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
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-> 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
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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.
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.
