Question

Two independent samples of 8 and 7 items respectively had the following values:
1.sample - 9 ,11 ,13 ,11 ,15 ,9 ,12 ,14
2.sample - 10 ,12 ,10 ,14 ,9 ,8 ,10
is the difference between the means of sample significant?​

Answer 1

Given:

Two independent samples of 8 and 7 items respectively had the following values:

1.sample - 9 ,11 ,13 ,11 ,15 ,9 ,12 ,14

2.sample - 10 ,12 ,10 ,14 ,9 ,8 ,10

To find:

Is the difference between the means of sample significant?​

Solution:

I have attached the data table as an image, as I was unable to fit that table in this tab.

Further, let's continue with the formulae.

Sample 1:

$n_1=8$

$\bar{x}= \dfrac {\sum x_i}{n_1}=\dfrac{98}{4}=11.75$

$s_1^2= \dfrac{\sum(x_i-\bar{x})}{n_1-1}=\dfrac{33.5}{8-1}=4.79$

$\therefore s_1=\sqrt{4.79}=2.188$

Sample 2:

$n_2=7$

$\bar{y}= \dfrac {\sum y_i}{n_2}=\dfrac{73}{7}=10.43$

$s_2^2= \dfrac{\sum(y_i-\bar{y})}{n_2-1}=\dfrac{23.71}{7-1}=3.95$

$\therefore s_2=\sqrt{3.95}=1.987$

The test hypothesis is given by,

$z=\dfrac{\bar{x_1}-\bar{x_2}}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$

substituting the obtained values in the above equation, we get,

$t=\dfrac{11.75-10.43}{\sqrt{\frac{4.79}{8}+\frac{3.95}{7}}}$

$|t|=1.215$

The critical value for t for a two-tailed test at 5% level of significance with  8 + 7 - 2 = 13 degrees of freedom is 1.771.

As 1.215 < 1.771

calculated value < tabulated value

∴ There is no significant difference between the means of samples.