# Sample Mean, Standard Deviation, Alpha

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Subject: sample mean/standard deviation/alpha

Details: An insurance company asserts that the mean amount paid for personal injury claims resulting from automobile accidents is $18,500. An actuary wants to check the accuracy of this assertion and is allowed to randomly sample 36 cases involving personal injury. The sample mean is $19,415, and past years' data indicates a standard deviation of $2,600. Test the company's claim with alpha=0.05.

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Subject: Formulate a null and an alternative hypothesis, for these situations.

Details: 1)The population mean IQ is 100. A psychologist wants to test the hypothesis that the mean IQ for alcoholics is different than 100.

2)An industrial engineer wants to test the hypothesis that the mean length of the steel beams being manufactured is 3.2 meters.

3)Last year the mean sales per week in a department store were $85,492. A manager wants to test the hypothesis that the weekly sales this year will be more.

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##### Solution Summary

Subject: sample mean/standard deviation/alpha

Details: An insurance company asserts that the mean amount paid for personal injury claims resulting from automobile accidents is $18,500. An actuary wants to check the accuracy of this assertion and is allowed to randomly sample 36 cases involving personal injury. The sample mean is $19,415, and past years' data indicates a standard deviation of $2,600. Test the company's claim with alpha=0.05.

=======

Subject: Formulate a null and an alternative hypothesis, for these situations.

Details: 1)The population mean IQ is 100. A psychologist wants to test the hypothesis that the mean IQ for alcoholics is different than 100.

2)An industrial engineer wants to test the hypothesis that the mean length of the steel beams being manufactured is 3.2 meters.

3)Last year the mean sales per week in a department store were $85,492. A manager wants to test the hypothesis that the weekly sales this year will be more.

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1

<br>Ho: M=18500

<br>Ha: M≠18500

<br>Calculate SE = SD/SQRT(N) = 2600/ SQRT(36) = 433.33

<br>Calculate t=(X-M)//SE = (19,415-18,500)/433.33= 2.11

<br>From a t-table, the two tailed 0.05 critical t value with df=35 is t*=2.02

<br>Since the calculated t > t*, we ...

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