Question

The factory owner claimed that their bottled fruit juice has the capacity of less than an average of 280ml. To test the claim, a group of consumers gets a sample of 80 bottles of the fruit juice, calculates the capacity, and then finds the mean capacity to be 265ml. The standard deviation is 8ml. Use a= 0.05 level of significance to test the claim.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge280$

$H_1:\mu<280$

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

4. Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a left-tailed test is $z_c = -1.6449.$

The rejection region for this left-tailed test is $R = \{z<-1.6449\}.$

5. The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{265-280}{8/\sqrt{80}}=-16.7705$$

6. Since it is observed that $z=-16.7705<-1.6449=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for left-tailed is $p=P(Z<-16.7705)=0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is less than 280, at the $\alpha = 0.05$ significance level.