 ___ were taken and the corresponding 98% confidence intervals constructed around the sample proportion, about __ 98 ___ of the intervals will contain the parameter p and about ___ 2 will not.
 For 98% of such intervals, the sample proportion would not differ from the actual population proportion by more than ___ .0116
___

2) CLINICAL TRIAL OF TAMIFLU
Clinical trials involved treating flu patients with Tamiflu, which is a medicine intended to attack the influenza virus and stop it from causing flu symptoms. Among 724 patients treated with Tamiflu, 72 experienced nausea as an adverse reaction.
Part I – Use a 0.01% significance level to test the claim that the rate of nausea is greater than the 6% rate experienced by flu patients given a placebo. Does nausea appear to be a concern for those given the
Tamiflu treatment?
Part II- Construct a 98% confidence interval estimate for the proportion of patients treated with Tamiflu who experience nausea as an adverse reaction.
PART I a) Describe in words the population and success attribute
Flu patients treated with Tamiflu
Experience nausea as an adverse reaction b) List all relevant statistics x = 72, n = 724, From this we get p-hat = 72/724 = 0.099 (9.9%)
α = 0.01, p = .06 (proportion experiencing nausea in the group receiving a placebo)
Note: the proportion experiencing nausea is higher in the sample of Tamiflu users; is that due to chance or could it be because of the drug? c) Verify assumptions
We are going to assume that patients taking Tamiflu experience nausea at the same rate as the ones taking the placebo, which is 6%
Using p = 0.06, we check the assumptions:
Np = 724*0.06 = 43.4, and n(1-p) = 724(.94) = 680.6
Since both are greater than 10, the distribution of p-hat (for samples of size 724) is approximately normal
The mean is p = 0.06 and the standard deviation of p-hat is SQRT( 0.06*0.94/724) = 0.008826 d) Set both hypothesis
H0: p = .06
Ha: p > .06 right tailed test e) Sketch graph, shade rejection region, label, and indicate possible locations of the point estimate in the graph.
You do this. The point estimate is p-hat = .099
****You should be wondering: Is p-hat =_ 0.099
__ higher than p = _ .06
_ by chance, or is it significantly higher? The p-value found below will help you in answering this. f) Run the test: 1-prop-ZTest: and obtain prop > 0.06, test statistic z = 4.47, p-value = P(p-hat > 0.099) = 0.0000039 < 0.01 (very strong evidence against the null hypothesis and in favor of the alternative hypothesis)
If the true proportion experiencing nausea is 6% it is very unlikely to observe 9.9% of the people in the sample experiencing nausea. P-hat of 0.099 is significantly higher than 0.06.
At the 1% significance level, there is strong evidence to support the claim that the proportion experiencing nausea is much higher in the group receiving Tamiflu. Nausea appears to be a concern for those taking Tamiflu.
Notice that the conclusion will be the same at any significance level because the p-value is very small.

g) Use formulas to find the test-statistic z
  
4.47
(Go to the z-table and get the area to the left of 4.47 which is almost 1) pq .06*.94
n 724 h) Find the p-value showing all steps
P(p-hat > .099) = P(z >4.47) = 1 – (almost 1) = almost zero < any significance level
Part II- Construct a 98% confidence interval estimate for the proportion of patients treated with
Tamiflu who experience nausea as an adverse reaction. a) Verify assumptions
For confidence intervals the assumptions are that np and n(1-p) are both greater than 15. This holds on true in this case
Np = 724*0.06 = 43.4, and n(1-p) = 724(.94) = 680.6 b) Construct the interval with the calculator:
Use 1-prop-ZInterval and get (.07357, .12532)
We are 98% confident that the proportion of Tamiflu patients experiencing nausea is between 0.074 and 0.125. Since all values in the interval are higher than 0.06, with 98% confidence we can say that the rate at of nausea is greater in the Tamiflu group than in the placebo group.
This conclusion agrees with the hypothesis testing conclusion. c) Now with the formulas: p
 z c p
  p )

.099

2.33
 n 724
.0731 < p < .1249 d) Complete the following:
(notice the margin of error = 0.0259 ~ 0.026 = 2.6%)
 We are __ 98 ___% confident that the percentage of patients taking Tamiflu and experiencing nausea is between ___7.3% and __ 12.5
%
 With __ 98 % confidence we can say that ___ 9.9
% of patients who take Tamiflu experience nausea, with a margin of error of __ 2.6 %
 The statement “98% confident” means that, if 100 samples of size __ 724 ___ were taken and the corresponding 98% confidence intervals were constructed around the sample proportion, about __ 98 ___ of the intervals will contain the parameter p and about ___ 2 will not.
 For 98% of such intervals, the sample proportion would not differ from the actual population proportion by more than ___ .0259
___

