Question

A sample of 500 observations taken from the first population gave x 1 = 305 . Another sample of 600 observations taken from the second population gave x 2 = 306 . Find the value of the test statistic z for p ^ 1 - p ^ 2 for H 0 : p 1 = p 2 versus H 1 : p 1 > p 2 . Will you reject the above mentioned null hypothesis at a significance level of 2.5 % (one-tailed test)? Round your answer to two 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 0.025. 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 = 0.5555
SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.030091
z = (p₁ - p₂) / SE

z = 3.32

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 3.32.

P-value = P(z > 3.323)

Use the calculator to determine the p-value.

P-value = 0.0004

P-value 0.00

Interpret results. Since the P-value (0.0004) is less than the significance level (0.025), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that P₁ > P₂.