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.
\[
p=0 \pm
\]
Answer
Final Answer: The 95% confidence interval for the true proportion of households that spend more than $125 a week on groceries is \(\boxed{(0.702, 0.788)}\).
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 \(\boxed{(0.702, 0.788)}\).
