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 98​% 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. 4 comma 350 4 comma 915 3 comma 836 4 comma 118 4 comma 096    4 comma 202 4 comma 116 4 comma 220 3 comma 967 4 comma 515 4 comma 050 3 comma 915 LOADING... Click the icon to view the table of​ Chi-Square critical values. The confidence interval estimate is nothing secless thansigmaless than nothing sec.

Answer 1

$\bar x$ = (4350 + 4915 + 3836 + 4118 + 4096 + 4202 + 4116 + 4220 + 3967 + 4515 + 4050 + 3915)/12 = 4191.67

s² = ((4350 - 4191.67)^2 + (4915 - 4191.67)^2 + (3836 - 4191.67)^2 + (4118 - 4191.67)^2 + (4096 - 4191.67)^2 + (4202 - 4191.67)^2 + (4116 - 4191.67)^2 + (4220 - 4191.67)^2 + (3967 - 4191.67)^2 + (4515 - 4191.67)^2 + (4050 - 4191.67)^2 + (3915 - 4191.67)^2)/11 = 86147.88

At 98% confidence level, the critical values are

x = x3 09.11 = 3.0535

Xk = 601,11 = 24.7250

The 98% confidence interval for population standard deviation is

sqrt((n - 1)s^2/$\chi^2_{R}$) < $\sigma$ < sqrt((n - 1)s^2/$\chi^2_{L}$)

= sqrt((11 * 86147.88)/24.7250) < $\sigma$ < sqrt((11 * 86147.88)/3.0535)

= 195.77 < $\sigma$ < 557.08