Answer to Question #350842 in Statistics and Probability for Bia

Question

A one sample +-test is conducted

on Ho: M = 81.6. The sample has a

mean of 84.1, s = 3.1, n = 25, and a =

01. What conclusion can be drawn?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=81.6$

$H_1:\mu\not=81.6$

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 $\alpha = 0.01,$ $df=n-1=24$ and the critical value for a two-tailed test is $t_c =2.79694.$

The rejection region for this two-tailed test is $R = \{t:|t|>2.79694\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{84.1-81.6}{3.1/\sqrt{25}}=4.0323$$

Since it is observed that $|t|=4.0323>2.79694=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=24$ degrees of freedom, $t=4.0323$ is $p= 0.000486,$ and since $p=0.000486<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 81.6, at the $\alpha = 0.01$ significance level.