I need help with all parts of this question 19 An experiment was conducted to compare the densities (in ounces per cubic inch) of cakes prepared from two different cake mixes. Six cake pans were filled with batter A, and six were filled with batter B. Expecting a variation in oven temperature, the experimenter placed a pan filled with batter A and another with batter B side by side at six different locations in the oven. The six paired observations of densities are as follows:| Location | 1 | 2 | 3 | 4 | 5 | 6 ||----------|-------|-------|-------|-------|-------|-------|| Batter A | 0.134 | 0.103 | 0.097 | 0.140 | 0.132 | 0.145 || Batter B | 0.128 | 0.119 | 0.113 | 0.151 | 0.134 | 0.162 |**(a) Do the data present sufficient evidence to indicate a difference between the average densities of cakes prepared using the two types of batter?** (Use α = 0.01. Use μ_Batter A − μ_Batter B = μ_d)**State the null and alternative hypotheses.**- \( H_0 : \mu_d = 0\) versus \( H_a : \mu_d > 0 \)- \( H_0 : \mu_d = 0\) versus \( H_a : \mu_d < 0 \)- \( H_0 : \mu_d = 0\) versus \( H_a : \mu_d \neq 0 \)- \( H_0 : \mu_d < 0\) versus \( H_a : \mu_d > 0 \)- \( H_0 : \mu_d = 0\) versus \( H_a : \mu_d \neq 0 \)**State the test statistic.** (Round your answer to three decimal places.)t = ____**State the rejection region.** (If the test is one-tailed, enter NONE for the unused region. Round your answers to three decimal places.)t > ____t < ____**State the conclusion.**- H0 is rejected. There is sufficient evidence to conclude that there is a difference between the average densities of cakes prepared using the two types of batter.- H0 is rejected. There is insufficient evidence to

Followings are the Explanation of the question Use the Excel and find the Mean and Stanard deviation for the difference Use the Paired t for the given data Calculated Mean Difference d = -0.0093 Standard deviation (S_d)= 0.00937 Sample size (n) = 6 α =1% = 0.01 HYPOTHESES: -NULL AND ALTERNATIVE The following null and alternative hypothesis Ho: μd=0Ha:∶μd≠0 This corresponds to a Two-tailed test, for which a t-test for one mean, Where, is the Population mean difference between [d= A-B ] Calculate the Critical value using t table with Corresponding Alpha and Degrees of Freedom (DF) Critical Value:- The significance level is α=0.05, and Degrees of Freedom (DF)= n-1 = 6--1=5 and the critical value for a Two-tailed test is Use the Excel and use the below formula to calculate the critical value for the two tailed test α = 0.01 and DF= 5 Formula :- =T.INV(α*2,DF) tC=±4.0321 t > -4.0321 t < 4.0321 (Note:- Put the location value of α and DF from the excel) TEST STATISTICS: - The t-statistic is computed as follows: t=(d¯-〖μ〗d)/(sd⁄√n)t=(-0.0093-0)/(0.0093⁄√6)t=-2.4389 Calculate the P-value :- P-value :- Use Excel and Find the P-value for the two tailed test Formula :- =T.DIST(X, DF, tails) = T.DIST(2.4389, 5, 2) ------ For a Example P-value = 0.0587 (Note:- Put the location value of test-stat and DF from the excel) (Note:- 2- Two tailed and for two tailed test-statistics values always Positive) DECISION ABOUT THE NULL HYPOTHESIS :- Using critical value :- Since it is observed that, |t|=2.4389<tC=4.0321 it is then concluded that the null hypothesis is not rejected. Using the P-value P-value = 0.05873 > α = 0.01 It is concluded that the null hypothesis not rejected. CONCLUSION :- It is concluded that the null hypothesis Ho is not rejected. Therefore, there is enough evidence to claim that the population mean difference is different than 0, at the 0.01 significance level.