**2. Recovering from Surgery:**A new postsurgical treatment was compared with a standard treatment. Seven subjects received the new treatment, while seven others (the controls) received the standard treatment. The recovery times, in days, are given below. Can you conclude that the mean recovery time for those receiving the new treatment is less than the mean for those receiving the standard treatment? Use the α = 0.05 level.| | Treatment | | | | | ||--------------|------------|-----------|-----------|----------|---------|---------|| New Treatment| 12 | 13 | 15 | 19 | 20 | 21 | 24 || Control | 18 | 23 | 24 | 30 | 32 | 35 | 39 |(a) **List the null and alternate hypotheses.**\[H_0 :\]\[H_1 :\](b) **List the distribution, the test statistic, the calculator program, and all input and output variables.**(c) **List the \( P \)-value.**(d) **Make the comparison.**We reject \[ H_0 \] / We do not reject \[ H_0 \] (circle one)(e) **State the final conclusion in the wording of the problem.**

From the information, given thatLet denotes the population mean recovery time receiving the new Treatment.Let denotes the population mean recovery time receiving the standard Treatment.The claim is to test whether the mean recovery time for those receiving the new treatment is less than the mean for those receiving the standard treatment.“Since you have posted a question with multiple sub-parts, we will solve first three sub-parts for you. To get remaining sub-part solved please repost the complete question andmention the sub-parts to be solved.”(a)Null hypothesis: That is, there is no evidence to conclude that mean recovery time for those receiving the new treatment is less than the mean for those receiving the standard treatment.Alternative hypothesis: That is, there is evidence to conclude that mean recovery time for those receiving the new treatment is less than the mean for those receiving the standard treatment.(b)The value of the test statistic is obtained as follows: since the difference between the two population means are being tested and moreover the population standard deviations are unknown, the type of distribution used is "t".Use TI-83 plus calculator to obtain the value of test statisticTI-83 plus calculator Procedure: · Press STAT>EDIT· Enter the variable Treatment in L1 and controls in L2. · Press STAT>TESTS· Choose 2-SampTest.· Choose Data.· Choose the alternative as “<”.· Select Pooled as Yes· Click Calculate.TI-83 plus calculator output:From the TI-83 plus calculator, the value of test statistic is –3.372Thus, the value of the test statistic is -3.372.. (c)From the TI-83 plus calculator, the P-value is 0.0028Thus, the P-value is 0.0028..(a)(b)Thus, the value of the test statistic is -3.372..(c)Thus, the P-value is 0.0028..