Question

In a survery of 125 households, a Food Marketing Institute found that 34 households spend more than $125 a week on groceries. Please find the 80% confidence interval for the true proportion of the households that spend more than $125 a week on groceries. a. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places. Confidence interval b. Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places. c. Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places. > Next Question

Answer 1

$\\ $ The sample data is $ \\ $ Sample size $ (n) = 125 \\ $ Number of successes $ (x) = 34 \\\\ $ Sample proportion $(\hat{p}) = \dfrac{x}{n} = \dfrac{34}{125} = 0.272 \\\\\\ $ Standard error $ (SE) = \sqrt{ \dfrac{ \hat{p}(\ 1 - \hat{p}\ )}{n}} \\\\ = \sqrt{ \dfrac{ 0.272(1-0.272)}{125}} \approx 0.0398 \\\\\\ $ Confidence Level $ = 80\% = 0.80 \\ $ So, significance level $ (\alpha) = 1 - 0.80 = 0.20 \\ $ Critical value $ : z_{c} = z_{\alpha/2} = z_{0.20/2} = z_{0.10} = 1.28 \\ ( P(\ z < -1.28\ ) $ is closest to 0.10 from Z table $)\\\\ $ Margin of error $ (ME) = z_{c} * SE \\ = 1.28 * 0.0398 \approx 0.051$

$\\ a) \\\\ $ Lower Limit $ = \hat{p} - ME = 0.272 - 0.051 = 0.221 \\ $ Upper Limit $ = \hat{p} + ME = 0.272 + 0.051 = 0.323 \\\\ $ So, 80\ \% confidence interval is $ \textbf{(0.221,0.323)} \\\\ b) \\\\ $ As lower limit is 0.221 and upper limit is 0.323 $ \\\\ $ so, 80\% confidence interval is $ \mathbf{0.221 < p < 0.323} \\\\ c) \\\\ $ As point estimate is $ \hat{p} = 0.272 \\ $ and margin of error is $ 0.051 \\\\ $ So, 80\% confidence interval is $ \mathbf{0.272 \pm 0.051}$