Construct a 95% confidence interval given each of the following sample means and sample sizes. Recall S that the format of a confidence interval for a mean is: * = (t *) SEest where and $ = the sample standard deviation. Assume conditions of the CLT are satisfied. To find to, use Table 4 in appendix A of our text book (or you can find a t-distribution table on the internet) and recall that df (degrees of freedom) = n - 1. 1 . x =90; 5 =2.5; n =10 t* SEEST. Interval: x =90; 5 =2.5; n=25 t* = 2. SEEST- Interval: 3 x =90; 5 =2.5; n =100 [* = SEEnt. Interval: 4 x =90; 5=3.5; n =100 t* SEEST Interval: 5 . x =4.5; 5 =0.75; n =25 t* = SEEM Interval: 6. x =4.5; 5 =0.75; n =100 t* =. SEEST. Interval: 7. x =4.5; s =0.75; n =500 t* = SEEST. Interval: 8. x =4.5; 5 =0.25; n =500 t* = SEES Interval: 9. What if: A shopper randomly selects four bags of oranges each bag labelled 10 pounds. The bags weighed 10.2, 10.5, 10.3, and 10.3 pounds. Assume the distribution of weights is Normal. Find a 95% confidence interval for the mean weight of all bags of oranges. Interpret your result. 10. In finding a confidence interval for a random sample of 30 students' GPAs, one interval was (2.60,3.20) and the other was (2.65,3.15) . One of them is a 95% interval and one is a 90% interval. Which is which and how do you know? b. If we used a large sample size (n =120 instead of n =30) would the 95% interval be wider or narrower than the one reported here? 11. State whether each of the following changes would make a confidence interval wider or narrower. (Assume nothing else changes.) a. Changing from a 90% confidence level to a 99% confidence level