Question

two independent samples of sizes 40 and 50 are randomly selected from two populations to test the difference between the population means . the sampling distribution of the sample mean difference is:

Answer 1

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 $n_x$ and $n_y$ respectively and the means be $x^-$ and $y^-$ respectively.

Now, we are assuming that the distribution of x and y are

$x\sim N($mewx, $\sigma x = \sigma$)

$y\sim N$ (mewy, $\sigma y = \sigma$)

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 $t\sim t_n_x+ny-2$ with the degrees of freedom given as df= $n_x+n_y-2$ =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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