Answer to Question #347249 in Statistics and Probability for Aira

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.