Question

The owner of the factory that sells a particular bottled fruit juice claimed that the

average capacity of their product is 250 ml. To test the claim, a consumer group gets

a sample of 100 such bottles, calculates the capacity of each bottle, and then finds

the mean capacity to be 248 ml. The standard deviation s is 5 ml. Use α = 0.03. Will

you reject or not to reject the claim.

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 t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.03,$ $df=n-1=99$ and the critical value for a two-tailed test is $t_c = 2.201819.$

The rejection region for this two-tailed test is $R = \{t:|t|> 2.201819\}.$

The t-statistic is computed as follows:

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

Since it is observed that $|t|=4> 2.201819=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=99$ degrees of freedom, $t=-4$ is $p=0.000122,$ and since $p=0.000122<0.03=\alpha,$ 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 $\alpha = 0.03$ significance level.