A consumer affairs investigator records the repair cost for 4 randomly selected washers. A sample mean of $52.63 and standard deviation of $22.01 are subsequently computed. Determine the 80% confidence interval for the mean repair cost for the washers. Assume the population is approximately normal. Find the critical value that should be used in constructing the confidence interval.

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The critical z value which should be used in determining the confidence interval is z=1.29 and the confidence interval is (38.43355,66.82645).

Given sample mean of $52.63 and standard deviation of $22.01.

We have to construct  the confidence interval of 80% for the mean repair cost for the washers.

μ=52.63

σ=22.01

n=4

First we have to find the value of z for the confidence level of 80% from z table and which is 1.29.

Margin of error is the difference between the calculated values and real values.

Margin of error=z*σ/[tex]\sqrt{n}[/tex]

where μ is mean

σ is standard deviation

n is sample size

z is z value for the confidence level

Margin of error=1.29*22.01/[tex]\sqrt{4}[/tex]

=14.19645

Confidence interval =mean ±margin of error

Upper level=mean +margin of error

=52.33+14.19645

=66.82645

Lower level=mean-margin of error

=52.33-14.19645

=38.43355

Hence the confidence interval is (38.43355,66.82645) and the critical value used is z=1.29.

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