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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 92 students in the high school and found a mean of 194 messages sent per day with a standard deviation of 63 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 9292 students in the high school and found a mean of 194194 messages sent per day with a standard deviation of 6363 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 9292 students in the high school and found a mean of 194194 messages sent per day with a standard deviation of 6363 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 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. 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 because it captures the central 95%95\% of the normal distribution.
  3. Calculate Margin of Error: Next, we will plug in the values for the standard deviation (σ=63\sigma = 63 messages) and the sample size (n=92n = 92 students) into the margin of error formula.
  4. Plug in Values: Now, we calculate the margin of error using the formula ME=z×(σ/n)=1.96×(63/92)ME = z \times (\sigma/\sqrt{n}) = 1.96 \times (63/\sqrt{92}).
  5. Calculate Square Root: We calculate the square root of the sample size, 92\sqrt{92}, which is approximately 9.59179.5917.
  6. Divide Standard Deviation: Now, we divide the standard deviation by the square root of the sample size: 639.5917≈6.5707\frac{63}{9.5917} \approx 6.5707.
  7. Multiply Z-Score: Finally, we multiply the z-score by this result to find the margin of error: 1.96×6.5707≈12.87861.96 \times 6.5707 \approx 12.8786.
  8. Round to Nearest Whole: Since we need to round to the nearest whole number, the margin of error is approximately 1313 messages.

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