Question

8-6 (Testing a Claim about a Standard Deviation or Variance) 1) Which T1-83. or TI-84 automatic calculator function will you use in this section? a) 1-PropZTest b ) T-Test c) X-Test (d) None of these. 2) When a SRS of 40 people used 'Diet A' for one year, their weight losses had standard deviation 49 lbs. Use a 0.01 significance level to test the claim that the population standard deviation of all users of "Diet A' is 6.0, which is the population standard deviation of use alternative hypotheses. a) Ho:0 S 6.0 b) Ho:0 = 6.0 Ho:o 6.0 Ho: G + 6.0 d) Ha:= 6.0 Ha: > 6.0 Ha: Ha: 0 < 6.0 6.0 3) See problem 2. State the critical value(s). a) 163.691 b) 22.164, 63.191 c) 20.707, 66.766 d) 22.154 4) See problem 2. Calculate the test statistic. c) 39.497 d) 41.783 a) 26.011 b) 58.476 5) See problem 2. State the technical conclusion based on the traditional method.

Answer 1

2) In this case, we want to test the claim that the population. standard deviation of all users of 'Diet A' TS 6.0, which means we have a thought of rejecting this claim..Genepally mull hypothesis is something which is rone reiection. perone, or should be something.consists the claim that is given. Thus in our case null hypothesis is :- He: 0-6.0 We do not have any prior information that the population. standand deviation should be smaller op greater than 6.0. So, the alternate hypothesis will be:- 0 6 .0, simply the opposite of the null hypothesis. Hence, the corneet option is :- (b) HO: 0= 6.0 "Haio 76.0 3) The population we have considered in this case, is normally distpibuted with unknown mean and standard deviation In such cases the test statistic.is - (n-1) s'2 where S' = sample variance 062 = (given value of populations.d) th sample size. It follomos 22 distribution under Hon with df (n-1). For the aHernative Haito the critical value is to be :- x 3-d/2, n-1 and X²a12 on-1- ol = level of significance Here we have, d=0.01. So, eritrean value will be uggs 59 and X3.005,139 X%.995,59 From table, 20.495,39 = 20.707 X0.60539 = 66.766 Hence, the correct option is :- ce) 20.707, 66.766 (e) - 4) Our test statistic is :- (n-1) s'2 we have , n-1 =39 812 = (4.9)2 = 24.01 (0 ) = (6.0)2 = 50.00 (n-1) 39X24.01 36.00 = 26.0108 3333 26.011 - Hence, the copnect option is :- ) 26.011

5) Ho being nejected if the computed value is smaller than x3.995, 39 or exceeds the X3.005359 value. Since 260 44 Ts:- 3.995, 59 96 10 11 < 1%. 005 39, hence the coppect technical conclusion will be: (d) The test statistic is not external to the critical values); FTR Ho. 6) Ho is beaglezeckornal at 10% levels of significance, it may happen accepted that Ho get rejected at other lower levels of significance. Hence, the appropriate non-technical conclusion is: (b) There is not sufficient evidence towarpant pejection of the claim .... 7) The correct option is clearly (a) Tn failing to reject that it is 6-0, it makes me think it probably pretty close to 6.0. 82. Since, we fail to Peject that A diet A has the Users added date more or less close 'value of standard deviation of weiaht losses compared to diet B users, but we do not have any information whether diet A' users has a bit higher value or lower value of s.d than 'diet B' users or not. So the apppoppiate answer - My evidence really did not address this question very well.