In: Statistics and Probability
A sample of 13 repair costs for smart phones have an average of $85.56 and a standard deviation of $10. Build the 99% confidence interval for the true average cost to repair a smart phone.
c )solution
Given that,
= 85.56
s =10
n = 13
Degrees of freedom = df = n - 1 = 13- 1 = 12
At 99% confidence level the t is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
t /2 df = t0.005,12 = 3.054 ( using student t table)
Margin of error = E = t/2,df * (s /n)
= 3.054* ( 10/ 13) = 8.47
The 99% confidence interval is,
- E < < + E
85.56 - 8.47 < < 85.56+ 8.47
(77.09 , 94.03)