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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 40 students in the high school and found a mean of 185 messages sent per day with a standard deviation of 87 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 4040 students in the high school and found a mean of 185185 messages sent per day with a standard deviation of 8787 messages. At the 95% 95 \% confidence level, find the margin of error for the mean, rounding to the nearest whole number. (Do not write ± \pm ).\newlineAnswer:

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 4040 students in the high school and found a mean of 185185 messages sent per day with a standard deviation of 8787 messages. At the 95% 95 \% confidence level, find the margin of error for the mean, rounding to the nearest whole number. (Do not write ± \pm ).\newlineAnswer:
  1. Identify z-score: To find the margin of error at the 95%95\% confidence level, we need to use the formula for the margin of error (ME) in a sample mean, which is ME=z×(σ/n)ME = z \times (\sigma/\sqrt{n}), where zz is the z-score corresponding to the confidence level, σ\sigma is the standard deviation, and nn is the sample size.
  2. Calculate margin of error: First, we need to determine the zz-score that corresponds to the 95%95\% confidence level. For a 95%95\% confidence interval, the zz-score is typically 1.961.96 because it captures the central 95%95\% of the normal distribution.
  3. Determine values: Next, we will plug in the values we have into the margin of error formula. We have the standard deviation (σ)(\sigma) of 8787 messages and the sample size (n)(n) of 4040 students.
  4. Calculate square root: Now, we calculate the margin of error using the formula ME=z×(σ/n)=1.96×(87/40)ME = z \times (\sigma/\sqrt{n}) = 1.96 \times (87/\sqrt{40}).
  5. Divide standard deviation: We calculate the square root of the sample size: 406.3246\sqrt{40} \approx 6.3246.
  6. Multiply by z-score: Now, we divide the standard deviation by the square root of the sample size: 876.324613.7563\frac{87}{6.3246} \approx 13.7563.
  7. Round to nearest whole: Finally, we multiply this result by the z-score to find the margin of error: 1.96×13.756326.96241.96 \times 13.7563 \approx 26.9624.
  8. Round to nearest whole: Finally, we multiply this result by the z-score to find the margin of error: 1.96×13.756326.96241.96 \times 13.7563 \approx 26.9624.Since we need to round to the nearest whole number, the margin of error rounded to the nearest whole number is 2727.

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