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Sagot :
The sampling distribution is t= 88 degrees of freedom.
To find the sampling distribution of the sample mean difference we need to do a two-sample t-test: equal variances.
When we do a two-sample t-test, we have to take into account the following assumptions:
1) the sample variances are normal.
2) the sample variances of both populations are equal.
Now, let the sample size be [tex]n_x[/tex] and [tex]n_y[/tex] respectively and the means be [tex]x^-[/tex] and [tex]y^-[/tex] respectively.
Now, we are assuming that the distribution of x and y are
[tex]x\sim N([/tex]mewx, [tex]\sigma x = \sigma[/tex])
[tex]y\sim N[/tex] (mewy, [tex]\sigma y = \sigma[/tex])
Accordingly, the system of hypothesis is defined by:
Null hypothesis: mewx=mewy
alternatively, it can also be mewx≠ mewy
Then, the random sample t is distributed as [tex]t\sim t_n_x+ny-2[/tex] with the degrees of freedom given as df= [tex]n_x+n_y-2[/tex] =40+50-2= 88
So, in this case answer will be 88 degrees of freedom.
To know more about sampling distribution: https://brainly.com/question/14090787
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