Question

9. (0 points) Assume that all cell phone prices are normally distributed. In a random sample of 20 cell phones, you find that the mean price was $500 with a standard deviation of $180. If you would like to construct a 95%-confidence interval to estimate the population mean, what distribution would you use to find the margin of error? Explain your answer
i would use a t distribution because its a sample and its normally distrubted, im not sure if thats correct

Answer 1

In this n= 20 Mean X = 500 Standard deviation S = 180 :d=1-Confidence 2= 1-0.95 n< 30 use t-distribution . 95y.confidence Interval critical value teeof df=n-1 df= 20-1 to.os 2 19 df = 19 Fo025, 19 = 2.093 Margin of error CE) = 180 LE) = 2.093 * A 20 Margin of error CE)=|84.24