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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 74 students in the high school and found a mean of 190 messages sent per day with a standard deviation of 99 messages. Determine a 95% confidence interval for the mean, rounding all values to the nearest whole number.

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 7474 students in the high school and found a mean of 190190 messages sent per day with a standard deviation of 9999 messages. Determine a 95% 95 \% confidence interval for the mean, rounding all values to the nearest whole number.

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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 7474 students in the high school and found a mean of 190190 messages sent per day with a standard deviation of 9999 messages. Determine a 95% 95 \% confidence interval for the mean, rounding all values to the nearest whole number.
  1. Identify Mean, Size, Deviation: Identify the sample mean (xˉ\bar{x}), sample size (nn), and sample standard deviation (ss). The sample mean (xˉ\bar{x}) is given as 190190 messages. The sample size (nn) is given as 7474 students. The sample standard deviation (ss) is given as 9999 messages.
  2. Calculate Standard Error: Determine the standard error of the mean (SEM). The standard error of the mean is calculated using the formula SEM=snSEM = \frac{s}{\sqrt{n}}. SEM=9974SEM = \frac{99}{\sqrt{74}} SEM≈998.6023SEM \approx \frac{99}{8.6023} SEM≈11.5106SEM \approx 11.5106 Round to the nearest whole number: SEM≈12SEM \approx 12
  3. Find Critical Value: Find the critical value for a 9595% confidence interval. For a 95%95\% confidence interval with a sample size of 7474, we use the t-distribution because the population standard deviation is unknown and the sample size is less than 3030. The degrees of freedom (df)(df) is n−1n - 1, which is 74−1=7374 - 1 = 73. Using a t-distribution table or calculator, we find the t-value that corresponds to a 95%95\% confidence level and 7373 degrees of freedom. The t-value is approximately 1.9931.993.
  4. Calculate Margin of Error: Calculate the margin of error (ME). The margin of error is calculated using the formula ME=t×SEMME = t \times SEM. ME=1.993×12ME = 1.993 \times 12 ME≈23.916ME \approx 23.916 Round to the nearest whole number: ME≈24ME \approx 24
  5. Determine Confidence Interval: Determine the confidence interval.\newlineThe confidence interval is calculated using the formula: (xˉ−ME,xˉ+ME)(\bar{x} - ME, \bar{x} + ME).\newlineLower bound = 190−24=166190 - 24 = 166\newlineUpper bound = 190+24=214190 + 24 = 214
  6. Round Confidence Interval: Round the confidence interval to the nearest whole number.\newlineThe lower bound is already a whole number: 166166.\newlineThe upper bound is already a whole number: 214214.

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