\(H_0:\) There is no significant
difference between the two methods.
\(H_1:\) There significant difference
between the two methods.
\(a=0.05\)
Given:
\(\bar{X}_1 = 85.9\)
\(\bar{X}_2 = 83\)
\(\sigma^2_1 = (5)^2 = 25\)
\(\sigma^2_2 = (6)^2 = 36\)
\(n_1 = 38\)
\(n_1 = 39\)
\[Z = \frac{\bar{X}_1-\bar{X}_2}{\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}}= \frac{85.9-83}{\sqrt{\frac{25}{38}+\frac{36}{39}}} = 2.306\]
Z-score = 2.306 equivalent to Z table probablity of p= 0.0102
Reject \(H_o\) (null hypothesis)
At 5% level of significance, the data is sufficient to conclude that the two methods are significantly different since p-value = 0.0102 is lesser than alpha = 0.05.