Answer to Question #344756 in Statistics and Probability for Gab

Question

Table Completion. Supply the missing part of the table using the problem stated below.

“A random sample of 200 business managers were administered a develop Managerial 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 𝛼 = 0.05

utilizing the 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:

Using the P-value approach:

P-probability of an event

TS-test statistic

ts-observed value of the test statistic calculated from your sample

cdf()-cumulative distribution function of the distribution of the test statistic (TS) under the null hypothesis

The p-value is equal to two times the p-value for the upper-tailed p-value since the value of the test statistic from the sample is positive.

$$p= 2 P(TS\ge 7.826 | H_0 \ is\ true) = 2(1 - cdf(7 826))$$

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.