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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 149 students in the high school and found a mean of 186 messages sent per day with a standard deviation of 99 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 149149 students in the high school and found a mean of 186186 messages sent per day with a standard deviation of 9999 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 149149 students in the high school and found a mean of 186186 messages sent per day with a standard deviation of 9999 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 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 calculator that has the capability to calculate the inverse of the cumulative distribution function for the standard normal distribution.
  3. Find square root: Next, we plug in the values into the margin of error formula: ME=1.96×(99149)ME = 1.96 \times (\frac{99}{\sqrt{149}}).
  4. Divide standard deviation: Now, we calculate the square root of the sample size, 149\sqrt{149}, which is approximately 12.206612.2066.
  5. Multiply z-score: We then divide the standard deviation, 9999, by the square root of the sample size, 12.206612.2066, to get 99/12.2066≈8.108799 / 12.2066 \approx 8.1087.
  6. Round to nearest whole: Finally, we multiply the z-score, 1.961.96, by the result from the previous step, 8.10878.1087, to get the margin of error: 1.96×8.1087≈15.89311.96 \times 8.1087 \approx 15.8931.
  7. Round to nearest whole: Finally, we multiply the z-score, 1.961.96, by the result from the previous step, 8.10878.1087, to get the margin of error: 1.96×8.1087≈15.89311.96 \times 8.1087 \approx 15.8931.We round the margin of error to the nearest whole number, which gives us 1616.

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