Question

A company claims that 10% of the users of a certain sinus drug experience drowsiness. In clinical studies of this sinus drug, 81 of the 900 subjects experienced drowsiness We want to test their claim and find out whether the actual percentage is not 10%. a. State the appropriate null and hypotheses. b. Compute the p-value of the test. Is there enough evidence at the 5% significance level to infer that the competitor is correct? c. Construct a 95% confidence interval estimate of the population proportion of the users of this allergy drug who experience drowsiness.

Answer 1

Solution:-

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

Null hypothesis: P = 0.10
Alternative hypothesis: P 0.10

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 = 0.01
z = (p - P) /S.D

z = - 1.0

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 -1.00 or greater than 1.00.

Thus, the P-value = 0.3173.

Interpret results. Since the P-value (0.3173) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that that 10% of the users of a certain sinus drug experience drowsiness.

c) 95% confidence interval estimate of the population proportion of the users of this allergy drug who experience drowsiness is C.I = (0.0804, 0.1196).

C.I = 0.10 + 1.96 × 0.01

C.I = 0.10 + 0.0196

C.I = (0.0804, 0.1196)