Question

ne 1: 80 81 82 78 83

Mine 2: 77 76 78 80 76 78

Use 0.05 level of significance to test whether the difference between the means of these two samples is significant.

Answer 1

X = Represent mine 1

Y = Represent mine 2

$Mean (\bar{X})=\frac{\Sigma X}{n_1}=\frac{404}{5}=80.8$

$Mean (\bar{Y})=\frac{\Sigma Y}{n_2}=\frac{465}{6}=77.5$

Sample standard deviation for each sample:

$S_X=\sqrt{\frac{(80-80.8)^2+(81-80.8)^2+(82-80.8)^2+(78-80.8)^2+(83-80.8)^2}{5-1}} =1.924$

$S_Y=\sqrt{\frac{(77-77.5)^2+(76-77.5)^2+(78-77.5)^2+(80-77.5)^2+(76-77.5)^2+(78-77.5)^2}{6-1}} =1.517$

Following steps are taken to test whether the difference between the means of these two samples is significant at  0.05 level of significance:

Null & Alternative Hypothesis:

$H_0:\mu_1=\mu_2$

$H_a:\mu_1\not =\mu_2$

Test Statistic:

Sample size is less than 30 for each of two samples. So, t-test for two-samples is used to test hypothesis.

$t=\frac{\bar{X}-\bar{Y}}{\sqrt{\frac{S_X}{n_1}+\frac{S_Y}{n_2}}}$

$t=\frac{80.8-77.5}{\sqrt{\frac{1.924}{5}+\frac{1.517}{6}}}=4.133$

Critical Value:

$df=n_1+n_2-2=5+6-2=9$

Critical value at 0.05 level of significance and degree of freedom is 9 would be:

$t_c=1.833$

Decision Making:

Hence, test statistic is greater than critical value, so the null hypothesis is rejected.

Conclusion:

Based on this, it can conclude that the difference between the means of these two samples is significant.