Arandom sample of 1500 home owners in a particular city found 465 home owners who had a swimming pool in their backyard. find a 95% confidence interval for the true percent of home owners in this city who have a swimming pool in their backyard. round your final answer to three decimal places.

The 95% confidence interval for the true percent of home owners in this city who have a swimming pool in their backyard is (0.287, 0.333).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence interval , we have the following confidence interval of proportions.

In which

Z is the zscore that has a pvalue of .

For this problem, we have that:

A random sample of 1500 home owners in a particular city found 465 home owners who had a swimming pool in their backyard. This means that and .

Find a 95% confidence interval for the true percent of home owners in this city who have a swimming pool in their backyard.

So , z is the value of Z that has a pvalue of = 0.975, so .

The lower limit of this interval is:

The upper limit of this interval is:

The 95% confidence interval for the true percent of home owners in this city who have a swimming pool in their backyard is (0.287, 0.333).

for b its (t) and c its (d)

step-by-step explanation:

any options?

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