Question

A researcher is wondering whether the smoking habits of young adults (18-25 years of age) in a certain city in the U.S. are
the same as the proportion of the general population of young adults in the U.S. A recent study stated that the proportion
of young adults who reported smoking at least twice a week or more in the last month was 0.16. The researcher collected
data from a random sample of 75 adults in the city of interest, report 6 of them smoke at least twice a week or more in the
last month.

What is p-value for the claim?

Answer 1

Solution:-

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

Null hypothesis: P = 0.16
Alternative hypothesis: P 0.16

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation (S.D) and compute the z-score test statistic (z).

S.D = sqrt[ P * ( 1 - P ) / n ]

S.D = 0.042332
z = (p - P) /S.D

z = - 1.89

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than -1.89 or greater than 1.89.

Thus, the P-value = 0.059.

Interpret results. Since the P-value (0.059) is greater than the significance level (0.05), we failed to reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim the smoking habits of young adults in a certain city in the U.S. are the smoking habits of young adults in a certain city in the U.S. are the same as the proportion of the general population of young adults in the U.S.