The Quinnipiac University Poll conducts nationwide surveys as a public service and for research. In one poll. participants were asked whether they thought eliminating the federal gas tax for the summer months is a good idea. The following problems are based on the results of that poll.

a. Of$611$Republicans, $275$thought it a good idea, and, of $872$Democrats, $366$thought it a good idea. Obtain a $90\%$confidence interval for the difference between the proportions of Republicans and Democrats who think that eliminating the federal gas tax for the summer months is a good idea.

b. Of $907$women,$417$thought it a good idea, and, of $838$men, $310$thought it a good idea. Obtain a$90\%$ confidence interval for the difference between the percentages of women and men who think that eliminating the federal gas tax for the summer months is a good idea.

Given in the question that, The Quinnipiac University Poll conducts nationwide surveys as a public service and for research. In one poll. participants were asked whether they thought eliminating the federal gas tax for the summer months is a good idea. The following problems are based on the results of that poll.

we need to obtain a $90\%$confidence interval for the difference between the proportions of Republicans and Democrats who think that eliminating the federal gas tax for the summer months is a good idea.

The given values are,$x_1=275,n_1=611,x_2=366,n_2=872$, and $90\%$confidence interval.

The formula for${\tilde{p}}_1$is given by,

${\tilde{p}}_1=\frac{x_1+1}{n_1+2}$

Substitute the values $x_1=275,n_1=611$We get

$=\frac{275+1}{611+2}$

$=0.450$

The formula for ${\tilde{p}}_2$is given by,

${\tilde{p}}_2=\frac{x_2+1}{n_2+2}$

$=\frac{366+1}{872+2}$

$=0.420$

Therefore, the sample proportions are $0.450$and $0.420$.

When both sample sizes are 5 or higher, the two proportions plus-four $z$confidence interval technique can be utilised.

Foe a confidence level of $(1-\alpha )$the plus-four $z$confidence interval for $P_1-P_2$is.

Calculate the value of $\alpha$,

$90=100(1-\alpha )$

$\alpha =0.1$

The value of $z$ at $\alpha /2$ from the $z$-score table is $1.645$.

For the difference between the two-population proportion, the needed confidence interval is determined as,

$\left({\tilde{p}}_1-{\tilde{p}}_2\right)\pm z_{\alpha /2}·\sqrt{\frac{{\tilde{p}}_1\left(1-{\tilde{p}}_1\right)}{n_1+2}+\frac{{\tilde{p}}_2\left(1-{\tilde{p}}_2\right)}{n_2+2}}=(0.450-0.420)\pm 1.945$

$.\sqrt{\frac{0.450(1-0.450)}{613}+\frac{0.420(1-0.420)}{874}}$

$=0.03\pm 0.043$

$=-0.013$to $0.073$

As a result, the difference between two proportions ranges from $-0.008$ to $0.064$.

Given in the question that, The Quinnipiac University Poll conducts nationwide surveys as a public service and for research. In one poll, participants were asked whether they thought eliminating the federal gas tax for the summer months is a good idea. The following problems are based on the results of that poll.

we need to Obtain a $90\%$confidence interval for the difference between the percentages of women and men who think that eliminating the federal gas tax for the summer months is a good idea.

The given values are, $x_1=417,n_1=907,x_2=310,n_2=838$, and $90\%$confidence interval.

The formula for ${\tilde{p}}_1$is given by,

${\tilde{p}}_1=\frac{x_1+1}{n_1+2}$

Substitute $x_1=417,n_1=907$

$=\frac{417+1}{907+2}$

$=0.460$

The formula for ${\tilde{p}}_2$is given by,

${\tilde{p}}_2=\frac{x_2+1}{n_2+2}$

Substitute $x_2=310,n_2=838$

$=\frac{310+1}{907+2}$

$=0.370$

Therefore, the sample proportions are $0.460$and $0.370$.

When both sample sizes are 5 or higher, the two proportions plus-four $z$confidence interval technique can be utilised.

For a confidence level of $(1-\alpha )$the plus-four$z$confidence interval for $P_1-P_2$is

Calculate the value of $\alpha$,

$90=100(1-\alpha )$

$\alpha =0.1$

The value of at $\alpha /2$from the z-score table is $1.645$

For the difference between the two-population proportion, the needed confidence interval is determined as,

$\left({\tilde{p}}_1-{\tilde{p}}_2\right)\pm z_{\alpha /2}·\sqrt{\frac{{\tilde{p}}_1\left(1-{\tilde{p}}_1\right)}{n_1+2}+\frac{{\tilde{p}}_2\left(1-{\tilde{p}}_2\right)}{n_2+2}}=(0.460-0.370)\pm 1.945.\sqrt{\frac{0.460(1-0.460)}{909}+\frac{0.370(1-0.370)}{840}}$

$=0.090\pm 0.039$

$=0.051$to $0.129$

As a result, the difference between two proportions ranges from$0.051$to 0.129.