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Math Help - Help with Confidence Intervals

  1. #1
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    Help with Confidence Intervals

    Can someone help me work out the answer to this word problems?

    1 - Based on information obtained from a sample of 40, a 95% confidence interval for the average profit level of regional banks is given by $65.83 million to $83.09 million. Determine the sample standard deviation of profit (in $mill correct to 2 decimal places).

    2 - The manufacturers of Good-O use two different types of machines to fill their 25 kg packs of dried dog food. On the basis of random samples of size 12 and 14 from output from machines 1 and 2 respectively, the mean and standard deviation of the weight of the packs of dog food produced were found to be 23.623 kg and 0.252 kg for machine 1 and 24.711 kg and 0.382 kg for machine 2. Hence, under the usual assumptions, determine a 98% confidence interval for the difference between the average weight of the output produced by machine 1 and by machine 2. Construct the interval as the average weight of machine 1 less that of machine 2, stating only the lower limit of the interval correct to 3 decimal places.
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  2. #2
    MHF Contributor matheagle's Avatar
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    I would use t with 39 degrees of freedom which is close to the z-score of 1.96, but not the same.
    Playing with Free p-Value Calculator for the Student t-Test
    I get 2.0227 as t_{39,.025}
    AND you should use t since this problem assumes we do not know the population variance.
    BUT I bet the problem wants 1.96 instead.

    anyhow, to do your problem, you shoudl take the difference of each interval.

    83.09 - 65.83 = 2ts/\sqrt{40}

    I would go with 83.09 - 65.83 = 2(2.0227)s/\sqrt{40}

    but most likely you're being asked to obtain s via 83.09 - 65.83 = 2(1.96)s/\sqrt{40}
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