Question

In: Statistics and Probability

In a random sample of ten cars, the mean minor repair cost was $150.00 and the...

In a random sample of ten cars, the mean minor repair cost was $150.00 and the standard deviation was $25.75. If the repair cost is assumed to be normally distributed, construct a 99% confidence interval for the population mean. Give your answer using two decimal digits:

1-the left end of the interval, LCL, is? ...

2- the right end of the interval, UCL, is?...

Expert Solution

solution

Given that,

= $150.00

s =$25.75

n = 10

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

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, 9= 3.250 ( using student t table)

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

= 3.250* (25.75 / $\sqrt$ 10) = 26.21

The 99% confidence interval estimate of the population mean is,

- E < < + E

150.00 - 26.21< <150.00 + 26.21

123.79 < <176.21

(LCL=123.79 ,UCL=176.21 )

orchestra answered 1 year ago

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