"Suppose two independent samples have been collected.Based on a sample of 26 from the first population, you observe 14 successes. The second sample of size 26 had 17 successes. What is the critical value for a 85% confidence interval that compares these two population proportions?"
I got -1.43953, is this correct?
To find the critical value for a confidence interval that compares two population proportions, we need to perform a hypothesis test.
Step 1: State the hypotheses.
Let p1 be the proportion of successes in the first population, and p2 be the proportion of successes in the second population.
The null hypothesis (H0) states that there is no difference between the proportions: p1 = p2.
The alternative hypothesis (Ha) states that there is a difference between the proportions: p1 ≠ p2.
Step 2: Determine the significance level (α).
In this case, the significance level is 1 - confidence level. Since we want a confidence interval at the 85% confidence level, the significance level is 1 - 0.85 = 0.15.
Step 3: Determine the test statistic.
For testing the difference between two population proportions, we use the Z-test statistic:
Z = [(p1 - p2) - 0] / sqrt[(p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2)]
Where p1 and p2 are the sample proportions, and n1 and n2 are the sample sizes.
In this case, p1 = 14/26 ≈ 0.538 and p2 = 17/26 ≈ 0.654.
Substituting these values into the formula:
Z = [(0.538 - 0.654) - 0] / sqrt[(0.538 * (1 - 0.538) / 26) + (0.654 * (1 - 0.654) / 26)]
Calculating the value of Z will give you the test statistic.
Step 4: Find the critical value.
To find the critical value for a two-tailed test, we have to divide the significance level by 2 and find the corresponding z-score from the standard normal distribution table.
For a significance level of 0.15/2 = 0.075, the critical value is approximately 1.43953.
Therefore, the critical value for a 85% confidence interval that compares these two population proportions is 1.43953.
Your answer is correct.