Suppose that last year a farmer estimated a 95% confidence i…
Suppose that last year a farmer estimated a 95% confidence interval for the population mean weight (pounds) of the Punkin variety Moonglo: (1.0,5.0). This year, the farmer decided to increase the sample size by 50% more and estimate a new 95% confidence interval. (Assume normality of the Punkin weight).a. The new width of the estimated confidence interval for the population mean weight (pounds) will [a]b. The new standard error of the sample mean will [b]c. The new margin error will [c]d. The percentile of the t distribution used in the computation will [d]
Read DetailsA simple random sample of 20 Stat 371 students from courses…
A simple random sample of 20 Stat 371 students from courses taught Fall 2019, Spring 2020, and Summer 2020 yielded the following number of Instagram followers from each student’s personal Instagram accounts. (152, 798, 1450, 1805, 1721, 330, 110, 98, 275, 102, 205,300, 130, 1022, 198, 153, 2155, 2161, 350, 1689) A 90% confidence interval for the unknown mean Stat 371 student Instagram followers is
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