Question

A recent drug survey showed an increase in the use of drugs and alcohol among local high school seniors as compared to the national percent. Suppose that a survey of 100 local seniors and 100 national seniors is conducted to see if the proportion of drug and alcohol use is higher locally than nationally. Locally, 67 seniors reported using drugs or alcohol within the past month, while 60 national seniors reported using them. Conduct a hypothesis test at the 5% level.

NOTE: If you are using a Student's t-distribution for the problem, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

a.) What is the test statistic? (If using the z distribution round your answer to two decimal places, and if using the t distribution round your answer to three decimal places.)

z = _____

b.) What is the p-value? (Round your answer to four decimal places.)

Answer 1

Solution:-

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

Null hypothesis: P₁< P₂
Alternative hypothesis: P₁ > P₂

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).

p = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂)

p = 0.635

SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.06805
z = (p₁ - p₂) / SE

z = 1.028

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 greater than 1.03.

Thus, the P-value = 0.1515.

Interpret results. Since the P-value (0.1515) is greater than the significance level (0.01), we failed to reject the null hypothesis.