In a public opinion survey, 80 out of a sample of 100 high-income voters and 55 out of a sample of 80 low-income voters supported the introduction of a new national security tax.
a/ Estimate, with 95% confidence level, the true proportion of low-income people who will vote for the introduction of the tax.
b/ Can we conclude at the 5% level of significance that the proportion of high-income voters favoring the new security tax is 10% higher than that of low-income voters?

Yes, we can conclude that at the 5% level of significance, the proportion of high income voters favoring the new security tax is 10% higher than that of low income voters using Test of Significance for Difference of Proportions.

What is confidence interval?

A confidence interval is the mean of your estimate plus and minus the variation in that estimate.

Proportion of high-income people who will vote for the introduction of the tax = p1 = [tex]\frac{80}{100} = \frac{4}{5}[/tex][tex]= 0.8[/tex]

Proportion of low-income people who will vote for the introduction of the tax = p2 = [tex]\frac{55}{80} = \frac{11}{16}[/tex] [tex]= 0.6875[/tex]

95% confidence interval of the true proportion of low income people who will vote for the introduction of the tax -

Upper confidence interval -

[tex]= p + 1.96 \sqrt{pq/n}[/tex]

[tex]=0.6875 + 1.96 \sqrt{(0.6875)(1-0.6875)/80} \\= 0.6875 + 1.96 \sqrt{0.00268} \\=0.789[/tex]

Lower confidence interval -

[tex]= p - 1.96 \sqrt{pq/n}[/tex]

[tex]=0.6875 - 1.96 \sqrt{(0.6875)(1-0.6875)/80} \\= 0.6875 - 1.96 \sqrt{0.00268} \\=0.586[/tex]

What is Test of Significance for Difference of Proportions?

Test of Significance for Difference of Proportions is used when we want to compare two distinct populations with respect to the prevalence of a certain attribute, say A, among their members.

n1 = 100

X1 = 80

p1 = 0.8

n2 = 80

X2 = 55

p2 = 0.6875

H0: the proportion of high-income voters favoring the new security tax is 10% higher than that of low-income voters.(P1 - P2 = 0.1)

H1: the proportion of high-income voters favoring the new security tax isn't 10% higher than that of low-income voters. (P1 - P2 != 0.1)

[tex]z = \frac{(p1 - p2) - (P1 -P2)}{(\frac{X1+X2}{n1+n2})(1-\frac{X1+X2}{n1+n2})(\frac{1}{n1}+\frac{1}{n2} ) }[/tex]

[tex]z = \frac{(0.8 - 0.6875)- 0.1}{\sqrt{0.75*0.25*0.0225} } \\\\z = \frac{0.0125}{0.0649} \\\\z = 0.1926[/tex]

Since z = 0.1926 < 1.96, null hypothesis cannot be rejected. Thus, the proportion of high-income voters favoring the new security tax is 10% higher than that of low-income voters.

Learn more about confidence interval here

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