Question

A chemical company alleged that the average weight of its bag of chemical is 50 kgs. with a standard deviation of 0.9 kg. A sample of 29 bags was taken and revealed a mean weight of 48.1 kgs. Shall we accept the allegation of the chemical company? Use a = 0.01.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=50$

$H_1:\mu\not=50$

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 $\alpha = 0.01,$ and the critical value for a two-tailed test is $z_c = 2.5758.$

The rejection region for this two-tailed test is $R = \{z: |z| > 2.5758\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{48.1-50}{0.9/\sqrt{29}}=-11.369$$

Since it is observed that $|z| = 11.369 > 2.5758=z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=2P(Z<-11.369)=0,$ p and since $p = 0 < 0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is different than 50, at the $\alpha = 0.01$ significance level.