A teacher surveyed her class to find out how many texts the students send in a week. She created this box plot to show the data

Assume that the paired data came from a population that is normally distributed. using a 0.05 significance level and dequalsxmin

Artemon [7]

"Assume that the paired data came from a population that is normally distributed. Using a 0.05 significance level and d = (x - y), find $\bar{d}$ , $s_{d}$ , the t-test statistic, and the critical values to test the claim that $\mu_{d} = 0$ "

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 $\bar{d}$
You are asked to find the mean difference between the two variables, which is given by the formula:
$\bar{d} = \frac{\sum (x - y)}{n}$

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:
 $\bar{d} = \frac{6}{8} = 0.75$ 

B) Find $s_{d}$
You are asked to find the standard deviation, which is given by the formula:
 $s_{d} = \sqrt{ \frac{\sum(d - \bar{d}) }{n-1} }$

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:
$s_{d} = \sqrt{ \frac{101.5}{8-1} }$
= √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:
$t = \frac{(\bar{x} - \bar{y}) - \mu_{d} }{SE}$

where SE = standard error is given by the formula:
$SE = \frac{ s_{d} }{ \sqrt{n} }$

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 $\mu_{d}$
6) divide by the standard error

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

The problem gives us $\mu_{d} = 0$ , therefore:
t = [(9.75 - 9) - 0] / 1.346
= 0.56

D) Find $t_{\alpha / 2}$
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 $t_{\alpha / 2} = 1.895$ 

Since our t-value is less than $t_{\alpha / 2}$ we can reject our null hypothesis!!

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He will water his cactus every 5th day.Just like before let’s count but this time by 5s.

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Your Answer is the 14th

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