Question

The following output from MINITAB presents the results of a hypothesis test. Test of mu=49 vs. not=49, The assumed standard deviation 10.4, N 58, Mean 46.32, SE 1.412088, 95%CI (43.5523077, 49.0878923, Z -1.962526, P 0,049701, Do you reject H0 at the

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=49$

$H_1:\mu\not=49$

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a two-tailed test is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z:|z|>1.96\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{46.32-49}{10.4/\sqrt{58}}=-1.962526$$

Since it is observed that $|z|=1.962526>1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(z<-1.962526)=0.049701,$ and since $p=0.049701<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 49, at the $\alpha = 0.05$ significance level.

$$SE=\dfrac{\sigma}{\sqrt{n}}=\dfrac{10.4}{\sqrt{58}}\approx1.365587$$

The corresponding 95% confidence interval is

$$CI=(\bar{x}-z_c\times SE, \bar{x}+z_c\times SE)$$

$$=(46.32-1.96\times 1.365587,$$

$$46.32+1.96\times 1.365587)$$

$$=(43.6434497, 48.9965503)$$