Consider the following hypothesis test. The following data are from matched samples taken from two populations. Population Element 1 2 1 21 20 2 28 26 3 18 18 4 20 20 5 26 24 Compute the difference value for each element. If your answer is "0", enter 0.a. Compose the difference valuefor each element.b. Compute d.c.Compute the standard deviation sd.d.Conduct a hypothesis test using α=.05.What is yourconclusion?

Answer:a) d: 1  2  0  0  2b) [tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{5}{5}=1[/tex] c) [tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =1[/tex] d) [tex]p_v =P(t_{(4)}>2.236) =0.0445[/tex] So the p value is lower than the significance level provided, so then we can conclude that we reject the null hypothesis that the difference mean between population 1 and 2 is less than 0. And we can say that the mean difference is higher than 0 at 5% of significance.  Step-by-step explanation:A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.   Let put some notation  x=Population 1 , y = population 2 x: 21  28  18  20  26 y: 20 26  18  20  24  a. Compose the difference valuefor each elementLet d =x-y, so the values for the difference are:d: 1  2  0  0  2b. Compute d.We need the mean for the difference. If we use the following formula we got:[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{5}{5}=1[/tex] c.Compute the standard deviation sd.For the standard deviation we can use the following formula:[tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =1[/tex] d.Conduct a hypothesis test using α=.05.What is yourconclusion?If we assume that the system of hypothesis for this case are: Null hypothesis: [tex]\mu_x- \mu_y \leq 0[/tex] Alternative hypothesis: [tex]\mu_x -\mu_y > 0[/tex] We can calculate the statistic given by : [tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{1 -0}{\frac{1}{\sqrt{5}}}=2.236[/tex] The next step is calculate the degrees of freedom given by: [tex]df=n-1=5-1=4[/tex] Now we can calculate the p value, since we have a right tailed test the p value is given by: [tex]p_v =P(t_{(4)}>2.236) =0.0445[/tex] So the p value is lower than the significance level provided, so then we can conclude that we reject the null hypothesis that the difference mean between population 1 and 2 is less than 0. And we can say that the mean difference is higher than 0 at 5% of significance.