Question

1)A university financial aid office polled a random sample of 824 male undergraduate students and 731 female undergraduate students. Each of the students was asked whether or not they were employed during the previous summer. 568 of the male students and 391 of the female students said that they had worked during the previous summer. Give a 90% 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 90%confidence interval. Round your answers to three decimal places.

2)The water works commission needs to know the mean household usage of water by the residents of a small town in gallons per day. They would like the estimate to have a maximum error of 0.14 gallons. A previous study found that for an average family the standard deviation is 2.3 gallons and the mean is 15 gallons per day. If they are using a 99% level of confidence, how large of a sample is required to estimate the mean usage of water? Round your answer up to the next integer

3)Given two independent random samples with the following results:

n1        7          n2        11

x1bar   143      x2bar   162

s1        12        s2        33

data: n1=7 x‾1=143    s1=12   n2=11 x‾2=162 s2=33

Use this data to find the 90% confidence interval for the true difference between the population means. Assume that the population variances are equal and that the two populations are normally distributed.

Step 1 of 3:

Find the point estimate that should be used in constructing the confidence interval.

Step 2 of 3:

Find the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places.

Step 3 of 3:

Construct the 90% confidence interval. Round your answers to the nearest whole number

Answer 1

Solution:

Question 1) Given: A university financial aid office polled a random sample of 824 male undergraduate students and 731 female undergraduate students. Each of the students was asked whether or not they were employed during the previous summer. 568 of the male students and 391 of the female students said that they had worked during the previous summer.

That is:

For Male :

n1 = 824

x1 = 568

then

and

For female:

n2 = 731

x2 = 391

Then

We have to find a 90% confidence interval for the difference between the proportions of male and female students who were employed during the summer.

Step 1) Find the point estimate that should be used in constructing the confidence interval.

Point estimate of difference between the proportions of male and female students who were employed during the summer is:

Step 2) Find the margin of error.

Now find Z_c value:

Z_c is z critical value for c = 90% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.90) / 2 = 1.90 / 2 = 0.9500

Look in z table for Area = 0.9500 or its closest area and find corresponding z value.

Area 0.9500 is in between 0.9495 and 0.9505 and both the area are at same distance from 0.9500

Thus we look for both area and find both z values

Thus Area 0.9495 corresponds to 1.64 and 0.9505 corresponds to 1.65

Thus average of both z values is : ( 1.64+1.65) / 2 = 1.645

Thus Z_c = 1.645

Thus

Step 3) Construct the 90%confidence interval.

Formula: