Question

Confidence interval for the mean: Hoping to lure more shoppers downtown, a city builds a new public parking garage in the central business district. The city plans to pay for the structure through parking fees. During a two month period (n = 44 days), daily fees collected average $126. Assume it is known that weekly fees have a standard deviation of - $15. a) It is not known that the daily collected fees follow a Normal Dist: Why is it appropriate to use the Normal Distribution for ? b) What is the critical value for a 90% CI what is the SE for mean: what is the ME E = Find the CI (using tables) Find the CI (using the calculator-Z interval): Write steps

Answer 1

Given no 44; X = 1266 - 15 (a) since n>, 90 and o is known Therefore we can use normal distribution by using Central limit theorem. of a (b) 2n critical at 902_, 2*=1:645 (from 2-table) (A+ 30.05) 2.2613 144 ME. 2* X SE 1645 X 2.2613 - 3. 7199 go% confidence interval is 7 + 2 * olun X I ME 126 3.7199 The 90 l. confidence interval is (122.2801, 129.7199) of (e) We are gol daily fee is confident between that the population mean 122. 2801 and 129.7199 cdo is yes the outside consultant was correct the confidence interval since 130 interval

data from given n = 6: 5.37; SS0.665 Ho: M = 4.8 mg vs HI M > 4.8 mg Here na 30 fr is unknown Therefore we use student's t distribution Test statistic, t. X- M 3/4 0. 5.37 - 4.8 0.665/ 6 0.57 0.2715 t = 2.10 critical value at x=0.05, df = n-1= 6-1=5 t-critical = 2.015 (from ta table) since, t > t-critical Therefore we reject Ho. There is that the 4.8 mg. sufficient patient's evidence mean level to in support fact the claim falls above