Question

In a study of red/green color blindness. 1000 men and 2800 women are randomly selected and tested Among the men. 85 have red/green color blindness. Among the women, 6 have red/green color blindness.1. Test the claim that men have a higher rate of red/green color blindness- The test statistic is _____- The P-Value is _____- Is there sufficient evidence to support the claim that men have a higher rate of red/green color blindness than women using the 0.05% significance level?A. Yes.B. No.2. Construct the 95% confidence interval for the difference between the color blindness rates of men and women_____ <(P1âP2)<_____

Answer 1

Answer:

1. Yes, there is sufficient evidence to support the claim that men have a higher rate of red/green color blindness than women.

2. The 95% confidence interval for the difference between the color blindness rates of men and women is [0.0656, 0.1004].

Step-by-step explanation:

We are given that in a study of red/green color blindness. 1000 men and 2800 women are randomly selected and tested

Among the men, 85 have red/green color blindness. Among the women, 6 have red/green color blindness.

Let $p_1$ = population proportion of men having red/green color blindness.

$p_2$   = population proportion of women having red/green color blindness.

So, Null Hypothesis, :      {means that men have a lesser or equal rate of red/green color blindness than women}

Alternate Hypothesis, :      {means that men have a higher rate of red/green color blindness than women}

(1) The test statistics that will be used here is Two-sample z-test statistics for proportions;

                             T.S.  =     ~  

N(0,1)

where, = sample proportion of men having red/green color blindness = = 0.085

$\hat p_2$ = sample proportion of women having red/green color blindness = $\frac{6}{2800}$ = 0.002

$n_1$ = sample of men = 1000

$n_2$ = sample of women = 2800

So, the test statistics =  $\frac{(0.085-0.002)-(0)}{\sqrt{\frac{0.085(1-0.085)}{1000}+\frac{0.002(1-0.002)}{2800} } }$  

                                     =  9.37    

The value of z-test statistics is 9.37.

Also, the P-value of the test statistics is given by;

               P-value = P(Z > 9.37) = Less than 0.0001

Since the P-value of our test statistics is less than the level of significance as 0.0001 < 0.05%, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we support the claim that men have a higher rate of red/green color blindness than women.

(2) The 95% confidence interval for the difference between the color blindness rates of men and women ($p_1-p_2$) is given by;

95% C.I. for ($p_1-p_2$) = $(\hat p_1-\hat p_2) \pm Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} }$

                                 = $(0.085-0.002) \pm (1.96 \times \sqrt{\frac{0.085(1-0.085)}{1000}+\frac{0.002(1- 0.002)}{2800} } )$

                                 = $0.083 \pm0.0174$

                                 = [0.0656, 0.1004]

Here, the crtical value of z at 2.5% level of significance is 1.96.

Hence, the 95% confidence interval for the difference between the color blindness rates of men and women is [0.0656, 0.1004].