Answer to Question #350104 in Statistics and Probability for rose

Question

 The owner of a factory that sells a particular bottled fruit juice claims that the average capacity of a bottle of their product is 250mL and standard deviation of 3 ml. A random sample of 50 bottles of fruit juice have a mean volume of 245 ml. At a level of significance of 5%, is there a sufficient evidence to the support the claim of the owner?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=250$

$H_1:\mu\not=250$

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:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{245-250}{3/\sqrt{50}}=-11.7851$$

Since it is observed that $|z|=11.7851>1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=2P(z<-11.7851)=0,$ 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 different than 250, at the $\alpha = 0.05$ significance level.