Question

A teacher conducted a study to know if blended leaming improves the students performances. A class of 25 students of Grade 11 was surveyed and found out that their mean score was 83 with a standard deviation of 3. A

study from other country revealed that = 80 with a standard deviation of 4. Test the hypothesis at 0.10 level of significance.

Answer 1

Difference of two independent normal variables

Let $X$ have a normal distribution with mean $\mu_X$ and variance $\sigma_X^2.$

Let $Y$ have a normal distribution with mean $\mu_Y$ and variance $\sigma_Y^2.$

If $X$ and $Y$are independent, then $X-Y$will follow a normal distribution with mean $\mu_X-\mu_Y$ and variance $\sigma_X^2+\sigma_Y^2.$

$$\mu_{X-Y}=83-80=3$$ $$s_{X-Y}=\sqrt{(3)^2+(4)^2}=5$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=0$

$H_1:\mu\not=0$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha=0.01,$

$df=n-1=25-1=24$ degrees of freedom, and the critical value for a two-tailed test is $t_c=2.79694.$

The rejection region for this two-tailed test is $R=\{t:|t|>2.79694\}$

The t-statistic is computed as follows:

$$t=\dfrac{\mu_{X-Y}-\mu}{s_{X-Y}/\sqrt{n}}=\dfrac{3-0}{5/\sqrt{25}}\approx0.12$$

Since it is observed that $|t|=0.12<2.79694=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed $\alpha=0.01,$ $df=24, t=0.12$ is $p=0.001071,$ and since $p=0.905482>0.01=\alpha,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 0, at the $\alpha=0.01$ significance level.

Therefore, there is not enough evidence to claim that blended leaming improves the students performances, at the $\alpha=0.01$ significance level.