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    Testing of hypothesis

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    23. Microwave Repair Costs A microwave oven repairer says that the mean repair cost for damaged microwave ovens is less than $100. You work for the repairer and want to test this claim. You find that a random sample of five microwave ovens has a mean repair cost of $75 and a standard deviation of $12.50. At a = 0.01,do you have enough evidence to support the repairer's claim? (Adapted from Consumer Reports)

    37. Gas Mileage A car company says that the mean gas mileage for its luxury sedan is at least 21 miles per gallon (mpg).You believe the claim is incorrect and find that a random sample of five cars has a mean gas mileage of 19 mpg and a standard deviation of 4 mpg. Assume the gas mileage of all of the company's luxury sedans is normally distributed. At a = 0.05,test the company's claim. (Adapted from Consumer Reports)

    10. Do You Eat Breakfast? A medical researcher estimates that no more than 55% of U.S. adults eat breakfast every day. In a random sample of 250 U.S. adults, 56.4% say they eat breakfast every day. At a = 0.01is there enough evidence to reject the researcher's claim? (Adapted from U.S. National Center for Health Statistics)

    15. Do Free Samples Work? You interview a random sample of 50 adults. The results of the survey show that 48% of the adults said they were more likely to buy a product when there are free samples. At a = 0.05,can you reject the claim that at least 52% of the adults are more likely to buy a product when there are free samples?

    17. Physical Science Assessment Tests On a physical science assessment test, the scores of a random sample of 22 eighth grade students have a standard deviation of 27.7. This result prompts a test administrator to claim that the standard deviation for eighth graders on the examination is less than 29.At a = 0.10, is there enough evidence to support the administrator's claim? (Adapted from National Center for Educational Statistics)

    24. Salaries An employment information service says that the standard deviation of the annual salaries for public relations managers is at least $14,500. The annual salaries for 18 randomly chosen public relations managers are listed. At a = 0.10,can you reject the claim? (Adapted from America's Career InfoNet)

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    1 23. Microwave Repair Costs A microwave oven repairer says that the mean repair cost for damaged microwave ovens is less than $100. You work for the repairer and want to test this claim. You find that a random sample of five microwave ovens has a mean repair cost of $75 and a standard deviation of $12.50. At a = 0.01,do you have enough evidence to support the repairer's claim? (Adapted from Consumer Reports)

    Data
    Hypothesized mean= $100
    Standard deviation= $12.50
    Sample mean= $75
    Sample size= 5
    Significance level= 0.01 0r 1%
    Population standard deviation (Known, Not Known)= Not Known

    1) Hypothesis
    Null Hypothesis: Ho: M = 100 (:Mean is 100 )
    Alternative Hypothesis: H1: M < 100 :( Mean is less than 100 )
    Significance level=alpha (a) = 0.01 or 1%
    No of tails= 1 ( Left tail )
    This is a 1 tailed ( Left tail ) test because we are testing that M < 100

    2) Decision rule
    sample size=n= 5
    Since sample size 5 is less than 30 and population standard deviation is 'Not Known' use t distribution
    Thus we use t distribution
    t at the 0.01 level of significance and 4 degrees of freedom (=n-1) and 1 tailed test= 3.7469
    t critical = -3.7469

    if sample statistic is <-3.7469 Reject Null Hypothesis, else Accept Null Hypothesis
    Alternatively,
    if p value is less than the significance level (= 0.01 ) Reject Null Hypothesis, else Accept Null Hypothesis

    3) Calculation of sample statistics

    Hypothesized Mean=M = 100 0
    Standard deviation =s= 12.5 0
    sample size=n= 5
    sx=standard error of mean=s/square root of n= 5.5902 = ( 12.5 /square root of 5)
    sample mean= 75

    t=(sample mean-M )/sx= -4.4721 =(75-100)/5.5902

    4) Compare sample statistic with critical value

    test statistic= -4.4721
    t critical = -3.7469

    therefore, sample statistic is outside acceptance region

    5) Decision

    Reject Null Hypothesis :( Mean is less than 100 )

    Alternatively, calculate p value
    Prob-value corresponding to t = -4.4721 is 0.55% which is less than the significance level of 1%
    Reject Null Hypothesis since probability value is less than the level of significance
    :( Mean is less than 100 )

    2 37. Gas Mileage A car company says that the mean gas mileage for its luxury sedan is at least 21 miles per gallon (mpg).You believe the claim is incorrect and find that a random sample of five cars has a mean gas mileage of 19 mpg and a standard deviation of 4 mpg. Assume the gas mileage of all of the company's luxury sedans is normally distributed. At a = 0.05,test the company's claim. (Adapted from Consumer Reports)

    Data
    Hypothesized mean= 21 mpg
    Standard deviation= 4 mpg
    Sample mean= 19 mpg
    Sample size= 5
    Significance level= 0.05 0r 5%
    Population standard deviation (Known, Not Known)= Not Known

    1) Hypothesis
    Null Hypothesis: Ho: M = 21 (:Mean is at least 21 mpg)
    Alternative Hypothesis: H1: M < 21 :( Mean is less than 21 mpg )
    Significance level=alpha (a) = 0.05 or 5%
    No of tails= 1 ( Left tail )
    This is a 1 tailed ( Left tail ) test because we are testing that M < 21

    2) Decision rule
    sample size=n= 5
    Since sample size 5 is less than 30 and population standard deviation is 'Not Known' use t distribution
    Thus we use t distribution
    t at the 0.05 level of significance and 4 degrees of freedom (=n-1) and 1 tailed test= 2.1318
    t critical = -2.1318

    if sample statistic is <-2.1318 Reject Null Hypothesis, else Accept Null Hypothesis
    Alternatively,
    if p ...

    Solution Summary

    Answers 6 questions on testing of hypotheses. -using t test, z test, Chi Square test,

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