Question

Twelve different video games showing substance use were observed and the duration of 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 sample data to construct a 95​% confidence interval estimate of sigma​, the standard deviation of the duration times of game play. Assume that this sample was obtained from a population with a normal distribution.

Answer 1

The 95​% confidence interval for the standard deviation of the duration times of game play is (279.76, 670.50).

Explanation:

The data for the twelve different video games showing substance use were observed and the duration of times of game play​ (in seconds) are listed below:

S = {4049, 3884, 3859, 4027, 4318, 4813, 4657, 4033, 5004, 4823, 4334, 4317}

It is assumed that the sample was obtained from a population with a normal distribution.

The confidence interval for population standard deviation is given by,

`\sqrt{((n-1)s^(2))/(\chi^(2)_(\alpha/2))}<\sigma<\sqrt{((n-1)s^(2))/(\chi^(2)_(1-\alpha/2))}`

Compute the sample variance s² as follows:

`\text{s}^(2) = (1)/(n - 1)\sum\limits_(i=1)^(n)(x_i - \overline{x})^(2)\\`

`=(1)/(12-1)* 1715527.67\\\\`

`=155957.061\\`

Compute the critical values of Ci-square for 95% confidence level and (n - 1) degrees of freedom as follows:

`\chi^(2)_(\alpha/2, (n-1))=\chi^(2)_(0.025, (12-1))=\chi^(2)_(0.025, 11)=21.92\\\chi^(2)_(1-\alpha/2, (n-1))=\chi^(2)_(1-0.025, (12-1))=\chi^(2)_(0.975, 11)=3.816`

*Use a Chi-square table.

Compute the 95​% confidence interval for the standard deviation of the duration times of game play as follows:

`\sqrt{((n-1)s^(2))/(\chi^(2)_(\alpha/2))}<\sigma<\sqrt{((n-1)s^(2))/(\chi^(2)_(1-\alpha/2))}`

`=\sqrt{((12-1)155957.061)/(21.92)}<\sigma<\sqrt{((12-1)155957.061)/(3.816)}`

`=279.7555<\sigma<670.4937`

`\approx279.76<\sigma<670.50`

Thus, the 95​% confidence interval for the standard deviation of the duration times of game play is (279.76, 670.50).