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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 96 students in the high school and found a mean of 160 messages sent per day with a standard deviation of 51 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 9696 students in the high school and found a mean of 160160 messages sent per day with a standard deviation of 5151 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 9696 students in the high school and found a mean of 160160 messages sent per day with a standard deviation of 5151 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 Formula: To find the margin of error at the 9595% confidence level, we need to use the formula for the margin of error (ME) for the mean, which is:\newlineME = z×(σ/n)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 level, 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=51\sigma = 51 messages\newlinen=96n = 96 students\newlinez=1.96z = 1.96\newlineME=1.96×(5196)ME = 1.96 \times (\frac{51}{\sqrt{96}})
  4. Calculate Denominator: Now, we calculate the denominator of the margin of error formula: 96≈9.798\sqrt{96} \approx 9.798
  5. Divide Standard Deviation: We then divide the standard deviation by the square root of the sample size: 519.798≈5.204\frac{51}{9.798} \approx 5.204
  6. Multiply by Z-Score: Finally, we multiply this result by the z-score to find the margin of error:\newlineME=1.96×5.204ME = 1.96 \times 5.204\newlineME≈10.200ME \approx 10.200
  7. Round to Nearest Whole Number: Since we need to round to the nearest whole number, the margin of error is approximately: ME≈10ME \approx 10

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