Assume that the paired data came from a population that is normally distributed. using a 0.05 significance level and dequalsxminusy, find d overbar, s subscript d, the t test statistic, and the critical values to test the claim that mu subscript dequals0. statcrunch

"Assume that the paired data came from a population that is normally distributed. Using a 0.05 significance level and d = (x - y), find , , the t-test statistic, and the critical values to test the claim that "

You did not attach the data, therefore I can give you the general explanation on how to find the values required and an example of a random paired data.

For the example, please refer to the attached picture.

A) Find 
You are asked to find the mean difference between the two variables, which is given by the formula:


These are the steps to follow:
1) compute for each pair the difference d = (x - y)
2) sum all the differences
3) divide the sum by the number of pairs (n)

In our example: 


B) Find 
You are asked to find the standard deviation, which is given by the formula:


These are the steps to follow:
1) Subtract the mean difference from each pair's difference 
2) square the differences found
3) sum the squares
4) divide by the degree of freedom DF = n - 1

In our example:

= √14.5
= 3.81

C) Find the t-test statistic.
You are asked to calculate the t-value for your statistics, which is given by the formula:


where SE = standard error is given by the formula:


These are the steps to follow:
1) calculate the standard error (divide the standard deviation by the number of pairs)
2) calculate the mean value of x (sum all the values of x and then divide by the number of pairs)
3) calculate the mean value of y (sum all the values of y and then divide by the number of pairs)
4) subtract the mean y value from the mean x value
5) from this difference, subtract  
6) divide by the standard error

In our example:
SE = 3.81 / √8
      = 1.346

The problem gives us , therefore:
t = [(9.75 - 9) - 0] / 1.346
  = 0.56

D) Find 
You are asked to find what is the t-value for a 0.05 significance level.

In order to do so, you need to look at a t-table distribution for DF = 7 and A = 0.05 (see second picture attached).

We find 

Since our t-value is less than  we can reject our null hypothesis!!

Answers4.) -1/205.) -1/406.) -55/127.) -3/4