While her husband spent 2½ hours picking out new speakers, a statistician decided to determine whether the percent of men who enjoy shopping

Question

While her husband spent 2½ hours picking out new speakers, a statistician decided to determine whether the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment. The population was Saturday afternoon shoppers. Out of 66 men, 23 said they enjoyed the activity. Eight of the 23 women surveyed claimed to enjoy the activity. Interpret the results of the survey. Conduct a hypothesis test at the 5% level. Let the subscript m = men and w = women.

State the distribution to use for the test. (Round your answers to four decimal places.)

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Kinsley 3 weeks 2021-09-25T11:45:09+00:00 1 Answer 0

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    2021-09-25T11:46:20+00:00

    Answer:

    z=\frac{p_{M}-p_{W}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}   (1)

    z=\frac{0.34848-0.34782}{\sqrt{0.34831(1-0.34831)(\frac{1}{66}+\frac{1}{23})}}=0.0057  

    p_v =P(Z>0.0057)=0.4977  

    The p value is a very high value and using the significance level \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the percent of men who enjoy shopping for electronic equipment is NOT significantly higher than the percent of women who enjoy shopping for electronic equipment.  

    Step-by-step explanation:

    1) Data given and notation  

    X_{M}=23 represent the number of men that said they enjoyed the activity of Saturday afternoon shopping

    X_{W}=8 represent the number of women that said they enjoyed the activity of Saturday afternoon shopping

    n_{M}=66 sample of male selected

    n_{W}=23 sample of demale selected

    p_{M}=\frac{23}{66}=0.34848 represent the proportion of men that said they enjoyed the activity of Saturday afternoon shopping

    p_{W}=\frac{8}{23}=0.34782 represent the proportion of women with red/green color blindness  

    z would represent the statistic (variable of interest)  

    p_v represent the value for the test (variable of interest)

    2) Concepts and formulas to use  

    We need to conduct a hypothesis in order to check if the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment , the system of hypothesis would be:  

    Null hypothesis:p_{M} \leq p_{W}  

    Alternative hypothesis:p_{M} > p_{W}  

    We need to apply a z test to compare proportions, and the statistic is given by:  

    z=\frac{p_{M}-p_{W}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}   (1)

    Where \hat p=\frac{X_{M}+X_{W}}{n_{M}+n_{W}}=\frac{23+8}{66+23}=0.34831

    3) Calculate the statistic

    Replacing in formula (1) the values obtained we got this:  

    z=\frac{0.34848-0.34782}{\sqrt{0.34831(1-0.34831)(\frac{1}{66}+\frac{1}{23})}}=0.0057  

    4) Statistical decision

    Using the significance level provided \alpha=0.05, the next step would be calculate the p value for this test.  

    Since is a one side right tail test the p value would be:  

    p_v =P(Z>0.0057)=0.4977  

    So the p value is a very high value and using the significance level \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the percent of men who enjoy shopping for electronic equipment is NOT significantly higher than the percent of women who enjoy shopping for electronic equipment.  

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