Question

Let p1 represent the population proportion of U.S. Senate and Congress (House of Representatives) democrats who are in favor of a new modest tax on "junk food". Let p2 represent the population proportion of U.S. Senate and Congress (House of Representatives) republicans who are in favor of a new modest tax on "junk food". Out of the 265 democratic senators and congressman 106 of them are in favor of a "junk food" tax. Out of the 275 republican senators and congressman only 57 of them are in favor of a "junk food" tax. Find a 95 percent confidence interval for the difference between proportions l and 2.

Answer 1

95 percent confidence interval for the difference between proportions l and 2 is [0.114 , 0.266].

Explanation:

We are given that out of the 265 democratic senators and congressman 106 of them are in favor of a "junk food" tax. Out of the 275 republican senators and congressman only 57 of them are in favor of a "junk food" tax.

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 democratic senators and congressman who are in favor of a "junk food" tax =
`(106)/(265)` = 0.40

`\hat p_2` = sample proportion of republican senators and congressman who are in favor of a "junk food" tax =
`(57)/(275)` = 0.21

`n_1` = sample of democratic senators and congressman = 265

`n_2` = sample of republican senators and congressman = 275

`p_1` = population proportion of U.S. Senate and Congress democrats who are in favor of a new modest tax on "junk food"

`p_2` = population proportion of U.S. Senate and Congress republicans who are in favor of a new modest tax on "junk food"

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

So, 95% confidence interval for the difference between population population, (
`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) } }`]

= [
`(0.40-0.21)-1.96 * {\sqrt{(0.40(1-0.40))/(265)+(0.21(1-0.21))/(275) } }`,
`(0.40-0.21)+1.96 * {\sqrt{(0.40(1-0.40))/(265)+(0.21(1-0.21))/(275) } }`]

= [0.114 , 0.266]

Therefore, 95% confidence interval for the difference between proportions l and 2 is [0.114 , 0.266].