Question

A machine at Juice Fresh is set to fill fruit juice in 250ml cartons. It is known that when the machine is working properly, the mean amount of juice-filled is 250ml. A supervisor from the quality control department took a random sample of cartons and found that the mean amount of juice-filled was 258ml with a standard deviation of 16ml. Using the 1% significance level, the supervisor wants to test if the machine is working properly. Assume that the amounts of juice-filled in all such cartons follow the normal distribution.

1) Find the rejection region(s) and calculate the test statistics. [6 marks]

2) Hence, can you conclude that the mean amount of fruit juice filled is different from 250ml? Does the machine need adjustment? [4 marks]

Answer 1

Hypothesis Test For Population Mean µ

$$H_0:\mu=\mu_0=250$$ $$H_1: \mu\not=\mu_0 (two-tailed \ test)$$

Hypothesis Test When σ Is Not Known:

The random variable

$${\bar{X}-\mu_0 \over s/\sqrt{n}}$$

has a Student's t-distribution with n-1 degrees of freedom. Reject $H_0$ if $t<-t_{\alpha/2;n-1}$ or $t>t_{\alpha/2;n-1}$.

Given that $\alpha=0.01, \mu_0=250, \bar{X}=258, s=16$

Let $n=25.$ Then $df=n-1=24.$

$$t_{\alpha/2;n-1}=t_{0.01/2;25-1}=2.796940$$

The rejection region for this two-tailed test is

$$R=\{t:|t|>2.796940\}$$

Let $n=30.$ Then $df=30-1=29.$

The rejection region for this two-tailed test is

$$R=\{t:|t|>2.756386\}$$

Let $n=31.$ Then $df=n-1=30.$

$$R=\{t:|t|>2.749996\}$$

The rejection region for this two-tailed test is

Let $n=36.$ Then $df=n-1=35.$

$$t_{\alpha/2;n-1}=t_{0.01/2;36-1}=2.723806$$

The rejection region for this two-tailed test is

$$R=\{t:|t|>2.723806\}$$

Let $n=49.$ Then $df=49-1=48.$

$$t_{\alpha/2;n-1}=t_{0.01/2;49-1}=2.682204$$

The rejection region for this two-tailed test is

$$R=\{t:|t|>2.682204\}$$

The t-statistic is computed as follows:

$${\bar{X}-\mu_0 \over s/\sqrt{n}}$$

If $n=25$

$$R=\{t:|t|>2.796940\}$$ $${\bar{X}-\mu_0 \over s/\sqrt{n}}={258-250 \over 16/\sqrt{25}}=2.5$$

Decision about the null hypothesis

Since it is observed that $|t|=2.5<2.796940,$ it is then concluded that the null hypothesis is not rejected.

Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 250, at the 0.01 significance level.

If $n=30.$

$$R=\{t:|t|>2.756386\}$$ $${\bar{X}-\mu_0 \over s/\sqrt{n}}={258-250 \over 16/\sqrt{30}}=2.738613$$

Decision about the null hypothesis

Since it is observed that $|t|=2.738613<2.756386,$ it is then concluded that the null hypothesis is not rejected.

Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 250, at the 0.01 significance level.

If $n=31.$

$$R=\{t:|t|>2.749996\}$$ $${\bar{X}-\mu_0 \over s/\sqrt{n}}={258-250 \over 16/\sqrt{31}}=2.783882$$

Decision about the null hypothesis

Since it is observed that $|t|=2.783882>2.749996,$ it is then concluded that the null hypothesis is rejected.

Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 250, at the 0.01 significance level.

If $n=36$

$$R=\{t:|t|>2.723806\}$$ $${\bar{X}-\mu_0 \over s/\sqrt{n}}={258-250 \over 16/\sqrt{36}}=3$$

Decision about the null hypothesis

Since it is observed that $|t|=3>2.723806,$  it is then concluded that the null hypothesis is rejected.

Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 250, at the 0.01 significance level.

We see that if $n\geq31,$ ,there is enough evidence to claim that the population mean $\mu$ is different than 250, at the 0.01 significance level. Therefore, the machine needs adjustment.