Question

A sample size 40 from a non-normal population yielded the sample mean X= 71 and S² =200. Test H₀: µ=72 against H1: µ≠72. Use α= 1

Answer 1

The central limit theorem states that if you take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large (usually $n\geq30$).

The following null and alternative hypotheses need to be tested:

$H_0:\mu=72$

$H_1:\mu\not=72$

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 $\alpha=0.01,$

$df=n-1=40-1=39$ ​degrees of freedom, and the critical value for a two-tailed test is $t_c=2.707913.$

The rejection region for this two-tailed test is $R=\{t:|t|>2.707913\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{71-72}{\sqrt{200}/\sqrt{40}}\approx-0.447214$$

Since it is observed that $|t|=0.447214<2.707913=t_c,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 72, at the $\alpha=0.01$ significance level.

Using the P-value approach: The p-value for two-tailed, $df=39, \alpha=0.01,$ $t=-0.447214$ is $p=0.657195,$ and since $p=0.657195>0.01=\alpha,$ 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 72, at the $\alpha=0.01$ significance level.