Question

The following sample of nine measurements was randomly selected from a normally distributed population:11,10,8,7,14,9,10,12

Test for significant difference between the sample mean and the population mean 10.Use a=0.05.

Answer 1

$$\mu=\dfrac{11+10+8+7+14+9+10+12}{8}=10.125$$

Variance

$$s^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n-1}=\dfrac{34.875}{7}$$ $$s=\sqrt{s^2}=\sqrt{\dfrac{34.875}{7}}\approx2.23207$$

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

$H_0:\mu=10$

$H_a:\mu\not=10$

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.

2. Based on the information provided, the significance level is $\alpha = 0.05,$ $df=n-1=7$ degrees of freedom, and the critical value for a two-tailed test is $t_c = 2.364619.$

3. The rejection region for this two-tailed test is $R = \{t: |t| > 2.364619\}.$

4. The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{10.125-10}{2.23207/\sqrt{8}}\approx0.158397$$

5. Since it is observed that $|t| =0.158397<2.364619=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed, $df=7$ degrees of freedom, $t=0.158397$ is $p= 0.878617,$ and since $p= 0.878617>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

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

is different than 10, at the $\alpha = 0.05$ significance level.