Question

Question:A candidate for one of Ohio's two U.S. Senate seats wishes to compare her support among registered voters in the northern half of the state with her support among registered voters in the southern half of the state. A random sample of 2000 registered voters in the northern half of the state is selected, of which 1062 support the candidate. Additionally, a random sample of 2000 registered voters in the southern half of the state is selected, of which 900 support the candidate. A 95% confidence interval for the difference in proportion of registered voters that support this candidate between the northern and southern halves of the state is:A. 0.050 to 0.112.B. 0.035 to 0.127.C. 0.040 to 0.122.D. 0.037 to 0.119.

Answer 1

Answer:

The correct option is (A).

Step-by-step explanation:

The (1 - α)% confidence interval for difference in proportion formula is,

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha /2}\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

The given information is:

n₁ = n₂ = 200,

X₁ = 1062,

X₂ = 900.

Compute the sample proportion as follows:

$\hat p_{1}=\frac{X_{1}}{n_{1}}=\frac{1062}{2000}=0.531\\\\\hat p_{2}=\frac{X_{2}}{n_{2}}=\frac{900}{2000}=0.45$

For the 95% confidence level, the z-value is,

z₀.₀₂₅ = 1.96

*Use a z-table.

Compute the 95% confidence interval for the difference in proportion as follows:

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha /2}\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

     $=(0.531-0.45)\pm1.96\sqrt{\frac{0.531(1-0.531)}{2000}+\frac{0.45(1-0.45)}{2000}}$

     $=0.081\pm 0.031\\=(0.050, 0.112)$

Thus, the 95% confidence interval for the difference in proportion of registered voters that support this candidate between the northern and southern halves of the state is (0.050, 0.112).

The correct option is (A).