Question

A machine which has been regulated dispenses an average of 330 ml fruit concentrate per bottle. A random sample of 49 bottles filled by the machine has a mean content of 320 ml and a standard deviation of 50 ml.”?

Answer 1

The parameter is average content of fruit concentrate per bottle.

The following null and alternative hypotheses need to be tested:

$H_0:\mu=330$

$H_1:\mu\not=330$

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 $\alpha = 0.05,$ $df=n-1=48$ and the critical value for a two-tailed test is $t_c =2.010635.$

The rejection region for this two-tailed test is $R = \{t:|t|>2.010635\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{320-330}{50/\sqrt{49}}=-1.4$$

Since it is observed that $|t|=1.4<2.010635=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed, $df=48$ degrees of freedom, $t=-1.4$ is $p=0.167944,$ and since $p=0.167944>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is different than 330, at the $\alpha = 0.05$ significance level.