Question

Aminah wish to perform the hypothesis testing H0: μ =1 versus H1: μ <1 versus with α=0.10. . The sample size 25 was obtained independently from a population with standard deviation 10. State the distribution of the sample mean given that null hypothesis is true and find the critical value, then calculate the values of sample mean if she reject the null hypothesis. Finally, compute the p-value, if the sample mean is -2.

Answer 1

Given that null hypothesis is true. $H_0: \mu=1$

Let $\mu=1$ be the population mean and $\sigma=10$ be the population standard deviation.

The sampling distribution of the mean will have:

the mean $\mu_{\overline{X}}$ :

$$\mu_{\overline{X}}=\mu=1$$

the variance  $\sigma_{\overline{X}}^2$:

$$\sigma_{\overline{X}}^2={\sigma^2 \over n}={10^2 \over 25}=4$$

The standard error of the mean is

$$\sigma_{\overline{X}}={\sigma \over \sqrt{n}}={10 \over \sqrt{25}}=2$$

Since the sample size $n=25<30$, we use t-score.

A test statistic is given by  

$$t-statistic={ \overline{X}-\mu\over {s/\sqrt{n}}}$$

where $\overline{X}$ is the sample mean, $\mu$ is the value of the population mean in the null hypothesis, $s$ is the sample standard deviation, and $n$ is the sample size.

$H_0: \mu=1, H_1: \mu<1$.

For this left-tailed test (lower-tail test) and $\alpha=0.10, n=25$, the critical value will be

$$t_c=-1.3178$$

The critical region will be to the left of -1.3178:

$${ \overline{X}-\mu\over {s/\sqrt{n}}}<-1.3178$$ $$\overline{X}<\mu-1.3178{10 \over \sqrt{25}}$$ $$\overline{X}<-1.6356$$

We reject the null hypothesis if  $\overline{X}<-1.6356.$

If the sample mean is -2, then

$$t-statistic={ -2-1\over {10/\sqrt{25}}}=-1.5,$$

compute the p-value for the left-tailed test with 25−1 degrees of freedom:

$$p-value=P(X\leq-2|\mu=1)=$$

$$=P(T\leq-1.5, df=25-1)=0.0733$$

$p-value=0.0733$ or $7.33 \%$ .