A) A university financial aid office polled a random sample of 829 male undergra

Question

A) A university financial aid office polled a random sample of 829 male undergraduate students and 759 female undergraduate students. Each of the students was asked whether or not they were employed during the previous summer. 622 of the male students and 527 of the female students said that they had worked during the previous summer. Give a 80% confidence interval for the difference between the proportions of male and female students who were employed during the summer.

Step 1 of 3: Find the point estimate that should be used in constructing the confidence interval. Round your answer to three decimal places.

Step 2 of 3: Find the margin of error. Round your answer to six decimal places.

Step 3 of 3: Construct the 80% confidence interval. Round your answers to three decimal places.

B) A student researcher compares the ages of cars owned by students and cars owned by faculty at a local state college. A sample of 126 cars owned by students had an average age of 7.44 years. A sample of 113 cars owned by faculty had an average age of 8.81 years. Assume that the population standard deviation for cars owned by students is 3.86 years, while the population standard deviation for cars owned by faculty is 2.86years. Determine the 90% confidence interval for the difference between the true mean ages for cars owned by students and faculty.

Step 1 of 3: Find the point estimate for the true difference between the population means.

Step 2 of 3: Calculate the margin of error of a confidence interval for the difference between the two population means. Round your answer to six decimal places.

Step 3 of 3:

Construct the 90% confidence interval. Round your answers to two decimal places.

Explanation / Answer

A) The statistical software output for this problem is:

Two sample proportion summary confidence interval:
p1 : proportion of successes for population 1
p2 : proportion of successes for population 2
p1 - p2 : Difference in proportions

80% confidence interval results:

Hence,

1. Point estimate = 0.056

2. Margin of error = 0.028817

3. 80% confidence interval: (0.027, 0.085)

B) The statistical software output for this problem is:

Two sample Z summary confidence interval:
1 : Mean of population 1 (Std. dev. = 3.86)
2 : Mean of population 2 (Std. dev. = 2.86)
1 - 2 : Difference between two means

90% confidence interval results:

Hence,

1. Point estimate = -1.37

2. Margin of error = 0.718175

3. 90% confidence interval: (-2.09, -0.65)

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