Question

 

A study examined parental influence on the decisions of teenagers from a certain large region to smoke. A randomly selected group of​ students, from the​ region, who had never smoked were questioned about their​ parents' attitudes toward smoking. These students were questioned again two years later to see if they had started smoking. The researchers found​ that, among the 256 students who indicated that their parents disapproved of kids​ smoking, 72 had become established smokers. Among the 30 students who initially said their parents were lenient about​ smoking,11 became smokers. Do these data provide strong evidence that parental attitude influences​ teenagers' decisions about​ smoking?

a) What kind of design did the researchers use?

b) Write the appropriate hypothesis

c) Are the assumptions and conditions satisfied?

d) Test the hypothesis and state the conclusion z score and p value

Reject or do not reject

Explain the p value

What type of error was committed

Create a 99% CI

Interpret interval

Answer 1

 

a) The researchers is using observational design study.

b)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P_Disapproved> P_Lenient
Alternative hypothesis: P_Disapproved < P_Lenient

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

c) Since np and n(1-p) are greater than 5, hence all the conditions are satisfied.

p = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂)

p = 0.29021

SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.087584
d)

z = (p₁ - p₂) / SE

z = - 0.98

where p₁ is the sample proportion in sample 1, where p₂ is the sample proportion in sample 2, n₁ is the size of sample 1, and n₂ is the size of sample 2.

Since we have a one-tailed test, the P-value is the probability that the z-score is less than - 0.98

Thus, the P-value = 0.1635

Interpret results. Since the P-value (0.1635) is greater than the significance level (0.01), we cannot reject the null hypothesis.

Do not reject the null hypothesis.