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# Statistics question; part of the chapter dont understand; Thank you

Twelve different video games showing substance use were observed and the duration times of game play (in seconds) are listed below (based on data from “Content and Ratings of Teen-Rated Video Games,” by Haninger and Thompson, Journal of the American Medical Association, Vol. 291, No. 7). The design of the study justifies the assumption that the sample can be treated as a simple random sample.  4049 3884 3859 4027 4318 4813 4657 4033 5004 4823 4334 4317

1. Use the sample data to construct a 95% confidence interval estimate of μ, the mean duration of game play.

2. Use the sample data to construct a 99% confidence interval estimate of σ, the standard deviation of the duration times of game play.

### 1 Answer by Expert Tutors

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1) CI for population mean mu is (sample mean) +/- (t)*s/sqrt(n)
sample mean & sd are 4,343.17 and 394.91
df = 11 and alpha = .05, so t=2.2010
margin of error (+/-) is 250.916
so the CI is (4,092.3, 4,594.1)

2) (n-1)s^2/(sigma^2) has a chi-squared distribution with df = n-1
One CI is based on percentiles of the chi-squared distribution (which are not equally distant from the mean).
Let Chi1 = (alpha/2 percentile) (smaller number)
Let Chi2 = (1-alpha/2 percentile) (larger number)

Thus there is 1-alpha probability between chi1 and chi2.

The lower CI limit (for population VARIANCE) = (n-1)*(sample variance)/chi2
Upper limit is (n-1)*(sample variance)/chi1

Sample variance = s^2 = 155,957
chi1 = 2.603
chi2 = 26.757

So, the CI for pop variance is (64,115, 659,002)
Taking sqrts gives CI for pop std dev = (253.2, 811.8)