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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 $32$ 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.

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Use normal distributions to approximate binomial distributions

Full solution

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

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.

Identify Mean, SD, Size: Identify the sample mean, standard deviation, and sample size. $\newline$ The sample mean ( $\bar{x}$ ) is given as $\$2900$ , the standard deviation ( $s$ ) is $\$1300$ , and the sample size ( $n$ ) is $32$ students.
Calculate SEM: Determine the standard error of the mean (SEM). The standard error of the mean is calculated using the formula $SEM = \frac{s}{\sqrt{n}}$ . $SEM = \frac{1300}{\sqrt{32}}$ $SEM = \frac{1300}{5.65685}$ (rounded to five decimal places) $SEM \approx 229.81$ Round SEM to the nearest whole number: $SEM \approx 230$ dollars.
Find Z-Score: Find the z-score that corresponds to a $95\%$ confidence level. $\newline$ For a $95\%$ confidence interval, the z-score is typically $1.96$ (this value comes from standard normal distribution tables).
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. $ME = z \times SEM$ $ME = 1.96 \times 230$ $ME \approx 450.8$ Round ME to the nearest whole number: $ME \approx 451$ dollars.
Calculate Confidence Interval: Calculate the lower and upper bounds of the $95$ % confidence interval. $\newline$ Lower bound = $\bar{x} - ME$ $\newline$ Lower bound = $2900 - 451$ $\newline$ Lower bound = $2449$ dollars. $\newline$ Upper bound = $\bar{x} + ME$ $\newline$ Upper bound = $2900 + 451$ $\newline$ Upper bound = $3351$ dollars. $\newline$ Round both values to the nearest whole number.
State Final Interval: State the final $95$ % confidence interval. $\newline$ The $95$ % confidence interval for the mean amount of money saved by each student is approximately $\$2449$ to $\$3351$ .

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