Question

It was determined that the sample mean weight of 10 randomly selected house cats was 8.4 pounds. We seek to create a 90% confidence interval for the population mean weight of house cats in the United States. Given t.05 found in the question above, and with the knowledge that s = .897 pounds, determine the 90% confidence interval for the population mean. Choose the best answer. Group of answer choices ( 7.760 , 9.040 ) ( 7.887 , 8.913 ) ( 7.880 , 8.920 ) ( 7.769 , 9.031 )

Answer 1

Given that,

= 8.4

s =0.897

n = 10

Degrees of freedom = df = n - 1 = 10- 1 = 9

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t0.05,9 = 1.833    ( using student t table)

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

= 1.833* (0.897 / 10) = 0.520

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

- E < < + E

8.4 - 0.520< < 8.4+ 0.520

7.880 < < 8.920

(7.880 , 8.920 )