As Director of Admissions at Garden Valley University, you believe that the average age of full-timestudents (Group 1) is less than the average age of part-time students (Group 2). You sample 45 full-timestudents, and the sample mean age is 22.3 with a standard deviation of 4.8. You sample 50 part-timestudents, and the sample mean age is 24.3 with a standard deviation of 5.6. Test the claim using a 2% levelof significance. Assume the population standard deviations are unequal and that both populations arenormally distributed. Where necessary, round results accurate to 4 decimal places.a. What are the correct hypotheses?Null:Ο Ho: μ = 0O Ho: ₁ = ₂Ho: M₁ M₂Ο Ho: μ = 22.3Alternative:Ο HA: Mı # μεΟ HA: μ. > μεHA: ₁2HA: < 0Ο HA: μ + 22.3b. What is the correct distribution to use for the test? Select an answerc. What is the test statistic?d. What is the p-value?p-value =e. The correct decision is to Select an answerf. INTERPRET the conclusion in the context of the problem; e.g., explain in words what the result of thetest means.There is evidence to support the claim that the average age of full-time students is less than theaverage age of part-time students. The data are statistically significant.O The evidence disproves the claim that the average age of full-time students is less than the averageage of part-time students. The data are statistically significant.O There is insufficient evidence to support the claim that the average age of full-time students is lessthan the average age of part-time students. The data are not statistically significant.O The test is inconclusive.

Followings are the Explanation of the question Use the Two sample independent t test when standard deviations are unequal ANSWER :- Given that Sample Mean X1 22.3 Standard Deviation (S_1) 4.8 Sample Size(n_1) 45 Sample Mean X2 24.3 Standard Deviation (S_2) 5.6 Sample Size (n_2) 50 Significance Level (α) 0.02 Null and Alternative Hypotheses :- The following null and alternative hypothesis Ho:μ1≥μ2Ha:μ1<μ2 Where the Population Mean for the full time and part time Respectively Which is the Left 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 =0.02, and Degrees of Freedom df= (n1 + n2 – 2) = (45+50-2) = 93 Use the Excel and use the below formula to calculate the critical value for the Left tailed test α = 0.02 and DF= 93 Formula :- =T.INV(α,DF) tC=-2.083 (Note:- Put the location value of α and DF from the excel) Test - Statistics:- Since it is assumed that the population variances are unequal, the t-statistic is computed as follows: t=((X1)¯-(X2)¯)/√((s12)/n1+(s22)/n2)t=(22.3-24.3)/√(4.82/45+5.62/50)t=-1.873 Calculate the P-value :- P-value :- Use Excel and Find the P-value for the left tailed test Formula :- =TDIST(X, DF, tails) = T.DIST( -1.873,93,1) ------ For a Example P-value = 0.032 (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=-1.873>tC=-2.083 It is then concluded that the null hypothesis not rejected. Using the P-value P-value = 0.0320 > α = 0.02 It is concluded that the null hypothesis not rejected. CONCLUSION:-There is insufficient evidence to support the claim that the average age of full-time studentsarts is less than the average age of part-time students. The data are not statistically significant. It is concluded that the null hypothesis Ho not rejected. Therefore, there is not enough evidence to claim that the population mean μ1 is less than μ2 at the 0.02 significance level.