Question

a.leadership skills test a random sample of 200 school managers were administered a developed leadership skills test. the sample mean and the standard deviation were 78 and 4.2 respectively. in the standardization of the test, the mean was 73 and the standard deviation was 8. test for significant difference using a = 0.05 utilizing the p value method. steps (p-value method)

Answer 1

Parameter: Difference of two independent normal variables $\mu_{X-Y}$

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 $\sigma_X^2+\sigma_Y^2.$

$$\mu_{X-Y}=78-73=5$$ $$s_{X-Y}=\sqrt{(4.2)^2+(8)^2}=9.0355$$

Statistic: $t-$ statistic

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 $\alpha=0.05,$

$df=n-1=200-1=199$ degrees of freedom, and the critical value for a two-tailed test is $t_c=1.972.$

The rejection region for this two-tailed test is $R=\{t:|t|>1.972\}$

The t-statistic is computed as follows:

$$t=\dfrac{\mu_{X-Y}-\mu}{s_{X-Y}/\sqrt{n}}=\dfrac{5-0}{9.0355/\sqrt{200}}\approx7.826$$

Using the P-value approach:

The p-value for two-tailed  $df=199, t=7.826$ is $p=0,$ and since $p=0<0.05=\alpha,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 0, at the $\alpha=0.05$ significance level.