Answer to Question #342403 in Statistics and Probability for gel

Question

A chemical company alleged that the average weight of the bag of chemical is 30kgs. with a standard deviation of 1.1 kgs. A sample of 26 bags was taken and revealed a mean weight of 28.9 kgs. Shall we accept the allegation of the chemical company? Use .01 level of significance

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=30$

$H_1:\mu\not=30$

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:

$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{28.9-30}{1.1/\sqrt{26}}\approx-5.09902$

Since it is observed that $|z|=5.09902>2.5758=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(z<-5.09902)=0,$ 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 30, at the $\alpha = 0.01$ significance level.