help please as soon as possible . I provided some notes . please use the formulas provided and provide step by step.
The data for a random sample of 10 paired observations are shown in the table below. Pair Sample from Population 1 Sample from Population 2 1 19 24 2 25 27 3 31 36 4 52 53 5 49 55 6 34 34 7 59 66 8 47 51 9 17 20 10 51 55 a. If you wish to test whether these data are sufficient to indicate that the mean for population 2 is larger than that for population 1, what are the appropriate null and alternative hypotheses? Define any symbols you use. Conduct the test, part a, using a = .10. c. Find a 90% confidence interval for M. Interpret this result. d. What assumptions are necessary to ensure the validity of this analysis? Paired - Difference Test of hypothesis for Md=( M- M2 ) One - sided Test Two-sided Test Ho : Md = Do . Ho : Ma = Do Aa . Md Do) Ha: Md # Do Large - Sample Case . Test Statistic Z = d - Do ~ d - Do Ed/ und sd/ und . Lej. Region . Lej Region Z c - Z h ( or Z > Z x if Ha: Md > Do) small - sample lase . Test Statistic t = d- Do sd / und . Rej. Region . Rej Region t ty if Ha: Md > Do) It 17 th/ z to , t1/2 are based on ( nd - 1 ) If Confidence Intervals Large Sample: d + 2 1/2 . Ed small - sample . d - to/2 . sad sna 1 to/ is based on (nd - 1)afConditions for Large Sample Inferences for Md 1 ) random sample of differences ? ) the sample size , nd , Is large ( nd 230 ) Conditions for Small S Sample Inferences for Md 1 ) random sample of differences 2 ) Population of differences is approx . normal Paired - Difference, Data Collection Table obs | Group 1 | Group z | Difference - Xu X 21 di = Xil X21 2 X 12 X 2 2 d2 = X 12 X 22 X 13 X 2 3 d3 = * 13 - x2 2 Search documents and file names for text n Xin X zn an = Kin - Xzn Example : You work in HR. You want to check if there is a difference in test scores after a training program. Construct a 901%0 conf. interval for the mean difference in test scores . Data name Before ( 1 ) After ( 2 ) Difference Sam 85 94 5 5-94 =-9 Brian 87 94-87 - 7Example : You work in HR. You want to check if there is a difference in test scores after a training program. Construct a 901%% conf. interval for the mean difference in test scores. Data name Before (1 ) After ( 2 ) Difference Sam 85 94 8 5- 94 =-9 Brian 94 87 94-87-7 Mike 78 79 78 - 79 =- 1 Jane 87 88 87 - 88 = -1 d = Edl =- 9+ 7 - 1- 1 =-1 na 4 sd = z(di- a ) = ( - 9 - (- 1)2 + ( 7 - 6 -1)3 + ( - + ( - 1) ) + ( - 1 -( ) )2 nd - 1 3 sd = 6 53 df = nd- 1= 4-1=3 190 " %% CI = 2 = . 1 => 2 /2 = . 05 t 0 05, 3 = 2.35 90 %% CI = d + th/ , sd = - 1 + 2. 35, 6.53 -= [-8.68, 6.68 Vnd 14Ex 2 : Marketing Department you would like to compare a client's calculator to competitor's . you sample of stores . At the 6.01 level of significance, does your client calculator sell for less than their competitors? Data store Client (i ) competitor [2) Difference (1)-12) 10 - 1 = -1 8- 11= -3 lo 7-10 =-3 12 3 OS J J US W N 11 O 10 13 - 3 12 - 3 10 2 d = Edi - - 1-3-3-3+0-3-3-2 - 2.25 na sd = 1.16 Ho and Haj 1. Client (1 ) less than comp (2) 2 . MI > MI - MZ ZO 3 . Define Md = MI - Mz => Maco 4 . Ha: Md CD Ho : Md = 0 Test Statistic t = d-Do - -2.15- 6--5416 Ha : Md
