You are part of a trivia team and have tracked your team's performance since you started playing, so you know that your scores are normally distributed with = 75 and = 12. Recently, a new person joined the team, and you think the scores have gotten better. Your goal is to answer the question: Are the new scores on average significantly better than 78? Follow the instructions below to determine the answer. Calculate the sample mean of the following 9 weeks' worth of score data: 80, 78, 60, 78, 89, 84, 60, 91, 70. X (Mean) = 80+78+60+78+89+84+60+91+70 / 9 690 / 9 X (Mean) = 76.67 Compute a 95% confidence interval for . (Show the process you used to determine the CI bounds) SE = /n Standard Error = SD/number of observations SE = 12 / 9 = 4 Z = (X - )/SE Z = (76.67 - 75) / 4 = 0.4175 CI = X Z(SE) CI = 76.67 1.96 4 CI = 76.67 + 7.84 = 84.51, 76.67 - 7.84 = 68.83 CI = 68.83, 84.51 We are 95% confident that the true population mean () is between 68.83 and 84.51 From the confidence interval you obtained, can we conclude that the team scores are significantly better since the new person joined the team? Explain your answer. From the confidence interval you obtained, 78 does fall within the confidence interval of 68.83 to 86.54. Therefore, based on this confidence interval, we cannot conclude that the team scores are significantly better since the new person joined the team. The null model is H0: = 75. The alternative model is H1: >