Question

A soft drink company is conducting research to select a new design for the can. The same drink is used, but is served in a red can and a blue can to each of the 16 participants. The participants are asked to rate each drink (red can and blue can) on a scale of 1 to 10. The difference between red and blue can ratings averaged - 1.125, with a standard deviation of 1.1475. Does this sample indicate that there is a difference in the ratings due to can color? Create a 90% confidence interval for the mean difference of can preference. CI--0.435 to 1.815 C1--0.138 to -2.112 C--1.102 to 1.148 C--1.63 to -0.62

Answer 1

90 % Confidence interval is d = to -1.125 1.1475 = 16 Sample mean diffrence d Standard deviation Sd Sample size n df=n-1 = 15 0.10 Critical Value ta2 = 1.75305 = 1.753 Lower limit = d - t2* =-1.125-1.753*1.1475716 = -1.6279 Upper limit = d + to2 * =-1.125+1.753*1.1475116 = -0.6221 90 % Confidence interval is =(-1.63 to -0.62) 4th option is correct