30.03.2023 - 05:14

# Suppose the amount of time teenagers spend weekly working at part-time jobs is normally distributed with a standard deviation of 40 minutes. A random sample of 15 teenagers was drawn, and each reported the amount of time spent at part-time jobs (in minute

Question:

Suppose the amount of time teenagers spend weekly working at part-time jobs is normally distributed with a standard deviation of 40 minutes. A random sample of 15 teenagers was drawn, and each reported the amount of time spent at part-time jobs (in minutes). These are listed below. Determine the 95% confidence interval estimate of the population mean.

180, 130, 150, 165, 90, 130, 120, 60, 200, 180, 80, 240, 210, 150, 125

Answers (1)
• April 15, 2023 в 06:10
Using the given sample data, we can calculate the sample mean as: (180+130+150+165+90+130+120+60+200+180+80+240+210+150+125)/15 = 150 Because the sample size is 15 and the population standard deviation is known (40 minutes), we can use a t-distribution with 14 degrees of freedom (15-1) to construct the confidence interval estimate. Using a t-table or calculator with a 95% confidence level and 14 degrees of freedom, we can determine the t-value to be \approx imately 2.145. The margin of error for the confidence interval estimate is then calculated as: 2.145 * (40/sqrt(15)) = 35.4 Finally, we construct the 95% confidence interval estimate of the population mean as: 150 +/- 35.4 or (114.6, 185.4) Therefore, we can be 95% confident that the true population mean of the time teenagers spend weekly working at part-time jobs lies within the interval of 114.6 to 185.4 minutes.
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