Question

e / Study / A company claims that the mean monthly residential electricity consumption in a certain region is more than 870 kilowatt-hours (kWh). You want to test this claim You find that a random sample of 68 residential customers has a mean monthly consumption of 890 kWh. Assume the population standard deviation is 125 kWh Ala010 can you support the claim? Complete parts (a) through (e) jestio (a) Identify Ho and H. Choose the correct answer below w transcribe kpert A Hy #870 O A. Ho > 890 (claim) OB. Ho > 870 (claim) HE 890 HO HS 870 0C. Họ & B90 OD. Ho #=870 (claim) Ha > 890 (claim) O E. HO HS 870 OF HO 1890 HA 870 (claim) HHX890 (claim) (b) Find the critical value(s) and identify the rojection region(s). Select the correct choice below and fill in the answer box within your choice. Use technology (Round to two decimal places as needed) OA. The critical value is B. The critical values are Identify the rejection region(s) Select the correct choice below OA. The rojection region is 2 > 128 p next fd OB. The rejection region is z<128 For the fo Click to select your answer(s). informatid whether dictribution arch יב a o

Answer 1

Let u be the mean monthly residential electricing consumption u is more than 870 kilojalt hours o Here we have to test the hypothesis, The company's daim thad Hoi u < 870 wls Hat u 870 (clain) i option is connect b) At a= 0.10 It is an one tail tast, Caitical value Za to the paind of normal distribution such that, PL 2 > 201 = a. PL2 > 2010] = 0,10 . From 2- table, we get. Critical value Zouto 1.28 option A is correct } | The Rejection region is 2 1.28 option A is connect. statistic is given os. 2 = X-M 20 is here n- sumple size = - sample mean = 800 kWh 0 - popullation Standaad deviation 126. 800 - 870 - 070x68 126/168 126 2 = -4.5812 de - Normal (6, 1) 68