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.)

Answer 1

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.