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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 61 students in the high school and found a mean of 176 messages sent per day with a standard deviation of 77 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 6161 students in the high school and found a mean of 176176 messages sent per day with a standard deviation of 7777 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:

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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 6161 students in the high school and found a mean of 176176 messages sent per day with a standard deviation of 7777 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. Margin of Error Formula: 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:\newlineME=z×(σ/n)ME = z \times (\sigma/\sqrt{n})\newlinewhere zz is the z-score corresponding to the desired confidence level, σ\sigma is the standard deviation, and nn is the sample size.
  2. Find Z-Score: First, we need to find the z-score that corresponds to the 95%95\% confidence level. For a 95%95\% confidence interval, the z-score is typically 1.961.96. This value can be found in standard z-score tables or by using a statistical calculator.
  3. Calculate Margin of Error: Next, we plug in the values we have into the margin of error formula:\newlineσ=77\sigma = 77 messages\newlinen=61n = 61 students\newlinez=1.96z = 1.96\newlineME=1.96×(7761)ME = 1.96 \times (\frac{77}{\sqrt{61}})
  4. Calculate Denominator: Now we calculate the denominator of the margin of error formula: 61≈7.81\sqrt{61} \approx 7.81
  5. Divide Standard Deviation: We then divide the standard deviation by the square root of the sample size: 777.81≈9.86\frac{77}{7.81} \approx 9.86
  6. Final Margin of Error Calculation: Finally, we multiply this result by the z-score to find the margin of error:\newlineME=1.96×9.86≈19.33ME = 1.96 \times 9.86 \approx 19.33\newlineSince we need to round to the nearest whole number, the margin of error is approximately 1919.

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