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

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. 8 7 7 8 7 11 6 10 9 9 10 10 11 10 11 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 it. (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

Solution :-

Given, n = 15

sigma, standard deviation = 1.6 minutes

Data - 8,7,7,8,7,11,6,10,9,9,10,10,11,10,11

Mean = sum of data / n = 134 / 15 = 8.93 or 8.9

(a)

90% Confidence Interval: 8.9 ± 0.68
that is - 8.22 to 9.58
as,
Margin of Error is 0.68
this is calculated using formula, Xbar + Z * (s / sqrt(n))

(b) Similarily, we can find for 99%

99% Confidence Interval: 8.9 ± 1.1
that is - 7.8 to 10

as, Margin of Error is 1.1

(c) to chech which one is wider,

For 90% - 9.58 - 8.22 = 1.36

For 99% - 10 - 7.8 = 2.2

Therefore, 99% CI is wider.

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