Answer to Question #117520 in Statistics and Probability for Angelica

Question

The Research & Development (R&D) team of a manufacturing firm discovered a new procedure that can be introduced into the production process that can substantially decrease production time. The firm has a limited fund allocated for process modifications that a standard is set for acceptability, that only a production time decrease of at least 8% is acceptable. Six production trials were undertaken showing a mean production time decrease of 8.4% with a standard deviation of 0.32%. Using a level of significance of (a) 0.01 and (b) 0.05, test the hypothesis that the process should be introduced. Use t-distribution

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq8$

$H_1:\mu<8$

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

Based on the information provided, the significance level is $\alpha=0.01,$ and the critical value for a left-tailed test is $t_c=-3.365$. The rejection region for this left-tailed test is  $R=\{t:t<-3.365.\}$

The t-statistic is computed as follows:

$$t={\bar{X}-\mu\over s/\sqrt{n}}={8.4-8\over 0.32/\sqrt{6}}\approx3.0619$$

Since it is observed that $t=3.0619>t_c=-3.365,$ it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu$  is less than 8, at the 0.01 significance level.

Using the P-value approach: The p-value is $p=0.986,$ and since $p=0.986>0.01,$ it is concluded that the null hypothesis is not rejected.

b) Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a left-tailed test is $t_c=-2.015.$ The rejection region for this left-tailed test is  $R=\{t:t<-2.015\}$

The t-statistic is computed as follows:

$$t={\bar{X}-\mu\over s/\sqrt{n}}={8.4-8\over 0.32/\sqrt{6}}\approx3.0619$$

Since it is observed that $t=3.0619>-2.015=t_c,$ it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu$  is less than 8, at the 0.05 significance level.

Using the P-value approach: The p-value is $p=0.986,$ and since $p=0.986>0.05,$ it is concluded that the null hypothesis is not rejected.

The process should not be introduced.