Question

in a random sample of 8 Laptops, the mean repair cost per laptop was $90.46 and the standard deviation was $19.30. Assume the variable is normally distributed and construct a 90% confidence interval forμ. Round your answer two decimal place

Answer 1

Given that,

= $90.46

s =$19.30

n = 8

Degrees of freedom = df = n - 1 =8 - 1 = 7

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t0.05,7 = 1.895    ( using student t table)

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

= 1.895* (19.30 / 8) = 12.93

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

- E < < + E

90.46 -12.93 < < 90.46+ 12.93

77.53 < <103.39

( 77.53 ,103.39 )