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The American Sugar Producers Association wants to estimate the mean yearly sugar consumption. A sample of 16 people reveals the mean yearly consumption to be 60 pounds with a standard deviation of 20 pounds. a. What is the value of the population mean? What is the best estimate of this value? b. Explain why we need to use the t distribution. What assumption do you need to make? c. For a 90 percent confidence interval, what is the value of t? d. Develop the 90 percent confidence interval for the population mean. e. Would it be reasonable to conclude that the population mean is 63 pounds? a. population mean is 60 pounds. The best estimator of population mean is sample mean, it is unbiased estimator. b. Since the population standard deviation is unknown and sample size is small(<30), we have to use t-distribution. The assumption is the population distribution is normal. c. Since the degree is 16-1 =15. t0.05(15) = 1.753 d. Interval is (sample mean – t0.05(15)*s/n1/2, sample mean + t0.05(15)*s/n1/2) =(60 – 1.753*20/4, 60 + 1.753*20/4) =(51.235, 68.765) e. null hypothesis: = 63, alternative hypothesis: 63 since |(sample mean – 63)/(s/n1/2)| = |(-3)/(20/4)| = 3/5 < t0.05(15), we can not reject null hypothesis. That is, we can conclude that the population mean is 63 with 90% confidence.