3) SMOKING AND COLLEGE EDUCATION
A survey showed that among 785 randomly selected subjects who completed four years of college, 18.3% smoke.
Part I – Use a 1% significance level to test the claim that the rate of smoking among those with four years of college is less than the 27% rate for the general population.
Part II – Construct a 98% confidence interval estimate for the proportion of smokers among students who completed four years of college.
DONE IN CLASS
4) - Nasonex
In clinical trials of Nasonex, 3774 adult adolescent allergy patients (patients 12 years and older) were randomly divided into two groups. The patients in group 1 (experimental group) received
200 mcg of Nasonex, while the patients in Group 2 (control group) received a placebo. Of the 2103 patients in the experimental group, 547 reported headaches as side effect. Of the 1671 patients in the control group, 368 reported headaches as a side effect.
Part 1 - Is there significant evidence to support the claim that the proportion of Nasonex users that experienced headaches as a side effect is greater than the proportion in the control group at the 0.05 significance level?
Part 2 – Use a feature of your calculator to construct a 90% confidence interval estimate for the difference between the two population proportions. What is the interval suggesting?
Population 1: Experimental group = allergy patients (12-years and older) who received 200 mcg of Nasonex Population 2: Placebo group = allergy patients (12-years and older) who received a placebo
Success attribute: experience headache p1 is the proportion of all Nasonex users who experience headaches) p2 is the proportion of all placebo (Non-Nasonex) users who experience headaches) n
1

2103 n
2

1671 x
1

547 x
2

368 p
1

547

.260
p
2

368

.220
2103 1671
First: Verify assumptions
Check that in each population np and nq are both 10 or more
Part 1 - Is there significant evidence to support the claim that the proportion of Nasonex users that experienced headaches as a side effect is greater than the proportion in the control group at the 0.05 significance level? a) Set both hypothesis p p
1
1

 p p
2
2

 p p
1
1

 p p
2
2


0
0
This is a right tailed test b) Sketch graph, shade rejection region, label, and indicate possible locations of the point estimate in the graph.

You do this
The point estimate is p
1
 p
2

.260 .220

.04
****You should be wondering: Is the proportion that experience headaches in the experimental group larger than the one in the control group by chance, or is it significantly higher? The p-value found below will help you in answering this. c) Use a feature of the calculator to test the hypothesis. Indicate the feature used and the results:
Use 2-Prop-ZTest and get
Test statistic z = 2.84
P( p
1
 p
2

.04
) = P(z > 2.84) = p-value = .0023 < .05 (significance level)
The p-value is very small. It is very unlikely (p = 0.023) to observe such a difference between sample proportions (p1 – p2 = 0.04) when the proportion experiencing headaches is the same in both populations. Such a difference between the sample proportions is significantly larger than zero and it is more likely to be observed when p1 is larger than p2.
At the 5% significance level we support the claim that proportion of Nasonex users that experienced headaches as a side effect is greater than the proportion in the control group
Part 2 – Use a feature of your calculator to construct a 90% confidence interval estimate for the difference between the two population proportions. What is the interval suggesting?
The point estimate is p
1
 p
2

.260 .220

.04
Is p
1
 p
2
by chance or significantly lower?
Note: In the experimental group, a higher percentage experience headache, could that be because of the drug?
Construct the interval by using 2-Prop-ZInterval and get .01695 < p
1
 p
2
< .0628
We are 90% confident that the difference between the proportion experiencing headaches in the
Nasonex and the placebo groups is between 0.01695 and 0.0628. Since the interval is completely above zero, it suggests that p
1
 p
2
> 0 which means that p
1
 p
2
. This is the same as the conclusion to the hypothesis testing part.
What is the margin of error?
0.0628 – 0.04 = 0.0228
(2) In view of your answers to part 1, could you reasonably conclude that Nasonex causes headaches as a side effect?
This was an experiment and in the case of an experimental study we could interpret statistical significance as a causal relationship.
In this case, the difference of 4% is statistically significant, but it may not have any practical significance.
Most parents would be willing to accept the additional “risk” of a headache in order to relieve their child’s allergy symptoms.

