Answer to Question #347190 in Statistics and Probability for Carol

Question

Activity 1: Draw Me

Directions: Given the following information, construct the rejection region. Show the solution in a step-by-step procedure.

1. H0 : = 84

Ha : 84

m= 87, s= 10, n = 35, alpha= 0.05

2. H0 : = 45

Ha : < 45

m= 40, s = 12, n = 32, alpha= 0.01

Answer 1

1. The following null and alternative hypotheses need to be tested:

$H_0:\mu=84$

$H_1:\mu\not=84$

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 $\alpha = 0.05,$ $df=n-1=34$ and the critical value for a two-tailed test is $t_c = 2.032244.$

The rejection region for this two-tailed test is $R = \{t:|t|> 2.032244\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{87-84}{10/\sqrt{35}}=1.7748$$

Since it is observed that $|t|=1.7748< 2.032244=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed, $df=34$ degrees of freedom, $t=-4$ is $p=0.084885,$ and since $p=0.084885>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is different than 84, at the $\alpha = 0.05$ significance level.

2. The following null and alternative hypotheses need to be tested:

$H_0:\mu=45$

$H_1:\mu<45$

This corresponds to a left-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 $\alpha = 0.01,$ $df=n-1=31$ and the critical value for a left-tailed test is $t_c = -2.452824.$

The rejection region for this left-tailed test is $R = \{t:t<-2.452824\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{40-45}{12/\sqrt{32}}=-2.3570$$

Since it is observed that $|t|=-2.3570> -2.452824=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for left-tailed, $df=31$ degrees of freedom, $t=-2.3570$ is $p=0.012459,$ and since $p=0.012459>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is less than 45, at the $\alpha = 0.01$ significance level.