A sample of 13 repair costs for smart phones have an average of $85.56 and a...

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.

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Expert Solution

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 / $\sqrt$ n)

= 3.054* ( 10/ $\sqrt$ 13) = 8.47

The 99% confidence interval is,

- E < < + E

85.56 - 8.47 < < 85.56+ 8.47

(77.09 , 94.03)

orchestra answered 1 year ago

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