A publisher wants to estimate the mean length of time (in minutes) all adults sp

Question

A publisher wants to estimate the mean length of time (in minutes) all adults spend reading newspapers. To determine this estimate, the publisher takes a random sample of 15 people and obtains the results below from past studies the publisher assumes sigma is 1 6 minutes and that the population of times is normally distributed. Construct the 90% and 99% confidence intervals for the population mean. Which interval is wider? If convenient, use technology to construct the confidence intervals. The 90% confidence interval is (Round to one decimal place as needed.) The 99% confidence interval is (Round to one decimal place as needed.) Which interval is wider? The 90% confidence interval The 99% confidence interval

Explanation / Answer

For the given data

Mean = 8.9333 , we calculated it by adding all numbers and divided it by 15

Varaince of population = 1.6 ( as given in the question)

so Mean of the population = mean of the sample

= 8.333 ; n = 15

so (i) 90 % confidence interval , for 90% confidence value, z must be +- 1.65

so 90% CI of mean = + - z /n = 8.333 +- 1.65 * 1.6/sqrt(15) = 8.333 + - 0.682 = ( 7.651, 9.012)

(ii) 95 % confidence interval , for 95% confidence value, z must be +- 1.96

so 90% CI of mean = + - z /n = 8.333 +- 1.96 * 1.6/sqrt(15) = 8.333 + - 0.810 = ( 7.523, 9.143)

the 95% interval is wider than 90 % interval.

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