Question

A 2009 holiday retail season , which kicked on november 27, 2009 had been marked by somewhat lower self supported consumer spending than was seen during the comparable period in 2008. To get an estimate of consumer spending, 436 randomly sampled american adults were surveyed. daily consumer spending for the six day period after thanksgiving, spanning the black friday weeked and cyber monday, averaged $84.71. a 95% confident interval based on this sample is ( $80.31, $89.11). determine if true or false and explain your reason . a) we are 95% confident that the average spending of all american adult is between $80.31 and $89.11. true or false. (b) the population mean spending is = $84.71 . true or false (c) the average spending of all american is $90.34 true or false (d) a 90% confident interval would be narrower than the 95% since we dont neeed to be sure about estimate. true or false..... please help

Answer 1

(A) TRUE because the 95% confidence interval is for all Americans or population and it is given in the question that the confidence interval is between 80.31 and 89.11

(B) FALSE, according to the question, the average estimate of $84.71 is for the sample size of n = 436, but not for the entire population. So, given mean spending is for the sample mean.

(C) False, 95% confidence interval is between 80.31 and 89.11, which means that we are 95% confident that the population mean must be within the range of confidence interval. So, 90.34 is outside the range, which makes it false

(D) TRUE because the z critical value will be smaller for 90% confidence interval as compared to 95% confidence interval. So, 90% confidence interval will be narrower due to smaller z critical value