Question

# A publisher wants to estimate the mean length of time? (in minutes) all adults spend reading...

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 standard deviation is 1.7 minutes and that the population of times is normally distributed.

9

10

6

8

11

7

10

8

7

12

12

11

9

9

9

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

From the given data, the following statistics are calculated:

n = sample size = 15 = sample mean = 138/15 = 9.2

Given: = Population SD = 1.7

SE = / = 1.7/ = 0.4389

(a) = 0.10

From Table, critical values of Z = 1.645

90 % Confidence Interval:  Z SE

= 9.2 (1.645 X 0.4389)

= 9.2 0.7221

= (8.4779 , 9.9221)

The 90 % confidence interval is:

8.4779 < < 9.9221

(b) = 0.01

From Table, critical values of Z = 2.576

99 % Confidence Interval is:

9.2 (2.576 X 0.4389)

= 9.2 1.1306

= (8.0694 , 10.3306)

The 99 % Confidence Interval is:

8.0694 < < 10.3306

(c)

99 % Confidence Interval is wider.

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