# A U.S. travel center survey reported that Americans stayed an average of 7.5 nights when they went on vacation. The sample size is 1500. Find a point estimate of the population mean. Find the 95% confidence interval of the true mean. Assume the population standard deviation was 0.8.

A U.S. travel center survey report states that Americans stayed an average of 7.5 nights when they went on vacation. The sample size is 1500.

The point estimate of the population mean is the average length of the holiday. It is equal to 7.5.

The standard deviation for the number...

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A U.S. travel center survey report states that Americans stayed an average of 7.5 nights when they went on vacation. The sample size is 1500.

The point estimate of the population mean is the average length of the holiday. It is equal to 7.5.

The standard deviation for the number of nights that the population was on the vacation is 0.8.

The 95% confidence interval is `mu+-1.96*sigma` , where `mu` is the average and `sigma` is the standard deviation.

Here, `mu` = 7.5 and `sigma` = 0.8, `mu+-1.96*sigma = 7.5+-1.568`

The required confidence interval is [5.932, 9.068]

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