In a study of monthly household expenditures for California households, a UCLA research team selected a random sample of 1049 households.Average household expenditure for the sample during the month of April was $2,754, with a sample standard deviation of $243.Show the 99% confidence interval estimate of average April household expenditures for the population California households.Report the upper bound for your interval.
A)2542.89
B)2773.36
C)2376.22
D)2711.36
E)2634.17

Question

Quizplus · Accepted Answer

To calculate the 99% confidence interval for the mean, we use the formula: $$ \bar{x} \pm z \frac{s}{\sqrt{n}} $$ where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, and $z$ is the z-score corresponding to the desired confidence level. For a 99% confidence level, the z-score is approximately 2.576.Given $\bar{x} = 2754$, $s = 243$, and $n = 1049$, the margin of error (ME) is calculated as: $$ ME = 2.576 \times \frac{243}{\sqrt{1049}} \approx 19.36 $$Thus, the confidence interval is: $$ 2754 \pm 19.36 $$The upper bound of this interval is: $$ 2754 + 19.36 = 2773.36 $$

In a Study of Monthly Household Expenditures for California Households