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Chapter 6: Problem 115

In the dataset StudentSurvey, 361 students recorded the number of hours of television they watched per week. The average is \(\bar{x}=6.504\) hours with a standard deviation of \(5.584 .\) Find a \(99 \%\) confidence interval for \(\mu\) and interpret the interval in context. In particular, be sure to indicate the population involved.

Short Answer

Expert verified
The 99% confidence interval for the population mean is (5.748, 7.26). This means we are 99% confident that the true average number of hours all students in the survey watch TV per week is between 5.748 hours and 7.26 hours.

Step by step solution

01

Find the Z-score

We need to determine the z-score associated with a 99% confidence interval. From standard z-tables, we find that the z-score for 99% confidence interval is 2.576.
02

Calculate the standard error

The standard error is calculated by dividing the standard deviation by the square root of sample size. Using the given standard deviation (蟽 = 5.584) and sample size (n = 361), the standard error is calculated as (5.584/鈭�361) = 0.294.
03

Calculate the margin of error

Margin of Error is calculated as the product of z-score and standard error. For this problem, it can be calculated as (2.576 * 0.294) = 0.756.
04

Calculate the confidence interval

The 99% confidence interval is calculated as: x虅 卤 margin of error. Using our computed value for x虅 (6.504) and margin of error (0.756), the confidence interval is (6.504 - 0.756, 6.504 + 0.756), which simplifies to (5.748, 7.26).

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Most popular questions from this chapter

A sample with \(n=12, \bar{x}=7.6,\) and \(s=1.6\)

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