SOLUTION: In a random selection of 64 of the 2400 intersections in a city, the mean number of car accidents per year was 3.2 and the sample standard deviation was 0.8. Obtain the 90% confide

Question 1203420: In a random selection of 64 of the 2400 intersections in a city, the mean number of car accidents per year was 3.2 and the sample standard deviation was 0.8. Obtain the 90% confidence interval for the mean number of accidents per intersection per year.

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
sample size is 64.
mean number of car accidents per year was 3.2 and sample standard deviation was .8.

t-score is indicated because the standard deviatiion is from the sample rather than from the population.

90% two tailed confidence interval has critical t-score with 63 degrees of freedom (sample size minus 1) equal to plus or minus t = 1.6694.

raw score is based on t-score formula of t = (x-m)/s.
t is the t-score.
m is the population mean.
s is the standard error.
solve for x to get the raw score.

standard error = sample standard deviation / sqrt(sample size) = .8/sqrt(64) = .8/8 = .1.

on the low side of the confidence interval, you get -1.6694 = (x-3.2)/.1.
solve for x to get x = .1 * -1.6694 + 3.2 = 3.03306.

on the high side of the confidence interval, you get 1.6694 = (x-3.2)/.1.
solve for x to get x = .1 * 1.6694 + 3.2 = 3.36694.

your 90% confidence interval is 33.03306 to 3.36694.

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