Answer to Question #345515 in Statistics and Probability for ken

Question

A. Solve the problems by following the steps and procedures in hypothesis testing. (20 points)

1. A school teacher suspects the claim that the mean number of students that use library 

materials in a certain school is at most 450. To check the claim, the professor checks a 

random sample of 100 library records and obtain that the mean number of students 

using library materials is 458 with a standard deviation of 9. What would the teacher's 

conclusion use 0.05 level of significance? (10 points)

2. prospective MBA student was made to estimate the difference in the salaries of 

professors in private and state universities. An independent study of simple random 

samples of the most recent MBA graduates of both universities revealed the following 

statistics: Conduct a test using 0.01 level of significance. (10 points)

Salary x¯ s n

Private 52,285/mo. 2,400 49

State 50,188/mo. 2,100 49

Answer 1

1.

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le 450$

$H_1:\mu>450$

This corresponds to a right-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 $\alpha = 0.05,$ $df=n-1=99$ and the critical value for a right-tailed test is $t_c = 1.660391.$

The rejection region for this right-tailed test is $R = \{t:t> 1.660391\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{458-450}{9/\sqrt{100}}=8.8889$$

Since it is observed that $t=8.8889> 1.660391=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed, $df=99$ degrees of freedom, $t=8.8889$ is $p=0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is greater than 450, at the $\alpha = 0.05$ significance level.

2.

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 $\sigma_X^2+\sigma_Y^2.$

$$\mu_{X-Y}=52285-50188=2097$$ $$s_{X-Y}=\sqrt{(2400)^2+(2100)^2}=3189$$

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 $\alpha=0.01,$

$df=n-1=49-1=48$ degrees of freedom, and the critical value for a two-tailed test is $t_c=2.682204.$

The rejection region for this two-tailed test is $R=\{t:|t|>2.682204\}$

The t-statistic is computed as follows:

$$t=\dfrac{\mu_{X-Y}-\mu}{s_{X-Y}/\sqrt{n}}=\dfrac{2097-0}{3189/\sqrt{49}}\approx4.603$$

Since it is observed that $|t|=4.603> 2.682204=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed  $df=48, t=4.603$ is $p=0.000031,$ and since $p=0.000031<0.01=\alpha,$ 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 $\alpha=0.01$ significance level.