In a survery c: 392 households, a Food Marketing Institute found that 292 households spend more than $\$ 125$ a week on groceries. To find the $95 \%$ confidence interval for the true proportion of the households that spend more than \$125 a week on groceries, you need to use which one of the following calculators? Two Independent Sample Means Comparison Given Statistics Confidence Interval for a Population Mean Given Data Two Independent Sample Means Comparison Given Data Hypothesis Test for a Population Mean Given Data Chi-Square Test for Goodness of Fit Two Dependent Sample Means Comparison Given Data Confidence Interval for a Population Proportion One-Way ANOVA Hypothesis Test for a Population Proportion Chi-Square Test for Independence Confidence Interval for a Population Mean Given Statistics Two Independent Proportions Comparison Hypothesis Test for a Population Mean Given Statistics 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. decimal places. c. Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places. \[ p=0 \pm \]
Solution
Step 1 ：The problem is asking for a confidence interval for a proportion. In this case, the proportion is the number of households that spend more than $125 a week on groceries out of the total number of households surveyed. The confidence interval will give us a range of values that we can be 95% confident contains the true proportion of households that spend more than $125 a week on groceries in the entire population.
Step 2 ：To calculate this, we can use the formula for a confidence interval for a proportion, which is: \(\hat{p} \pm z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) where: \(\hat{p}\) is the sample proportion (in this case, 292/392), \(z\) is the z-score for the desired confidence level (for a 95% confidence level, the z-score is approximately 1.96), \(n\) is the sample size (in this case, 392).
Step 3 ：Substitute the given values into the formula: \(n = 392, x = 292, confidence\_level = 0.95, z = 1.96, \hat{p} = 0.7448979591836735, se = 0.022017212051010517, ci\_lower = 0.7017442235636929, ci\_upper = 0.7880516948036541\).
Step 4 ：Final Answer: The 95% confidence interval for the true proportion of households that spend more than $125 a week on groceries is \(\boxed{(0.702, 0.788)}\).
