Question

A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In a random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder. Compute a 95% confidence interval for the difference between the proportions of males and females who have the blood disorder.

Answer 1

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

P.Q. =
`\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+ (\hat p_2(1-\hat p_2))/(n_2)} }` ~ N(0,1)

where,
`\hat p_1` = sample proportion of males having blood disorder=
`(250)/(1000)` = 0.25

`\hat p_2` = sample proportion of females having blood disorder =
`(275)/(1000)` = 0.275

`n_1` = sample of males = 1000

`n_2` = sample of females = 1000

`p_1` = population proportion of males having blood disorder

`p_2` = population proportion of females having blood disorder

Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.

So, 95% confidence interval for the difference between the population proportions, (
`p_1-p_2`) is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 <
`\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+ (\hat p_2(1-\hat p_2))/(n_2)} }` < 1.96) = 0.95

P(
`-1.96 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+ (\hat p_2(1-\hat p_2))/(n_2)} }` <
`{(\hat p_1-\hat p_2)-(p_1-p_2)}` <
`1.96 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+ (\hat p_2(1-\hat p_2))/(n_2)} }` ) = 0.95

P(
`(\hat p_1-\hat p_2)-1.96 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+ (\hat p_2(1-\hat p_2))/(n_2)} }` < (
`p_1-p_2`) <
`(\hat p_1-\hat p_2)+1.96 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+ (\hat p_2(1-\hat p_2))/(n_2)} }` ) = 0.95

95% confidence interval for (
`p_1-p_2`) =

[
`(\hat p_1-\hat p_2)-1.96 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+ (\hat p_2(1-\hat p_2))/(n_2)} }`,
`(\hat p_1-\hat p_2)+1.96 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+ (\hat p_2(1-\hat p_2))/(n_2)} }`]

= [
`(0.25-0.275)-1.96 * {\sqrt{(0.25(1-0.25))/(1000)+ (0.275(1-0.275))/(1000)} }`,
`(0.25-0.275)+1.96 * {\sqrt{(0.25(1-0.25))/(1000)+ (0.275(1-0.275))/(1000)} }` ]

= [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].