Mathematics, 05.05.2020 20:03 voidnyaah

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.

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Answer from: miya2579

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

Step-by-step 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. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

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

$\hat p_2$ = sample proportion of republican senators and congressman who are in favor of a "junk food" tax = $\frac{57}{275}$ = 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 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ < 1.96) = 0.95

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

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

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

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

= [0.114 , 0.266]

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

Answer from: Quest

y=x+9

step-by-step explanation:

Answer from: Quest

It's like discovering the answer to a puzzle. the purpose in solving an equation is to find the value or values of the variable that makes it a true statement. any value of the variable that makes the equation true is called a solution to the equation.

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