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A study of a local high school tried to determine the mean number of text messages that each student sent per day. The study surveyed a random sample of 95 students in the high school and found a mean of 199 messages sent per day with a standard deviation of 68 messages. At the
95% confidence level, find the margin of error for the mean, rounding to the nearest whole number. (Do not write
+- ).
Answer:

A study of a local high school tried to determine the mean number of text messages that each student sent per day. The study surveyed a random sample of $95$ students in the high school and found a mean of $199$ messages sent per day with a standard deviation of $68$ messages. At the $95 \%$ confidence level, find the margin of error for the mean, rounding to the nearest whole number. (Do not write $\pm$ ). $\newline$ Answer:

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Interpret confidence intervals for population means

Full solution

Q. A study of a local high school tried to determine the mean number of text messages that each student sent per day. The study surveyed a random sample of

95

students in the high school and found a mean of

199

messages sent per day with a standard deviation of

68

messages. At the

95 \%

confidence level, find the margin of error for the mean, rounding to the nearest whole number. (Do not write

\pm

).

\newline

Answer:

Identify Given Information: Identify the given information. $\newline$ We have a sample mean $\bar{x}$ of $199$ messages, a standard deviation $\sigma$ of $68$ messages, and a sample size $n$ of $95$ students. We want to calculate the margin of error at the $95\%$ confidence level.
Determine Critical Value: Determine the critical value (z-score) for the $95\%$ confidence level. $\newline$ For a $95\%$ confidence level, the z-score typically used is $1.96$ . This value is obtained from a standard normal distribution table or z-score table.
Calculate SEM: Calculate the standard error of the mean (SEM). The standard error of the mean is calculated using the formula $SEM = \frac{\sigma}{\sqrt{n}}$ , where $\sigma$ is the standard deviation and $n$ is the sample size. $SEM = \frac{68}{\sqrt{95}}$ $SEM \approx \frac{68}{9.7468}$ $SEM \approx 6.976$
Calculate Margin of Error: Calculate the margin of error. $\newline$ The margin of error ( $E$ ) is calculated using the formula $E = z \times SEM$ , where $z$ is the z-score and $SEM$ is the standard error of the mean. $\newline$ $E = 1.96 \times 6.976$ $\newline$ $E \approx 13.67296$
Round Margin of Error: Round the margin of error to the nearest whole number. $\newline$ Rounding $13.67296$ to the nearest whole number gives us $14$ .

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