The only issue I have is finding the sample standard deviation.... I've got everything else figured out. Please show the steps to getting the answer to s.

Question

Twelve different video games showing substance use were observed and the duration times of game play (in seconds) are listed below. The design of the study justifies the assumption that the sample can be treated as a simple random sample. Use the data to construct a 99% confidence interval estimate of μ, the mean duration of game play. 4044 3891 3558 4030 4328 4803 4653 4036 4998 4832 4341 4317 What is the confidence interval estimate of the population mean μ? ? < μ < ?

 

Answer 1

This one is tricky!

1.) Find the mean by adding all the data points and divide by N (12) to get 4,319.25

2.) Subtract the mean from all data points and square them - It is important to note that even if you get a negative after subtracting and squaring, to ignore it and simply make it positive.

4,044-4,319.25^2 = 75,762.56

3,981-4,319.25^2 = 183,398.06

3,558-4,319.25^2 = 574,501.56

4,030-4,319.25^2 = 83,665.56

4,328-4,319.25^2 = 76.56

4,803-4,319.25^2 = 234,014.06

4,653-4,319.25^2 = 111,389.06

4,036-4,319.25^2 = 80,230.56

4,998-4,319.25^2 = 460,701.56

4,832-4,319.25^2 = 262,912.56

4,341-4,319.25^2 = 12,488.06

4,317-4,319.25^2 = 5.06

For a total of 2,079,145.22

3.) Now we need to find the estimated SD

The formula for that is sqrt of (sum of squares/n-1)

sqrt of (2,079,145.22/11) = 434.756485654

Estimated SD = 434.756485654

4.) Now to find the confidence interval. Because the SD is unknown, and we have to use the estimated SD, we need to use a t-statistic rather than a z-statistic.

A 99% confidence interval tcrit is 3.106.

The formula is xbar+-t*estimatedSD/Sqrt of N

XBar is the mean: 4,319.25

TCrit:3.106

Estimated SD: 434.756485654

N = 12

1. The square root of 12 is 3.46410161514
2. 434.756485654/3.46410161514 = 125.503387012

3.106 * 125.503387012 = 389.813520059

4,319.25 - 389.813520059 = 3929.43647994

4,319.25 + 389.813520059 = 4709.06352006

So approximately give or take a few decimals and rounding, the confidence interval is

3,929.43 < u < 4,709.06