5) – Vasectomies and Prostate Cancer
Approximately 450,000 vasectomies are performed each year in the U.S. In this surgical procedure for contraception, the tube carrying sperm from the testicles is cut and tied. Several studies have been conducted to analyze the relationship between vasectomies and prostate cancer. The results of one such study by E. Giovannucci et al. appeared in the paper “A
Retrospective Cohort Study of Vasectomy and Prostate Cancer in U.S. Men”. Of 21,300 men who had not had a vasectomy, 69 were found to have prostate cancer; of 22,000 men who had had a vasectomy, 113 were found to have prostate cancer.
Part 1 - At the 1% significance level, do the data provide sufficient evidence to conclude that men who have had a vasectomy are at greater risk of having prostate cancer?
Part 2 – Use the calculator to determine a 98% confidence interval for the difference between the prostate cancer rates of men who have had a vasectomy and those who have not.
Population 1: men without vasectomy
Population 2: men with vasectomy
Success attribute: have prostate cancer
Let p1 be the proportion of men without a vasectomy who have prostate cancer
Let p2 be the proportion of men with a vasectomy who have prostate cancer
NO VASECTOMY n
1

21300
VASECTOMY n
2

22000 x
1

69 x
2

113 p
1

69

.0032
p
2

113

.0051
21300 22000 p-hat(1) = 0.0032 is our first estimate for p1 p-hat (2) = 0.0051 is our first estimate for p2 p-hat(1) – p-hat(2) = 0.0032 – 0.0051= -0.0019 is our first estimate for p1 – p2
0.32% of men in the NO-vasectomy group and 0.51% in the vasectomy group had prostate cancer.
Is the percent higher in the vasectomy group by chance or is it significantly higher? Is the proportion of men with prostate cancer lower in the group of men without a vasectomy?
We are going to assume that all assumptions are met and we can perform a test with the methods learned in class.
Part 1 - At the 1% significance level, do the data provide sufficient evidence to conclude that men who have had a vasectomy are at greater risk of having prostate cancer? a) Set both hypothesis p
1
 p
2
 p
1
 p
2
 p
1
 p
2
 p
1
 p
2

0
0
This is a left tailed test b) Sketch graph, shade rejection region, label, and indicate possible locations of the point estimate in the graph.
You do this.
The point estimate is p
1
 p
2

.0032 .0051
 
.0019

****You should be wondering: Is the proportion of men with prostate cancer in the group without vasectomy lower than in the group with vasectomy by chance, or is it significantly lower? The p-value found below will help you in answering this. c) Use a feature of the calculator to test the hypothesis. Indicate the feature used and the results:
Use 2-Prop-ZTest and get
Test statistic z = - 3.05
P( p
1
 p
2
 
.0019
) = P(z < - 3.05) = p-value = .001 < .01 (significance level)
It is unlikely to observe such a difference (-0.0019) or a more extreme one between the sample proportions when you select samples from two populations that have the same proportions?
The proportion of men with prostate cancer in the group without vasectomy is significantly lower than the proportion in the group with vasectomy.
At the 1% significance level, the data provide sufficient evidence to conclude that men who have had a vasectomy are at greater risk of having prostate cancer.
Part 2 – Use the calculator to determine a 98% confidence interval for the difference between the prostate cancer rates of men who have had a vasectomy and those who have not. Are the results consistent with the results of the hypothesis test? Explain.
The point estimate is p
1
 p
2

.0032 .0051
 
.0019
Is p
1
 p
2
by chance or significantly lower?
Construct the interval by using 2-Prop-ZInterval and get -.0033 < p
1
 p
2
< -.0005
Since the interval is completely below zero, it suggests that
This is the same as p
1
 p
2
< 0 which means p
1
 p
2 p
2
 p
1
. We get the same conclusion as in the hypothesis testing part.
With 98% confidence the data provide sufficient evidence to conclude that men who have had a vasectomy are at greater risk of having prostate cancer.
Some other related questions:
(1) Is this study a designed experiment or an observational study?
Observational
(2) In view of your answers to part 1, could you reasonably conclude that having a vasectomy causes an increased risk of prostate cancer?
No, for an observational study, it is not reasonable to interpret statistical significance as a causal relationship. In the case of an experimental study we could interpret statistical significance as a causal relationship.