Identify the class width, class midpoints, and class boundaries for the given frequency distribution.Daily Low Tempurature Frequ
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"<span>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 </span><span>
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:
</span><span>
</span>
B) Find <span>
</span><span>You are asked to find the standard deviation, which is given by the formula:
</span>
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 <span>, therefore:
t = [(9.75 - 9) - 0] / 1.346</span>
= 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 <span>
</span>
Since our t-value is less than <span>
</span> we can reject our null hypothesis!!
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 </span><span>
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:
</span><span>
B) Find <span>
</span><span>You are asked to find the standard deviation, which is given by the formula:
</span>
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 <span>
t = [(9.75 - 9) - 0] / 1.346</span>
= 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 <span>
Since our t-value is less than <span>