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 the 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 flu-like 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
Ha:p > .019right 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 = .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
(use the z-table to find the area to the left of z = 0.65 which is 0.7422)
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:
.0104 < p < .0336(notice the margin of error = 0.0116 ~ 0.012 = 1.2%)
d) Complete the following:
- 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 > .06right 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
(Go to the z-table and get the area to the left of 4.47 which is almost 1)
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:
.0731 < p < .1249(notice the margin of error = 0.0259 ~ 0.026 = 2.6%)
d) Complete the following:
- 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