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

Answer:

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

Step-by-step 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,

$\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2}}}$

Compute the sample variance s² as follows:

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

   $=\frac{1}{12-1}\times 1715527.67$

   $=155957.061$

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:

$\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2}}}$

$=\sqrt{\frac{(12-1)155957.061}{21.92}}$

$=279.7555$

$\approx279.76$

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