Answer to Question #350614 in Statistics and Probability for joy

Question

Sixteen oil tins are taken at random from an automatic filling machine. The mean weight of the tins is 14.2 kg, with a standard deviation of 0.40 kg. Can we conclude that the filling machine is wasting oil by filling more than the intended weight of 14 kg, at a significance level of 5%? (5 marks)

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le 14$

$H_1:\mu>14$

This corresponds to a right-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 $\alpha = 0.05,$ $df=n-1=15$ and the critical value for a right-tailed test is $t_c =1.75305.$

The rejection region for this right-tailed test is $R = \{t:t>1.75305\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{14.2-14}{0.40/\sqrt{16}}=2.5$$

Since it is observed that $t=2.5>1.75305=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed, $df=15$ degrees of freedom, $t=2.5$ is $p=0.012253,$ and since $p=0.012253<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than 14, at the $\alpha = 0.05$ significance level.