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25 July, 11:30

A supervisor records the repair cost for 14 randomly selected refrigerators. A sample mean of $79.20 and standard deviation of $10.41 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the refrigerators. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

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  1. 25 July, 13:23
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    ($74.623, $83.777)

    The 90% confidence interval is = ($74.623, $83.777)

    Critical value at 90% confidence = 1.645

    Step-by-step explanation:

    Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

    The confidence interval of a statistical data can be written as.

    x+/-zr/√n

    Given that;

    Mean x = $79.20

    Standard deviation r = $10.41

    Number of samples n = 14

    Confidence interval = 90%

    Using the z table;

    The critical value that should be used in constructing the confidence interval.

    z (α=0.05) = 1.645

    Critical value at 90% confidence z = 1.645

    Substituting the values we have;

    $79.20+/-1.645 ($10.42/√14)

    $79.20+/-1.645 ($2.782189528308)

    $79.20+/-$4.576701774067

    $79.20+/-$4.577

    ($74.623, $83.777)

    The 90% confidence interval is = ($74.623, $83.777)
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