Answer to Question #93443 in Statistics and Probability for mimi

Question

mean amount of juice
is 250ml.
a random sample of cartons and found that the mean amount of juice
258ml with a standard deviation of 16ml. Using the 1% significance level
Find the rejection region(s) and calculate the test statistics

Answer 1

Hypothesis Test For Population Mean $\mu$.

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

Since $\sigma$ is unknown, then we use $t-$statistic. The random variable ${\bar{X}-\mu \over s/\sqrt{n}}$ has a Student's t-distribution with $df=n-1$ degrees of freedom. We reject $H_0$ if $t<-t_{\alpha/2;n-1}$ or $t>t_{\alpha/2;n-1}$.

Given that $\alpha=0.01, \mu=250, \bar{X}=258, s=16.$ What is the value of n?

Let $n=25.$ Then $df=n-1=25-1=24.$ Define the critical value from the table

$$t_{\alpha/2;n-1}=t_{0.01/2;25-1}=t_{0.005;24}=2.797$$

The rejection region for this two-tailed test is

$$R=\{t:|t|>2.797\}$$ $${\bar{X}-\mu \over s/\sqrt{n}}={258-250 \over 16/\sqrt{25}}=2.500$$

Decision about the null hypothesis

Since it is observed that $|t|=2.500<2.797,$ it is then concluded that the null hypothesis is not rejected.

Conclusion

It is concluded that the null hypothesis $H_0$ 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.

From the analysis of the table of critical values we see that if we increase $n$ (and the number of degrees of freedom) and don't change the value of $\alpha$, then the critical value decreases.

On the other hand if we increase $n$ and don't change $\bar{X}, \mu, s$ the random variable ${\bar{X}-\mu \over s/\sqrt{n}}$ increases.

Let $n=30.$ Then $df=n-1=30-1=29.$ Define the critical value from the table

$$t_{\alpha/2;n-1}=t_{0.01/2;30-1}=t_{0.005;29}=2.756$$

The rejection region for this two-tailed test is

$$R=\{t:|t|>2.756\}$$ $${\bar{X}-\mu \over s/\sqrt{n}}={258-250 \over 16/\sqrt{30}}=2.739$$

Decision about the null hypothesis

Since it is observed that $|t|=2.739<2.756,$ it is then concluded that the null hypothesis is not rejected.

Conclusion

It is concluded that the null hypothesis $H_0$ 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.

Let $n=31.$ Then $df=n-1=31-1=30.$ Define the critical value from the table

$$t_{\alpha/2;n-1}=t_{0.01/2;31-1}=t_{0.005;30}=2.750$$

The rejection region for this two-tailed test is

$$R=\{t:|t|>2.750\}$$ $${\bar{X}-\mu \over s/\sqrt{n}}={258-250 \over 16/\sqrt{31}}=2.784$$

Decision about the null hypothesis

Since it is observed that $|t|=2.784>2.750,$ it is then concluded that the null hypothesis is rejected.

Conclusion

It is concluded that the null hypothesis $H_0$ is rejected. Therefore, 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.

If we take $n\leq30$ there is not enough evidence to claim that the population mean $\mu$ is different than 250, at the 0.01 significance level.

If we take $n\geq31$ 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.