In a survey of 28 teenagers who were asked how many hours per week they spend...

In a survey of 28 teenagers who were asked how many hours per week they spend watching T.V., the sample mean was 13 hours and the population standard deviation is 5.8 hours. Find a 99% Confidence Interval for the population mean number of hours teenagers watch T.V.

Write the interval below. Write a sentence interpreting this. (Round answer to 2 decimal places)

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Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 13

Population standard deviation = = 5.8
Sample size = n =28

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

Margin of error = E = Z/2* ( / $\sqrt$ n)

= 2.576 * (5.8 / $\sqrt$ 28)

= 2.82

At 99% confidence interval estimate of the population mean is,

- E < < + E

13-2.82 < <13+2.82

10.18< < 15.82

sample mean lies between thease two intervals 10.18 to 15.82

orchestra answered 1 year ago

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