Question

in a study of red/green color blindness, 600 men and 2150 women are randomly selected and tested. Among the men, 56 have red/green color blindness. Among the women, 5 have red/green color blindness. Test the claim that men have a higher rate of red/green color blindness. (Note: Type ‘‘p_m′′ for the symbol pm , for example p_mnot

Answer 1

Answer:

z=13.36 (Statistic)

$p_v =P(Z>13.36)\approx 0$

The p value is a very low value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of men with red/green color blindness is significant higher than the proportion of female with red/green color blindness .

Step-by-step explanation:

1) Data given and notation

$X_{MCB}=56$ represent the number of men with red/green color blindness

$X_{WCB}=5$ represent the number of women with red/green color blindness

$n_{MCB}=600$ sample of male selected

$n_{WCB}=600$ sample of demale selected

$p_{MCB}=\frac{56}{600}=0.093$ represent the proportion of men with red/green color blindness

$p_{WCB}=\frac{5}{2150}=0.0023$ 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 proportion for men with red/green color blindness is a higher than the rate for women  , the system of hypothesis would be:

Null hypothesis:$p_{MCB} \leq p_{WCB}$

Alternative hypothesis:$p_{MCB} > \mu_{WCB}$

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

$t=\frac{p_{MCB}-p_{WCB}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{MCB}}+\frac{1}{n_{WCB}})}}$   (1)

Where $\hat p=\frac{X_{MCB}+X_{WCB}}{n_{MCB}+n_{WCB}}=\frac{56+5}{600+2150}=0.0221$

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

3) Calculate the statistic

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

$z=\frac{0.093-0.0023}{\sqrt{0.0221(1-0.0221)(\frac{1}{600}+\frac{1}{2150})}}=13.36$  

4) Statistical decision

For this case we don't have a significance level provided $\alpha$, but we can calculate the p value for this test.  

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

$p_v =P(Z>13.36)\approx 0$

So the p value is a very low value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of men with red/green color blindness is significant higher than the proportion of female with red/green color blindness .