Question

Walmart Target 896 844 624 585 493 458 691 813 955 1008 876 832 694 616 644 606 622 670 941 906 833 837 1021 957 768 816 862 777 831 862 876 851 740 747 1042 1037 831 781 892 807

Answer 1

Solution:

Here we use two sampel t test as below

Solution ; Now for simplicity we use walmart as Trecetment I Target as Trecetment 2. Now t-test as given and a DA Trecetment 2, Ans, g). Ho: Mw=lt Versus Ho: Mindle

b)

Treatment 1 (X) 896 585 493 813 955 876 694 644 670 941 833 1021 816 862 832 876 740 1042 831 Diff(X-M) 80.40 -230.60 -322.60 -2.60 139.40 60.40 - 121.60 - 171.60 -145.60 125.40 17.40 205.40 0.40 46.40 16.40 60.40 - 75.60 226.40 15.40 76.40 Sq. Diff(x - M2 6464.16 53176.36 104070.76 6.76 19432.36 3648.16 14786.56 29446.56 21199.36 15725.16 302.76 42189.16 0.16 2152.96 268.96 3648.16 5715.36 51256.96 237.16 5836.96 892 Treatment 2(X) 844 624 458 691 1008 832 616 606 622 906 837 957 768 777 862 851 747 1037 781 807 Diff(X-M) 62.45 - 157.55 - 323.55 -90.55 226.45 50.45 - 165.55 - 175.55 - 159.55 124.45 55.45 175.45 - 13.55 -4.55 80.45 69.45 -34.55 255.45 -0.55 25.45 Sq. Diff(x - 2)2 3900.00 24822.00 104684.60 8199.30 51279.60 2545.20 27406.80 30817.80 25456.20 15487.80 3074.70 30782.70 183.60 20.70 6472.20 4823.30 1193.70 65254.70 0.30 647.70

Difference Scores Calculations Treatment 1 N : 20 df = N - 1 = 20 - 1 = 19 MY: 815.6 551:379564.8 521 = SS/(N - 1) = 379564.8/(20-1) = 19977.09 Treatment 2 N2:20 df2 = N - 1 = 20 - 1 = 19 M2: 781.55 SS2: 407052.95 22 = 552/(N - 1) = 407052.95/(20-1) = 21423.84 T-value Calculation s?p= ((df/(df4 + df) * 5?1) + ((df2/(df2 + df2) * 522) = ((19/38) * 19977.09) + ((19/38) * 21423.84) = 20700.47 SPM, = 52/ N1 = 20700.47/20 = 1035.02 SPM, = 52 / N2 = 20700.47/20 = 1035.02 t= (M- M2)/1+ sẽm2) = 34.05/72070.05 =

Therefore value of test statistic is 0.7484 where test statistic is formula for či- žz t= V 32 (netto) (23-)= 2 ( 12; -) ? n₂ 5? nit ₂-2 Now o find and Critical value when L = 0.05 duf - ni+n2 -2 = 20720-2 = 38

Now find value Correspounding to 38 dif and x = 0.05 in the t-table given below - Critical point when 2=0.05 is 2.0244 Noo Here. test statistic < t-table value Hence we accept Ho d) u fail to • fail to reject Ho ie. the claim that Uwe is equal to ly because the test statistic is less than the critical point. Thank you!! please like it...