A recent survey of 31 students reported that the average amount of time they spent listening...

A recent survey of 31 students reported that the average amount of time they spent listening to music was 11.5 hours per week, with a sample standard deviation of 9.2 hours. Which of the following is a 90% confidence interval for the mean time per week spent listening to the radio?

Is this confidence interval for a population mean or population proportion? When do you use a z-test and when do you use a t-test?

Solutions

Expert Solution

Given that,

= 11.5

s =9.2

n = 31

Degrees of freedom = df = n - 1 =31 - 1 = 30

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t_/2,df = t_0.05,30 = 1.697 ( using student t table)

Margin of error = E = t_/2,df * (s / $\sqrt$ n)

= 1.697 * ( 9.2 / $\sqrt$ 31) = 2.8041

The 90% confidence interval estimate of the population mean is,

- E < < + E

11.5 - 2.8041 < < 11.5+ 2.8041

8.6959 < < 14.3041

(8.6959 , 14.3041 )

orchestra answered 1 year ago

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