Stastics exam

MATH 105 - Worksheet Week 8 Names:

1. The drug Ziac is used to treat hypertension. In a clincial test, 3.2% of 221 Ziac users experienced

dizziness. Calculate and interpret a 95% confidence interval for the proportion of Ziac users who will

experience dizziness.

2. In a study designed to test the effectiveness of echinacea for treating upper respiratory tract infections

in children, 337 children were treated with echinacea and 370 other children were given a placebo.

The number of days of peak severity of symptoms for the echinacea group had a mean of 6.0 days

and a standard deviation of 2.3 days. The numbers of days of peak severity of symptoms for the

placebo group had a mean of 6.1 days and a standard deviation of 2.4 days.

(a) Construct the 95% confidence interval for the mean number of days of peak severity of symp-

toms for those who receive echinacea treatment. Interpret the confidence interval.

(b) Construct the 95% confidence interval for the mean number of days of peak severity of symp-

toms for those who are given a placebo. Interpret the confidence interval.

(c) Compare the two confidence intervals. What do the results suggest about the effectiveness of

echinacea? Justify your answer based on the confidence intervals.

3. A researcher claimed that less than 20% of adults smoke cigarettes. A Gallup survey of 1016

randomly selected adults showed that 17% of the respondents smoke. Is this evidence to support the

researcher’s claim?

(a) What is the sample proportion, p̂, for this problem?

(b) State the null and alternative hypothesis.

H0 : p =

H1 : p <

Note that you should have written the same number in the null and alternative hypotheses.

This is the value that the researcher is claiming. The sample proportion should never go in the

null or alternative hypotheses. Therefore,the number you have in (a) should not be in the null

or alternative hypothesis.

(c) Calculate the test statistic using the following formula:

z = p̂− p√

pq n

=

where p is the value you stated in the null hypothesis and q is 1− p.

(d) Now, using either your calculator or the normal distribution table, look up the area that cor-

responds to the z-score you calculated in (c). You don’t need to subtract from 1 because the

alternative hypothesis has < in it. This value is called the p-value.

p-value=

The p-value represents the probability of getting a p̂ of .17 or smaller if the null hypothesis

is true (p=.20). Therefore, if this probability is small then we would doubt that the null hy-

pothesis is true (i.e. if it is not very likely to get p̂ if the true proportion is 20% then this

gives us reason to doubt that p = .2). Thus, small p-values support the alternative hypothesis.

P -values less than .10 or .05 are generally considered small.

(e) Circle the correct answer:

i. The p-value was small which gives us evidence that the actual percent of adults who smoke

is less than 20%.

ii. The p-value was large so we do not have evidence that the actual percent of adults who

smoke is less than 20%.

The answer your circled in (e) is the conclusion to the hypothesis test. We always state conclusions in

context of the problem and we state whether we did or did not have evidence for the alternative hypothesis.

In this case, we did have evidence for the alternative hypothesis that the percentage of adult smokers is

less than 20%.