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.

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. = [tex]\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex] ~ N(0,1)

where, [tex]\hat p_1[/tex] = sample proportion of democratic senators and congressman who are in favor of a "junk food" tax = [tex]\frac{106}{265}[/tex] = 0.40

[tex]\hat p_2[/tex] = sample proportion of republican senators and congressman who are in favor of a "junk food" tax = [tex]\frac{57}{275}[/tex] = 0.21

[tex]n_1[/tex] = sample of democratic senators and congressman = 265

[tex]n_2[/tex] = sample of republican senators and congressman = 275

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

[tex]p_2[/tex] = 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, ([tex]p_1-p_2[/tex]) 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 < [tex]\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex] < 1.96) = 0.95

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

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

95% confidence interval for [tex](p_1-p_2)[/tex] = [[tex](\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex] , [tex](\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex]]

= [ [tex](0.40-0.21)-1.96 \times {\sqrt{\frac{0.40(1-0.40)}{265}+\frac{0.21(1-0.21)}{275} } }[/tex],[tex](0.40-0.21)+1.96 \times {\sqrt{\frac{0.40(1-0.40)}{265}+\frac{0.21(1-0.21)}{275} } }[/tex]]

= [0.114 , 0.266]

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