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    Diaper Hypothesis - Statistics

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    14.1 A diaper company is considering 3 different filler materials for their disposable diapers. Eight diapers were tested with each of the 3 filler materials and 24 toddlers were randomly given a diaper to wear. As the child played, fluid was injected into the diaper every 10 minutes until the product failed (leaked). The amount of fluid (in grams) at the time of failure was recorded for each diaper. The data are shown below:
    Material 1 Material 2 Material 3
    791 809 828
    789 818 814
    796 803 855
    802 781 844
    810 813 847
    790 808 848
    800 805 836
    790 811 873
    (a) What is the response variable and what is the factor?
    (b) How many levels of the factor are being studied?

    (c) Is there any difference in the average amount of fluid the diaper can hold using the 3 different filler materials? If so, which ones are different?

    (d) What is your recommendation to the company and why?

    14.27 The manager is concerned about the accuracy of her employees after many hours of intensive computer work. At the same time as the data to complete the task was taken, a measure of accuracy (% correct) was also recorded. These data are shown below:
    After 2 h After 4 h After 6 h
    Beginning of Day Intensive Work Intensive Work Intensive Work
    Right Left Right Left Right Left Right Left
    95.88 89.70 93.60 90.78 87.96 82.54 79.88 92.38
    91.94 90.00 98.00 94.17 96.39 87.96 84.70 86.58
    92.79 85.68 90.10 94.17 98.59 95.76 93.68 94.17
    95.76 95.53 97.00 89.70 96.00 91.00 93.92 95.00
    91.46 81.97 87.96 96.84 84.35 83.24 97.76 96.84
    92.72 90.15 95.53 90.78 96.39 95.00 74.68 84.48
    93.60 92.72 96.84 92.72 94.90 94.61 86.40 89.23
    94.61 83.45 80.76 91.00 94.61 87.96 99.00 90.00
    99.00 98.00 95.53 92.19 79.38 79.60 94.61 86.11
    97.76 88.30 92.00 88.82 89.70 85.00 76.65 82.91
    100.00 93.60 96.39 97.17 93.68 96.84 98.59 94.17
    89.18 89.23 80.90 85.58 98.59 90.00 81.13 90.00
    87.47 97.00 85.68 94.61 92.93 91.75 86.96 95.88
    94.17 96.39 93.29 91.94 85.68 94.90 92.72 83.00
    91.40 91.94 96.00 84.97 93.60 87.27 91.94 94.90
    95.53 84.19 94.90 90.15 90.57 89.70 80.69 94.61
    93.68 96.39 94.34 89.00 96.39 84.70 91.75 87.79
    94.90 84.44 92.19 87.79 87.96 95.53 95.00 95.00
    97.76 94.17 95.00 91.40 94.90 93.92 85.58 82.97
    97.00 93.29 98.59 91.40 95.53 90.57 96.84 88.69

    (a) Is there any difference in the average accuracy level after differing amounts of computer intensive work?

    (b) Is there any difference in the average accuracy level to complete the task for the right hand versus the left hand?

    (c) Is there a significant interaction effect?

    (d) What is your recommendation to this manager?

    17.1 The administration of a university has been using the following distribution to classify the ages of their students:
    Estimated % of
    Age Group Student Population
    Less than 18 2.7
    18-19 29.9
    20-24 53.4
    Older than 24 14

    A recent student survey provided the following data on age of students:

    Age Group Frequency
    Less than 18 6
    18-19 118
    21-24 102
    Older than 24 26

    (a) Set up a table that compares the expected and observed frequencies for each group.

    (b) Based on the table, do you think that the data represent the estimated distribution?

    (c) Set up the hypotheses for the chi-square goodness of fit test.

    (d) Perform the goodness of fit test at the 0.05 level of significance.

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    https://brainmass.com/statistics/hypothesis-testing/diaper-hypothesis-statistics-5813

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    14.1 A diaper company is considering 3 different filler materials for their disposable diapers. Eight diapers were tested with each of the 3 filler materials and 24 toddlers were randomly given a diaper to wear. As the child played, fluid was injected into the diaper every 10 minutes until the product failed (leaked). The amount of fluid (in grams) at the time of failure was recorded for each diaper. The data are shown below:
    Material 1 Material 2 Material 3
    791 809 828
    789 818 814
    796 803 855
    802 781 844
    810 813 847
    790 808 848
    800 805 836
    790 811 873
    (a) What is the response variable and what is the factor?
    *The response variable is the amount of fluid (in grams) and the factor is Materials (1,2,3)
    (b) How many levels of the factor are being studied?
    *The number of levels of a factor is equal to the number of variations of that factor that were used in the experiment. Therefore, three levels of the factor are studied
    (c) Is there any difference in the average amount of fluid the diaper can hold using the 3 different filler materials? If so, which ones are different?
    *We can simply use three t-tests to test the difference of the sample means.
    first we can calculate the mean and variance for each group:
    the formula is mean Mi=ΣXi/N and variance Vi=Σ(Xi-Mi)2/(N-1)
    we then have the following table:
    Child Material 1 V1 Material 2 V2 Material 3 V3
    1 791 25 809 9 828 228.766
    2 789 49 818 144 814 848.266
    3 796 0 803 9 855 141.016
    4 802 36 781 625 844 0.766
    5 810 196 813 49 847 15.016
    6 790 36 808 4 848 23.766
    7 800 16 805 1 836 50.766
    8 790 36 811 25 873 892.516
    Total 6368 394 6448 866 6745 2200.875
    Mean 796.0 56.286 806.0 123.7 843.125 314.411
    Then we compare M1 and M2 first:
    H0: M1=M2 or Md=M2-M1=0 vs. H1: M1<>M2 or Md=M2-M1<>0
    We have Md= M2-M1=806-796=10
    And the estimated standard error of the difference between means
    Smd=SQRT[(V1+V2)/N]=SQRT((56.286+123.7)/8)=4.743
    Then t1=(Md-0)/Smd=10/4.743=2.108
    Usually, we use 95% level of confidence. The degree of freedom is df=8-1=7.
    Since this is a two-tailed distribution, we should find use 0.975 t value (with 0.025 on each side). From the t-table, we can find that t*(0.975,7)=2.365
    Since t1<t*, we can accept H0 that there's no difference in the mean of M1 and M2.

    In the same way, we test M1 and M3:
    H0: M1=M3 or Md=M3-M1=0 vs. H1: M1<>M3 or Md=M3-M1<>0
    We have Md= M3-M1=843.125-796=47.125
    And the estimated standard error of the difference between means
    Smd=SQRT[(V1+V3)/N]=SQRT((56.286+314.411)/8)=6.807
    Then t2=(Md-0)/Smd=47.125/6.807=6.923
    Again, t*(0.975,7)=2.365
    Since t2>t*, we can NOT accept H0, so there is difference in the mean of M1 and M3 or M3 is significantly greater than M1.

    In the same ...

    Solution Summary

    The solution sets up hypothesis tests for chi-squared goodness of a fit next and performs the goodness of fit test at the .05 level of significance.

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