Question

The claim is that the proportion of adults who smoked a cigarette in the past week is less than 0.30?, and the sample statistics include n equals 1271 subjects with 394 saying that they smoked a cigarette in the past week. Find the value of the test statistic.

Answer 1

Solution:-

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

Null hypothesis: P > 0.30
Alternative hypothesis: P < 0.30

Note that these hypotheses constitute a one-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 = 0.01285
z = (p - P) / S.D

z = 0.78

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 one-tailed test, the P-value is the probability that the z-score is greater than 0.78.

Thus, the P-value = 0.218.

Interpret results. Since the P-value (0.218) is greater than the significance level (0.05), we have to accept the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that the proportion of adults who smoked a cigarette in the past week is less than 0.30.