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In a study of red/green color blindness, 550 men and 2400 women are randomly selected and tested. Among the men, 48 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 than women. Also, construct a 98% confidence interval for the difference between color blindness rates of men and women. Does there appear to be a significant difference?

Answer :

Answer: 98% confidence interval is (0.06,0.11).

Step-by-step explanation:

Since we have given that

Number of men = 550

Number of women = 2400

Number of men have color blindness = 48

Number of women have color blindness = 5

So, [tex]p_M=\dfrac{48}{550}=0.087\\\\p_F=\dfrac{5}{2400}=0.0021[/tex]

So, at 98% confidence interval, z = 2.326

so, interval would be

[tex](p_M-P_F)\pm z\sqrt{\dfrac{p_M(1-p_M)}{n_M}+\dfrac{p_F(1-p_F)}{n_F}}\\\\=(0.087-0.0021)\pm 2.326\sqrt{\dfrac{0.087\times 0.912}{550}+\dfrac{0.0021\times 0.9979}{2400}}\\\\=0.0849\pm 2.326\times 0.012\\\\=(0.0849-0.027912, 0.0849+0.027912)\\\\=(0.06,0.11)[/tex]

Hence, 98% confidence interval is (0.06,0.11).

No, there does not appears to be a significant difference.

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