Question

Problem 5 (7 marks) A survey of 36 retail stores revealed that the average price of a tablet was $375 with a standard deviation of $20. a) What is the 95% confidence interval to estimate the true cost of the tablet? (4 marks) b) What sample size would be needed to estimate the true average price of a tablet with an error of +5$ and a 99% confidence? (3 marks) Problem 6 (10 marks) Rainbow Trout, Inc. feeds fingerling trout in special ponds and markets them when they attain a certain weight. A sample of 9 trouts were isolated in a pond and fed a special food mixture called "Grow Em Fast". At the end of the experimental period, the weights of the trouts were in grams): 124, 125, 123, 118, 125, 127, 125, 126 and 121. a) What is the mode? Justify properly your answer. (1 mark) b) What is the arithmetic mean? (2 marks) c) What is the median? Justify properly your answer. (2 marks) d) Based on the values of the arithmetic mean, median, and mode, what is the most likely shape of the distribution? Justify properly your answer. (1 mark) e) Compute the standard deviation of the sample. (4 marks)

Answer 1

Friday

31/07/2020

05 A) Given that X = 375 Therefore, a/2 = 0.025 B) Given. Lower limit = = X-E = 375-6.767 = 368.233 S = 20 n= 36 Here, da 20 E= 5 c196 = 99 CI = 95 df = 36-1 = 35 tc = = t0.025,35 = 2.030 At 99% confidence levels Upper limits = R +E = 375+6.767 = 381.767 Standard error = s/vn = 20/736 = 3.3333 Now, Margin of error = = ta/2.,df * std error = 2.033.3333 = 6.7667 C.I. - 368.233<381.767 a = 1-9996 = 1-0.99 = 0.01 Therefore, a/2 = 0.005 & Za /2 = 2.58 At 95% confidence levelt is = 1-9596 = 1-0.95 = 0.05 Therefore, the 95% confidence interval of the population mean is = Now, Sample size = n = 12*o/E)^2 = (2.58*20/5) 2 = 106.50 = 107 ... rounding to nearest whole number.