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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 52 students in the high school and found a mean of 194 messages sent per day with a standard deviation of 57 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 5252 students in the high school and found a mean of 194194 messages sent per day with a standard deviation of 5757 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 5252 students in the high school and found a mean of 194194 messages sent per day with a standard deviation of 5757 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 Margin of Error Formula: To find the margin of error at the 9595% 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 for 9595% Confidence: 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 we have into the margin of error formula. We have the standard deviation (σ\sigma) as 5757 messages and the sample size (nn) as 5252 students.
  4. Plug in Values: Now we calculate the margin of error using the formula: ME=z×(σ/n)=1.96×(57/52)ME = z \times (\sigma/\sqrt{n}) = 1.96 \times (57/\sqrt{52}).
  5. Calculate Square Root: We calculate the square root of the sample size: 52≈7.2111\sqrt{52} \approx 7.2111.
  6. Divide Standard Deviation: Now we divide the standard deviation by the square root of the sample size: 577.2111≈7.9036\frac{57}{7.2111} \approx 7.9036.
  7. Multiply Z-Score: Finally, we multiply the z-score by the result from the previous step to find the margin of error: ME=1.96×7.9036≈15.4901ME = 1.96 \times 7.9036 \approx 15.4901.
  8. Round to Nearest Whole Number: Since we are asked to round to the nearest whole number, we round 15.490115.4901 to 1515.

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