Question 5: Which statement is true: (a) A 90% confidence interval for μ is wider than...

Solution:-

7) (c)

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 0.05. 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 = 0.48
SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.09992
z = (p₁ - p₂) / SE

z = - 2.40

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 less than -2.40.

Thus, the P-value = 0.0082

Interpret results.

Since the P-value (0.0082) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that new drug is effective.

8) (b) 95% confidence interval for the difference between the two proportions is C.I = ( - 0.43, - 0.05).

C.I = (0.36 - 0.60) + 1.96 × 0.0969

C.I = - 0.24 + 0.190

C.I = ( - 0.43, - 0.05)