Question

A survey of 500 shoppers was taken in a large metropolitan area to determine various information about consumer behavior. Among the questions was, "Do you enjoy shopping for clothing?" Of 240 males, 136 answered yes. Of 260 females, 224 answered yes.
a) Is there evidence of a significant difference between males and females in the proportion who enjoy shopping for clothing at the 0.01 level of significance?b) Find the p-value in (a) and interpret its meaningc) Construct and interpret a 95% confidence interval estimate of the difference between the proportion of males and females who enjoy shopping for clothing.d) What are your answers to (a) through (c) if 206 males enjoyed shopping for clothing?

Answer 1

Answer:

a) Yes (See below)

b) $p_v =2*P(Z<-7.339)\approx 2.15x10^{-13}$  

c) We are confident at 95% that the difference between the two proportions is between $-0.3704 \leq p_M -p_F \leq -0.2196$

d) $p_v =2*P(Z<-0.1287)= 0.898$  

We are confident at 95% that the difference between the two proportions is between $-0.065 \leq p_M -p_F \leq 0.0569$

Step-by-step explanation:

Data given and notation  

$X_{M}=136$ represent the number of males that answer yes

$X_{F}=224$ represent the number of female that answer yes

$n_{M}=240$ represent the number of males selected

$n_{F}=260$ represent the number of females selected  

$p_{M}=\frac{136}{240}=0.567$ represent the proportion of males that answer yes

$p_{F}=\frac{224}{260}=0.862$ represent the proportion of female that answer yes

z would represent the statistic (variable of interest)  

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

$\alpha=0.01$ significance level given

Part a

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:  

Null hypothesis:$p_{M} - p_{F}=0$  

Alternative hypothesis:$p_{M} - \mu_{F} \neq 0$  

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

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

Where $\hat p=\frac{X_{M}+X_{F}}{n_{M}+n_{F}}=\frac{136+224}{240+260}=0.72$  

Calculate the statistic  

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

$z=\frac{0.567-0.862}{\sqrt{0.72(1-0.72)(\frac{1}{240}+\frac{1}{260})}}=-7.339$    

Part b

Statistical decision

Since is a two sided test the p value would be:  

$p_v =2*P(Z<-7.339)\approx 2.15x10^{-13}$  

Comparing the p value with the significance level given $\alpha=0.01$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that we have enough evidence to conclude that we have a significant difference in the two proportions.  

Part c

The confidence interval for the difference of two proportions would be given by this formula  

$(\hat p_M -\hat p_F) \pm z_{\alpha/2} \sqrt{\frac{\hat p_M(1-\hat p_M)}{n_M} +\frac{\hat p_F (1-\hat p_F)}{n_F}}$  

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.96$  

And replacing into the confidence interval formula we got:  

$(0.567-0.862) - 1.96 \sqrt{\frac{0.567(1-0.567)}{240} +\frac{0.862(1-0.862)}{260}}=-0.3704$  

$(0.567-0.862) + 1.96 \sqrt{\frac{0.567(1-0.567)}{240} +\frac{0.862(1-0.862)}{260}}=-0.2196$  

And the 95% confidence interval would be given (-0.3704;-0.2196).  

We are confident at 95% that the difference between the two proportions is between $-0.3704 \leq p_M -p_F \leq -0.2196$

Part d

Everything change. The new proportion for males would be:

$p_{M}=\frac{206}{240}=0.858$ represent the proportion of males that answer yes

$z=\frac{0.858-0.862}{\sqrt{0.86(1-0.86)(\frac{1}{240}+\frac{1}{260})}}=-0.1287$    

$p_v =2*P(Z<-0.1287)= 0.898$  

Comparing the p value with the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we don't have significant difference between the two propotions.

$(0.858-0.862) - 1.96 \sqrt{\frac{0.858(1-0.858)}{240} +\frac{0.862(1-0.862)}{260}}=-0.065$  

$(0.858-0.862) + 1.96 \sqrt{\frac{0.858(1-0.858)}{240} +\frac{0.862(1-0.862)}{260}}=0.0569$  

And the 95% confidence interval would be given (-0.065;0.0569).  

We are confident at 95% that the difference between the two proportions is between $-0.065 \leq p_M -p_F \leq 0.0569$