Answer to Question #339018 in Statistics and Probability for buddy

Question

Assume the service life is distributed normally, a random sample of several electronic stuff with a service life (years): 6, 7, 7, 7, 6, 8, 8, 7, 6.

Does this indicate that the average lifespan of a shaker from Onix is not the same as 7

year? Do hypothesis testing with a significance level:

a. 1%

b. 5%

c. 10%

Answer 1

$\bar{x}=\dfrac{6+7+7+7+6+8+8+7+6}{9}$

$=\dfrac{62}{9}\approx6.8889$

$s^2=\dfrac{\sum _i(x_i-\bar{x})^2}{n-1}\approx0.611111$

$s=\sqrt{0.611111}\approx0.7817$

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

$H_0:\mu=7$

$H_a:\mu\not=7$

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 $\alpha = 0.01,$ $df=n-1=8$ degrees of freedom, and the critical value for a two-tailed test is $t_c = 3.355361.$

The rejection region for this two-tailed test is $R = \{t: |t| >3.355361\}.$

The t-statistic is computed as follows:

$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{6.8889-7}{0.7817/\sqrt{9}}\approx-0.426378$

Since it is observed that $|t|= 0.426378 < 3.355361=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed test, $df=8$ degrees of freedom, $t=-0.426378$ is $p = 0.681074,$ and since $p =0.681074 \ge 0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

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

is different than 7, at the $\alpha = 0.01$ significance level.

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

$H_0:\mu=7$

$H_a:\mu\not=7$

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 $\alpha = 0.05,$ $df=n-1=8$ degrees of freedom, and the critical value for a two-tailed test is $t_c = 2.306002.$

The rejection region for this two-tailed test is $R = \{t: |t| >2.306002\}.$

The t-statistic is computed as follows:

$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{6.8889-7}{0.7817/\sqrt{9}}\approx-0.426378$

Since it is observed that $|t|= 0.426378 < 2.306002=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed test, $df=8$ degrees of freedom, $t=-0.426378$ is $p = 0.681074,$ and since $p =0.681074 \ge 0.05=\alpha,$ 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 7, at the $\alpha = 0.05$ significance level.

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

$H_0:\mu=7$

$H_a:\mu\not=7$

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 $\alpha = 0.10,$ $df=n-1=8$ degrees of freedom, and the critical value for a two-tailed test is $t_c =1.859547.$

The rejection region for this two-tailed test is $R = \{t: |t| >1.859547\}.$

The t-statistic is computed as follows:

$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{6.8889-7}{0.7817/\sqrt{9}}\approx-0.426378$

Since it is observed that $|t|= 0.426378 < 1.859547=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed test, $df=8$ degrees of freedom, $t=-0.426378$ is $p = 0.681074,$ and since $p =0.681074 \ge 0.10=\alpha,$ 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 7, at the $\alpha = 0.10$ significance level.