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A study of a local high school tried to determine the mean amount of money that each student had saved. The study surveyed a random sample of 32 students in the high school and found a mean savings of 2900 dollars with a standard deviation of 1300 dollars. Use the normal distribution/empirical rule to estimate a 95% confidence interval for the mean, rounding all values to the nearest whole number.

A study of a local high school tried to determine the mean amount of money that each student had saved. The study surveyed a random sample of 3232 students in the high school and found a mean savings of 29002900 dollars with a standard deviation of 13001300 dollars. Use the normal distribution/empirical rule to estimate a 95% 95 \% confidence interval for the mean, rounding all values to the nearest whole number.

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Q. A study of a local high school tried to determine the mean amount of money that each student had saved. The study surveyed a random sample of 3232 students in the high school and found a mean savings of 29002900 dollars with a standard deviation of 13001300 dollars. Use the normal distribution/empirical rule to estimate a 95% 95 \% confidence interval for the mean, rounding all values to the nearest whole number.
  1. Identify Mean, SD, Size: Identify the sample mean, standard deviation, and sample size.\newlineThe sample mean (xˉ\bar{x}) is given as $2900\$2900, the standard deviation (ss) is $1300\$1300, and the sample size (nn) is 3232 students.
  2. Calculate SEM: Determine the standard error of the mean (SEM). The standard error of the mean is calculated using the formula SEM=snSEM = \frac{s}{\sqrt{n}}. SEM=130032SEM = \frac{1300}{\sqrt{32}} SEM=13005.65685SEM = \frac{1300}{5.65685} (rounded to five decimal places) SEM≈229.81SEM \approx 229.81 Round SEM to the nearest whole number: SEM≈230SEM \approx 230 dollars.
  3. Find Z-Score: Find the z-score that corresponds to a 95%95\% confidence level.\newlineFor a 95%95\% confidence interval, the z-score is typically 1.961.96 (this value comes from standard normal distribution tables).
  4. Calculate Margin of Error: Calculate the margin of error (ME). The margin of error is found by multiplying the z-score by the standard error of the mean. ME=z×SEMME = z \times SEM ME=1.96×230ME = 1.96 \times 230 ME≈450.8ME \approx 450.8 Round ME to the nearest whole number: ME≈451ME \approx 451 dollars.
  5. Calculate Confidence Interval: Calculate the lower and upper bounds of the 9595% confidence interval.\newlineLower bound = xˉ−ME\bar{x} - ME\newlineLower bound = 2900−4512900 - 451\newlineLower bound = 24492449 dollars.\newlineUpper bound = xˉ+ME\bar{x} + ME\newlineUpper bound = 2900+4512900 + 451\newlineUpper bound = 33513351 dollars.\newlineRound both values to the nearest whole number.
  6. State Final Interval: State the final 9595% confidence interval.\newlineThe 9595% confidence interval for the mean amount of money saved by each student is approximately $2449\$2449 to $3351\$3351.

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