8. A random sample of $ n = 25 $ individuals is selected from a population with $ \mu = 20 $, and a treatment is administered to each individual in the sample. After treatment, the sample mean is found to be $ M = 22.2 $ with $ SS = 384 $.a. How much difference is there between the mean for the treated sample and the mean for the original population? (*Note: In a hypothesis test, this value forms the numerator of the $ t $ statistic.*)b. If there is no treatment effect, how much difference is expected between the sample mean and its population mean? That is, find the standard error for $ M $. (*Note: In a hypothesis test, this value is the denominator of the $ t $ statistic.*)c. Based on the sample data, does the treatment have a significant effect? Use a two-tailed test with $ \alpha = .05 $.

The following defined test statistic t is used to compare the difference between population mean and sample mean and it is used when the population standard deviation(σ) is unknown The test statistic is t=M-μSSn where M is the sample mean μ is the population mean SSn is the Standard error.a. Finding the difference between population mean and sample mean: Given, population mean i.e., original population mean μ = 20 sample mean i.e., treated sample mean M = 22.2 Now, difference is (M - μ) = 22.2 - 20 = 2.2 Here, the value 2.2 is also numerator of the t statistic. Hence, there is a 2.2 difference between the mean for the treated sample and mean for the original population. b. Finding the standard error for M: Given, SS = 384 and n = 25 we have, standard error =SSn put the values, SS = 384 and n = 25 in the above standard error. We get, standard error =38425 = 3845 = 76.8 Hence, the standard error for M is 76.8 c. Is the treatment having a significant effect (or) not: Making the Hypotheses: Null Hypothesis H0 : μ = 20 i.e., the treatment doesn't having a significant effect. Alternative Hypothesis Ha : μ ≠ 20 i.e., the treatment having a significant effect. Test statistic: The test statistic is t=M-μSSn From parts a and b, we have (M - μ) = 2.2 and SE = 76.8 Now, test statistic is t=2.276.8 = 0.0286 Critical value: tα/2 is the critical value with (n-1) = 24 df at 0.05 level of significance is -2.0639 to +2.0639 [refer from two tailed t-table values] Decision Rule: Reject null hypothesis H0 , if test statistic lies outside the critical region of (-2.0639 to +2.0639). Do not reject null hypothesis H0 , if test statistic lies within the critical region of (-2.0639 to +2.0639). Conclusion: The test statistic (0.0286) lies within the critical region of (-2.0639 to +2.0639). So, results are not statistically significant at 0.05 level of significance i.e., we are accepting null hypothesis H0 at 0.05 level of significance. Hence, the treatment doesn't having a significant effect.