Suppose that a committee is studying whether there is waste of time in our judicial system. It is interested in the mean amount of time individuals waste at the courthouse waiting to be called for jury duty. The committee randomly surveyed 81 people who recently served as jurors. The sample mean wait time was eight hours with a sample standard deviation of four hours. Construct a 95% confidence interval for the population mean time wasted. Which distribution should you use for this problem

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Answer:

The t-distribution is used, as we have the standard deviation of the sample.

The 95% confidence interval for the population mean time wasted is between 7.12 hours and 8.88 hours.

Step-by-step explanation:

We have the standard deviation for the sample, which meas that the t-distribution should be used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 81 - 1 = 80

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 80 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975[/tex]. So we have T = 1.99

The margin of error is:

[tex]M = T\frac{s}{\sqrt{n}} = 1.99\frac{4}{\sqrt{81}} = 0.88[/tex]

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 8 - 0.88 = 7.12 hours.

The upper end of the interval is the sample mean added to M. So it is 8 + 0.88 = 8.88 hours.

The 95% confidence interval for the population mean time wasted is between 7.12 hours and 8.88 hours.