Question #350614

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)


1
Expert's answer
2022-06-16T09:50:55-0400

The following null and alternative hypotheses need to be tested:

H0:μ14H_0:\mu\le 14

H1:μ>14H_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 α=0.05,\alpha = 0.05, df=n1=15df=n-1=15 and the critical value for a right-tailed test is tc=1.75305.t_c =1.75305.

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

The t-statistic is computed as follows:


t=xˉμs/n=14.2140.40/16=2.5t=\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=tc,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=15df=15 degrees of freedom, t=2.5t=2.5 is p=0.012253,p=0.012253, and since p=0.012253<0.05=α,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 μ\mu is greater than 14, at the α=0.05\alpha = 0.05 significance level.


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