Given Pbar_1 = 0.85, n_1 = 400, Pbar2 = 0.90, n_2 = 350. Use Table 1. Construct

Question

Given Pbar_1 = 0.85, n_1 = 400, Pbar2 = 0.90, n_2 = 350. Use Table 1. Construct a 90% confidence interval for the difference between the population proportions. (Negative values should be indicated by a minus sign. Round intermediate calculations and final answer to 4 decimal places.) Is there a statistical difference between the population proportions at the 10% significance level? Yes, since the confidence interval includes the value 0. No, since the confidence interval includes the value 0. Yes, since the confidence interval does not include the value 0. No, since the confidence interval does not include the value 0.

Explanation / Answer

Solution;

90% Ci for difference in population proportion is:

z alpha/2 for 90% is 1.645 from the table given below:

90% CI for diff in population proportion is:

(0.85-0.90)±1.645 sqrt{([0.85)(1-0.85)/400] +[0.90(1-0.90)/350]}

(0.85-0.90)±1.645   (0.02399)

-0.05±0.0395

-0.05-0.03948<p1 -p2<-0.05+0.03948

-0.08948<p1 -p2< -0.01052

For (B )

OptionD since it does not contain zero

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