Question

A random sample of ten measurements were obtained from a normally

distributed population wit mean μ = 8.5. The sample values are x̄= 6.2 and s=4

a. Test the null hypothesis that the mean of the population is 8.5 against the

alternative hypothesis, μ <8.5. Use α =0.05.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=8.5$

$H_1:\mu<8.5$

This corresponds to a left-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 $\alpha = 0.05,$ $df=n-1=10-1=9,$ and the critical value for a left-tailed test is $t_c = -1.833113.$

The rejection region for this left-tailed test is $R = \{t: t < -1.833113\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{6.2-8.5}{4/\sqrt{10}}\approx-1.8183$$

Since it is observed that $t = -1.8183 \ge-1.833113= t_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for left-tailed, $df=9$ degrees of freedom, $t=-1.8183,$ is $p = 0.051189,$ and since $p = 0.051189 \ge 0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

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

is less than 8.5, at the $\alpha = 0.05$ significance level.