Question

4. [7 marks] Consider two independent populations. We took two SRS of sizes 10 from each population and x = 525.751, sı = 107.121, X2 = 373.269 and s2 = 67.498 Is there enough evidence to reject the null hypothesis in the following test? (a = 1%) H:H1 - H2 = 0 Ha: M1 - M2 > 0

Answer 1

Q4) The standard error here is computed as:

$SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{107.121^2 + 67.498^2}{10}} =40.0386$

The test statistic here is computed here as:

$t^* = \frac{\bar X_1 - \bar X_2}{SE} = \frac{525.751 - 373.269}{40.0386} = 3.8084$

For n₁ + n₂ - 2 = 18 degrees of freedom, as this is a one tailed test, the p-value here is computed from the t distribution tables as:
p = P( t₁₈ > 3.8084) = 0.0006

As the p-value here is 0.0006 < 0.01 which is the level of significance, therefore the test is significant here and we can reject the null hypothesis here. Therefore we have sufficient evidence here that the difference in means is more than 0 here.