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The means of two random samples of size 9 and 7 are 196.42 and 198.82 respectively. The sums of the squares of the deviation from the means are $26.94$ and $18.73$ respectively.
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$n_1 = 9$ and $n_2 = 7$ [\lt 30 so it is a small sample] $$\overline x_i=196.42 \space and \space \overline x_2=198.82$$

$\sum (x_{1i}-\overline x_1)^2=26.940 \\ \sum (x_{2i}-\overline x_2)^2=18.73$

Step 1:

Null Hypothesis $(H_0): µ_1 = µ_2$ [i.e the samples are drawn from the same population]

Alternative Hypothesis $(H_a): µ_1 != µ_2$ [i.e The samples are not drawn from the same population] [Two tailed test]

Step 2:

LOS = 5% [two tailed test]

Degree of freedom = $n_1 + n_2 – 2 = 9 + 7 – 2 = 14$

Critical value $(t_x) = 2.145$

Step 3:

Since samples are small

$S_p=\sqrt{\dfrac {\sum (x_{1i}-\overline x_1)^2 + \sum (x_{2i}-\overline x_2)^2}{n_1+n_2-2}} \\ =\sqrt{\dfrac {(26.94)+(18.73)}{14}} \\ =1.8061 \\ S.E=S_p\sqrt{\dfrac 1{n_1}+\dfrac 1{n_2}} \\ =1.8061 \sqrt{\dfrac 19+\dfrac 17} \\ =0.9102$

Step 4: Test statistics

$t_{cal}=\dfrac {\overline x_1-\overline x_2}{S.E}\\ =\dfrac {196.42-198.82}{0.9102}\\ =-2.6368$

Step 5: Decision

Since $|t_{cal}| \gt t_x; H_0$ is rejected

Samples cannot be considered to have been drawn from same population