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A study was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of 62 residents and found the mean weight to be 189 pounds with a standard deviation of 26 pounds. At the
95% confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest tenth. (Do not write
+- ).
Answer:

A study was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of $62$ residents and found the mean weight to be $189$ pounds with a standard deviation of $26$ pounds. At the $95 \%$ confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest tenth. (Do not write $\pm$ ). $\newline$ Answer:

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Interpret confidence intervals for population means

Full solution

Q. A study was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of

62

residents and found the mean weight to be

189

pounds with a standard deviation of

26

pounds. At the

95 \%

confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest tenth. (Do not write

\pm

).

\newline

Answer:

Identify Given Information: Identify the given information and the formula to calculate the margin of error at the $95\%$ confidence level using the normal distribution. $\newline$ Given: $\newline$ - Sample mean ( $\bar{x}$ ) = $189$ pounds $\newline$ - Standard deviation ( $\sigma$ ) = $26$ pounds $\newline$ - Sample size ( $n$ ) = $62$ residents $\newline$ - Confidence level = $95\%$ $\newline$ The formula for the margin of error ( $E$ ) when using the normal distribution is: $\newline$ $E = Z \times (\sigma/\sqrt{n})$ $\newline$ Where $\bar{x}$ $0$ is the Z-score corresponding to the desired confidence level. For a $95\%$ confidence level, the Z-score is approximately $\bar{x}$ $2$ .
Calculate Margin of Error: Calculate the margin of error using the formula. $\newline$ First, calculate the standard error (SE) which is the standard deviation divided by the square root of the sample size. $\newline$ $SE = \frac{\sigma}{\sqrt{n}}$ $\newline$ $SE = \frac{26}{\sqrt{62}}$ $\newline$ $SE \approx \frac{26}{7.874}$ $\newline$ $SE \approx 3.3$ $\newline$ Now, calculate the margin of error (E) by multiplying the Z-score by the standard error. $\newline$ $E = Z \times SE$ $\newline$ $E = 1.96 \times 3.3$ $\newline$ $E \approx 6.468$ $\newline$ Round the margin of error to the nearest tenth. $\newline$ $E \approx 6.5$

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