Question

. In order to determine how many hours per week freshmen college students watch television, a random sample of 25 students was selected. It was determined that the students in the sample spent an average of 19.5 hours with a sample standard deviation of 4.2 hours watching TV per week. Please answer the following questions:

(a) Provide a 95% confidence interval estimate for the average number of hours that all college freshmen spend watching TV per week.

(b) Assume that a sample of 36 students was selected (with the same mean and the sample standard deviation). Provide a 95% confidence interval estimate for the average number of hours that all college freshmen spend watching TV per week.

Answer 1

Solution :

a ) Given that,

= 19.5

s = 4.2

n = 25

Degrees of freedom = df = n - 1 = 25 - 1 = 24

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,24 =2.064

Margin of error = E = t/2,df * (s /n)

= 2.064 * (4.2/ 25)

= 1.73

Margin of error = 1.73

The 95% confidence interval estimate of the population mean is,

- E < < + E

19.5 - 1.73 < < 19.5 + 1.73

7.77 < < 11.23

(7.77, 11.23 )

b ) Given that,

= 19.5

s = 4.2

n = 36

Degrees of freedom = df = n - 1 = 36 - 1 = 35

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,35 =2.030

Margin of error = E = t/2,df * (s /n)

= 2.030 * (4.2/ 36)

= 1.42

Margin of error =1.42

The 95% confidence interval estimate of the population mean is,

- E < < + E

19.5 - 1.42 < < 19.5 + 1.42

18.08 < < 20.92

(18.08 , 20.92 )