For the following data:

	Size	Mean	Standard deviation
Sample 1	4	40	8
Sample 2	5	50	10

The sample statistic t follows t-distribution with 'm' degrees of freedom, then m is equal to

Given information:

Sample 1: $n_1 = 4$, $\bar{x}_1 = 40$, $s_1 = 8$

Sample 2: $n_2 = 5$, $\bar{x}_2 = 50$, $s_2 = 10$

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When comparing two samples with unequal variances, Welch's formula for degrees of freedom is used:

$m = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{(s_1^2/n_1)^2}{n_1-1} + \frac{(s_2^2/n_2)^2}{n_2-1}}$

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Calculate the variance for each sample:

$s_1^2 = 8^2 = 64$

$s_2^2 = 10^2 = 100$

Calculate variance over sample size:

$\frac{s_1^2}{n_1} = \frac{64}{4} = 16$

$\frac{s_2^2}{n_2} = \frac{100}{5} = 20$

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Calculate the numerator:

$\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2 = (16 + 20)^2$

$= (36)^2$

$= 1296$

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Calculate the denominator:

$\frac{(s_1^2/n_1)^2}{n_1-1} = \frac{(16)^2}{4-1}$

$= \frac{256}{3}$

$= 85.33$

$\frac{(s_2^2/n_2)^2}{n_2-1} = \frac{(20)^2}{5-1}$

$= \frac{400}{4}$

$= 100$

Denominator $= 85.33 + 100 = 185.33$

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Calculate degrees of freedom:

$m = \frac{1296}{185.33}$

$= 6.99$

$\approx 7$

Therefore, $m = 7$ degrees of freedom.