Answer to Question #345874 in Statistics and Probability for Bugay

Question

The owner of a factory that sells a particular bottled fruit juice claims that the average capacity of their products is 250 ml.to test the claim, a consumer group gets a sample of 100 such bottles, calculate the capacity of each bottle, and then the finds mean capacity to be 248 ml. The standard deviation is 5 ml. Is the claim true? Conduct a hypothesis testing using a=0.05.

1. Describe the population parameter of intersect.

2. Formulate the null and alternative hypothesis.

3.what is the appropriate form of test statistic to be used.

4. Identify the appropriate rejection region.

5.compute for the test statistic.(show solution)

6.Make a decision based on the computed value of z and the critical region.

Answer 1

1.  the population parameter of intersect is the average capacity.

2. The following null and alternative hypotheses need to be tested:

$H_0:\mu=250$

$H_1:\mu\not=250$

3. This corresponds to a two-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 two-tailed test is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z:|z|> 1.96\}.$

5. The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{248-250}{5/\sqrt{100}}=-4$$

6. Since it is observed that $|z|=4>1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed is $p=2P(Z<-4)=0.000063,$ and since $p=0.000063<0.05=\alpha,$ 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.