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:

Answer 1

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 (0.05, 0.112).

Explanation:

Before building the confidence interval, the central limit theorem and subtraction of normal variables is explained.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
`\mu = p` and standard deviation
`s = \sqrt{(p(1-p))/(n)}`

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Northern half:

1062 out of 2000, so:

`p_N = (1062)/(2000) = 0.531`

`s_N = \sqrt{(0.531*0.469)/(2000)} = 0.0112`

Southern half:

900 out of 2000, so:

`p_S = (900)/(2000) = 0.45`

`s_S = \sqrt{(0.45*0.55)/(2000)} = 0.0111`

Distribution of the difference:

`p = p_N - p_S = 0.531 - 0.45 = 0.081`

`s = √(s_N^2 + s_S^2) = √(0.0112^2 + 0.0111^2) = 0.0158`

Confidence interval:

The confidence interval is:

`p \pm zs`

In which

z is the z-score that has a p-value of
`1 - (\alpha)/(2)`.

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a p-value of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower bound of the interval is:

`p - zs = 0.081 - 1.96*0.0158 = 0.05`

The upper bound of the interval is:

`p + zs = 0.081 + 1.96*0.0158 = 0.112`

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 (0.05, 0.112).