Question

Ten individuals are chosen at random from a normal population and their heights are found to be 63, 63, 66, 67, 68, 69, 70, 70, 71, 71 inches. Test if the sample belongs to the population whose mean height is 66 inches. [Given t at 9 degrees of freedom at 5% level of significance is 2.62]

Answer 1

$$63, 63, 66, 67, 68, 69, 70, 70, 71, 71$$

$n=10$

$$mean=\bar{x}=\dfrac{1}{n}\sum _i x_i=\dfrac{1}{10}(63+63+66+67$$ $$+68+69+70+70+71+71)=\dfrac{678}{10}=67.8$$

$$Var(X)=s^2=\dfrac{1}{n-1}\sum (x_i-\bar{x})^2$$

$$=\dfrac{1}{10-1}((63-67.8)^2+(63-67.8)^2$$

$$+(66-67.8)^2+(67-67.8)^2+(68-67.8)^2$$

$$+(69-67.8)^2+(70-67.8)^2+(70-67.8)^2$$

$$+(71-67.8)^2+(71-67.8)^2=\dfrac{81.6}{9}$$

$$s=\sqrt{s^2}\approx\sqrt{\dfrac{81.6}{9}}\approx3.01109$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=66$

$H_1:\mu\not=66$

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 $\alpha = 0.05,$

$df=n-1=8,$ and the critical value for a two-tailed test is $t_c= 2.262156.$

The rejection region for this two-tailed test is $R = \{t: |t| > 2.262156\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{67.8-66}{3.01109/\sqrt{10}}\approx1.89038$$

Since it is observed that $|t| = 1.89038 \le t_c = 2.262156,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed $df=9$ degrees of freedom, $t=1.89038$ is $p = 0.091281,$ and since $p=0.091281\leq0.05=\alpha,$ 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 $66,$ at the $\alpha = 0.05$ significance level.

Therefore, there is enough evidence to claim that the sample belongs to the population whose mean height is 66 inches, at the $\alpha = 0.05$ significance level.