Question

Question 5: Which statement is true: (a) A 90% confidence interval for μ is wider than a 95% confidence interval for μ (b) The width (Le, the length) of a confidence interval is equal to its margin of error (c) Decreasing the sample size will make the confidence wider (d) The standard deviation of the sample mean is larger than the population standard deviation o. Question 6. A test of H0 : μ-128 against Ha : μ > 12.8 has test statistic z-1956. Is this test statistically significant at the 5% level ( . 0.05) ? Is statistically significant at the 1% level. ( 0.01)? (a) It is statistically significant at 5% level only (b) It is statistically significant at 19% level only (c) It is statistically significant at both 5% and 1 % levels (d) It is not statistically significant at either 5% or 1% levels. The next two questions refer to the following setup: An experiment was conducted to test the effect of a new drug on a viral infection. The infection was induced in 100 mice, and the mice were randomly split into two groups of 50. The first group, the control group, received no treatment for the infection. The second group received the drug after a 30 days period. The proportions of survivors pi and pz in the two groups were found to be 0.36 and 0.60 respectively Question 7: What is the value of the test statistic for testing Ho:pi p ? ( Select the closest value) (a) 1.85 (b)-1.85 (c) -2.40 (d) 2.84 Question 8: what is the 95% confidence interval for the actual difference in the cure rates for the treatment versus the control group? ( () 043,-0.05) ( (a) (0.27, 0.51) ( (0.15,0.015) (c) (-0.051, 0.015)

Answer 1

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).
$C.I = \left (p_{1}-p_{2} \right )\pm z_{\alpha /2}\times \sqrt{\frac{p_{1}(1-p_{1})}{n_{1}}&plus;\frac{p_{2}(1-p_{2})}{n_{2}}}$

C.I = (0.36 - 0.60) + 1.96 × 0.0969

C.I = - 0.24 + 0.190

C.I = ( - 0.43, - 0.05)