Question

A gubernatorial candidate wants to analyze how she is performing with different age groups. Her campaign staff surveyed 150 random constituents in the 18–24 age range and 175 constituents in the 65+ age range. Of those constituents, 67 of the people 18–24 years old and 90 of the people 65+ years old said they planned on voting for the candidate for governor. Find the lower bound of the 99% confidence interval for the difference in population proportions of people 18–24 years old who plan to vote for the candidate for governor and people 65+ years old who plan to vote for the candidate for governor. Let the people 18–24 years old be sample 1 and let the people 65+ years old be sample 2. Assume that the samples are independent. Enter your answer as a decimal rounded to three decimal places.

Answer 1

Given that,

Sample 1 : n1 = 150, x1 = 67 and $\hat p_1 = 67/150 = 0.4467$

For sample 2 : n2 = 175, x2 = 90 and

= 90/175 0.5143 P2

A 99% confidence level has significance level of 0.01 and critical value is, $Z_{\alpha/2} = 2.575$

The 99% confidence interval for the difference in population proportions of people 18–24 years old who plan to vote for the candidate for governor and people 65+ years old who plan to vote for the candidate for governor.

$(\hat p_1 - \hat p_2) \pm Z_{\alpha/2} *\sqrt {\frac {\hat p_1 *(1-\hat p_1)}{n_1} +\frac {\hat p_2*(1-\hat p_2)}{n_2}}$

$(0.4467 - 0.5143) \pm 2.575 *\sqrt {\frac {0.4467 *(1-0.4467)}{150} +\frac {0.5143*(1-0.5143)}{175}}$

$-0.0676\pm 0.1428$

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Where, lower bound = -0.210 and upper bound = 0.075