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,
[tex]\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2}}}<\sigma<\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2}}}[/tex]
Compute the sample variance s² as follows:
[tex]\text{s}^{2} = \dfrac{1}{n - 1}\sum\limits_{i=1}^{n}(x_i - \overline{x})^{2}\\[/tex]
[tex]=\frac{1}{12-1}\times 1715527.67\\\\[/tex]
[tex]=155957.061\\[/tex]
Compute the critical values of Ci-square for 95% confidence level and (n - 1) degrees of freedom as follows:
[tex]\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[/tex]
*Use a Chi-square table.
Compute the 95% confidence interval for the standard deviation of the duration times of game play as follows:
[tex]\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2}}}<\sigma<\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2}}}[/tex]
[tex]=\sqrt{\frac{(12-1)155957.061}{21.92}}<\sigma<\sqrt{\frac{(12-1)155957.061}{3.816}}[/tex]
[tex]=279.7555<\sigma<670.4937[/tex]
[tex]\approx279.76<\sigma<670.50[/tex]
Thus, the 95% confidence interval for the standard deviation of the duration times of game play is (279.76, 670.50).
