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the answer is as following

Question 2 A The real estate assessor for the Denmark Shire wants to study various characteristics concerning single-family houses in the Shire. A random sample of 70 houses reveals the following: . Heated area of the house (in square feet): =1 ,759, S: 380 . 42 houses have air conditioning. (a) Set up a 99% condence interval estimate of the population average heated area of the house. (Marks = 8) (b) Set up a 95% condence interval estimate of the population proportion of houses that have air conditioning. (Marks = 8) Given the computer printout below that relates to a model for predicting weekly sales based on the number of customers making purchases over a 15 week period: - m- - .- -_I_ _l -_ M a) Is the number of customers a signicant independent variable in this model? Explain why or why not? (Marks = 2) b) What is the value of the coefcient of determination here? What does it mean for this problem? (Marks = 2) c) What is the value for the standard error of estimate? (Marks = 2) Data Null H othesis -= Level of Si . nicance Number of Successes am . Ie Size Intermediate Calculations Sample Proportion 0.125 Standard Error 0.06846532 WW Answers to this rt should be trans erred to our examination booklet. Using the Ph Stat 2 printout above, answer the following questions. (i). Was this a test of a mean or a proportion (circle your response) Mean Proportion (Marks = 1) (ii). Write the null and alternative hypothesis (iv). Was it a two or a one-tail test? (circle your response) Two One (1 mark) ( v). Explain how the critical value and the p-value both lead to the same decision about Ho. (3 marks) [Marks = 16 + 6 + 8]S QUESTION 2 0.1 10129497 10,657445214 Adwind R Squince 0.631094972 A 70 = 1759 ; 2 =42 StandON ENTEE 9.236036641 IST S = 380 ANDY (one " Z" due to DecisionTree, ACCT. 1.1604314 31.1604324 0.000243105 Raldual 11.39014069 0.87416-4669 33 230573) (a ) 2 380 = 45. 4187 1) Coolielena Shadard Error( Siet 16.0321934 3 310167093 019150493 0.009569 0.930760721 0.006 158139 4,99501 1643 0,000245 SOL 9 9 / CI for u = 2128 Weekly Sales = f (No. of customers) POSED SOLUTIONS = 1759 1 2.575 (45. 418#) ' = - 16.032/936 + 0.030760238 X D = 1759 + 716 . 9631' ( a) .yes, The at stat in large a riquepursutate 16412 08168 = 1 = 1875. 9631 to 4.995 , 102 = 15-2 = 13 df. OPOSE ie , between 1642 and 18"to square fet 0/ 8 top to . ol, Baf = 3.012 . ( ofuse : (: 26490, 1683.69 5/ = 1899.31 ) - Deduct 2 mark's) also , the very small povalue 0.00 0245 (8) p =m X = 12 = 0.60 O suggests acting repection of it: pso. ( bo) of " = 0.65744 .... PR 65-74 % of The variation in bale's can be : 95 / CIffer TI = $ 205 of customers . " = 0.60 196 (0.0586) 0.4852 = TT = 0.7148 ( C ) 'S yx = 0.93603..... it, between 48 % and 72% of houses. 0/ 8 Ibendicales the standard deviation of the observations around the regressionSAMPLE - PROPOSED SOLUTIONS ( i) Proportion ( 11 ) Ho : D= 0.25 (2 ) H , : MCO . 25 (2) ( ili ) 2lest (1) (in 1 tail test p = 0 . @339 ... 1 1 = 0.05 Dence pvalue (0.0339 ) "