19 **Study Overview**A study was conducted using a treatment group and a placebo group. The results are shown in the accompanying table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below using a 0.05 significance level for both parts.**Table Information**- Treatment Group (\( \mu_1 \)): - Sample Size (\( n \)): 28 - Sample Mean (\( \bar{x} \)): 2.39 - Sample Standard Deviation (\( s \)): 0.84- Placebo Group (\( \mu_2 \)): - Sample Size (\( n \)): 30 - Sample Mean (\( \bar{x} \)): 2.67 - Sample Standard Deviation (\( s \)): 0.51**Tasks and Questions****a. Hypothesis Testing**Test the claim that the two samples are from populations with the same mean.- **Hypothesis Options:** - A. \( H_0: \mu_1 \neq \mu_2 \); \( H_1: \mu_1 < \mu_2 \) - B. \( H_0: \mu_1 = \mu_2 \); \( H_1: \mu_1 \neq \mu_2 \) - C. \( H_0: \mu_1 < \mu_2 \); \( H_1: \mu_1 \geq \mu_2 \) - D. \( H_0: \mu_1 = \mu_2 \); \( H_1: \mu_1 > \mu_2 \)- **Results:** - The test statistic, \( t \), is \([ \text{Round to two decimal places as needed.} ]\). - The P-value is \([ \text{Round to three decimal places as needed.} ]\).- **Conclusion Options:** - A. Fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean. - B. Fail to reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that the two samples are from populations with

From the information, given thatThe claim is to test whether the two samples are from populations with the same mean or not.(a)That is, there is no evidence to conclude that there are differences between the two population means.That is, there is evidence to conclude that there are differences between the two population means.Correct option: Option BThe value of the test statistic is obtained below as follows : The required value is, The value of the test statistic is –1.52.The degrees of freedom formula for unequal variance is:The required value is,Thus, the degrees of freedom is 43.Use Excel to obtain the probability value.Follow the instructions to obtain the P-value: 1. Open Excel2. Go to Formula bar.3. In formula bar enter the function as“=TDIST”4. Enter the test statistics as ABS(–1.52).5. Enter the degrees of freedom as 436. Enter the tails as 2.7. Click enterFrom the Excel sheet, the P-value is 0.1358Thus, the P-value is 0.136. conclusion:Use the level of significance as 0.05The P-value is greater than the level of significance.By, the rejection rule, fail to reject the null hypothesis.That is, there is no evidence to conclude that there are differences between the two population means.Correct option: Option A(b)The 95% confidence interval for the difference between two population means is obtained below as follows: The formula for two sample t-test confidence interval is, Use EXCEL Procedure for finding the critical value of t.Follow the instructions to obtain the critical value of t: 1. Open Excel2. Go to Formula bar.3. In formula bar enter the function as“=TINV”4. Enter the probability as 0.055. Enter the degrees of freedom as 43.6. Click enter7. EXCEL output: From the EXCEL , the critical value of t at 0.05 level of significance with the 43 degrees of freedom is 2.016692 The required confidence interval is,The 95% confidence interval for the difference between two population means is –0.651<µ1- µ2<0.091. (a)Correct option: Option BThe value of the test statistic is –1.52.Thus, the P-value is 0.136.That is, there is no evidence to conclude that there are differences between the two population means.Correct option: Option A(b)The 95% confidence interval for the difference between two population means is –0.651<µ1- µ2<0.091.