Use the given statistics to complete parts (a) and (b). Assume that thepopulations are normally distributed.(a) Test whether μ₁ > μ₂ at the α = 0.01 level of significance for the given sampledata.(b) Construct a 99% confidence interval about μ₁ −μ₂.(a) Identify the null and alternative hypotheses for this test.A. Ho: M₁ µ₂.There is not sufficient suidenan at the ~-004 level of significance to conclude that ..

Followings are the Explanation of the question Use the Two sample independent t test Given that, Sample Mean X1 50.4 Standard Deviation (S_1) 5.5 Sample Size(n_1) 22 Sample Mean X2 44.3 Standard Deviation (S_2) 10.5 Sample Size (n_2) 18 Significance Level (α) 0.01 HYPOTHESES:- NULL AND ALTERNATIVE The following null and alternative hypotheses need to be tested: H0∶μ1=μ2H1∶μ1>μ2 Where the Population Mean for the Population 1 and Population 2 Respectively Which is the Right tailed test of hypothesis Critical Value:- Calculate the Critical value using t table with Corresponding Alpha and Degrees of Freedom (DF) The significance level is α =0.01, and Degrees of Freedom df= (n1 + n2 – 2) = (22+18-2) = 38 Use the Excel and use the below formula to calculate the critical value for the right tailed test α = 0.01 and DF= 38 Formula :- =T.INV(α*,DF) tC=2.428 (Note:- Put the location value of α and DF from the excel)Test Statistics :- Assume the population variances are equal, t=(x¯1-x¯2)/(√(S2*[1/n1)+1/n2])Where,S2=(n1-1*s12+n2-1*s12)/(n1+n2-2)S2=((22-1)*〖5.5〗2+(18-1)*〖10.5〗2)/(22+18-2)S2=66.04Therefore,Teststatistic,tstat=(50.4-44.3)/(√(66.04*[1/22)+1/18])tstat=2.36Test-statistics=2.36 Calculate the P-value :- P-value :- Use Excel and Find the P-value for the two tailed test Formula :- =TDIST(X, DF, tails) = T.DIST( 2.36, 38,1) ------ For a Example P-value = 0.012 (Note:- Put the location value of test-stat and DF from the excel) (Note:- 2- Two tailed and for two tailed or Right tailed test-statistics values give always Positive) DECISION ABOUT THE NULL HYPOTHESIS :- Using critical value :- Since it is observed that t=2.36<tC=2.428 It is then concluded that the null hypothesis not rejected. Using the P-value P-value = 0.012 > α = 0.01 It is concluded that the null hypothesis not rejected. CONCLUSION :- It is concluded that the null hypothesis Ho not rejected. Therefore, there is not enough evidence to claim that the population mean μ1 is Greater than μ2 at the 0.01 significance level.b) Construct the 99% Confidence interval for the μ1 - μ2 Calculate Pooled Standard deviation :- Sp=√(((n1-1)s12+(n2-1)s22)/(n1+n2-2))Sp=√(((22-1)5.52+(18-1)10.52)/(22-18-2))Sp=66.04 Calculate Standard Error :- SE=Sp√(1/n1+ 1/n1) SE=66.04*√(1/22+ 1/18) SE= 2.58 Construct the 99% confidence interval for the difference between the population means C.I=[(X¯1-X¯2)-TC*Se,(X¯1-X¯2)+TC*Se)]C.I=[(50.4-44.3)-2.0244*2.58,(50.4-44.3)+2.0244*2.58]C.I=(0.871,11.328) Therefore, the 99 % confidence interval for the difference between the population means 0.871<μ1-μ2<11.328