Answer to Question #338813 in Statistics and Probability for Sarah

Question

. A new machine is being considered to replace the old machine being used. This new machine was tested for 10 consecutive hours with the following output: 119, 122, 118, 122, 120, 124, 126, 125, 125, and 124. If the average output per hour using the old machine is 120 units, is the management justified in stating that the output per hour can be increased with the new machine? Use a 0.01 level of significance.

Answer 1

$$\bar{x}=\dfrac{1}{10}(119+122+118+122+120$$

$$124+126+125+125+124)=122.5$$

$$s^2=\dfrac{\sum _i(x_i-\bar{x})^2}{n-1}=\dfrac{68.5}{9}$$

$$s=\sqrt{\dfrac{68.5}{9}}\approx2.7588$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le120$

$H_a:\mu>120$

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 $\alpha = 0.01,$ $df=n-1=9$ degrees of freedom, and the critical value for a right-tailed test is $t_c = 2.821433.$

The rejection region for this right-tailed test is $R = \{t: t >2.821433\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{122.5-120}{2.7588/\sqrt{10}}\approx2.8656$$

Since it is observed that $t = 2.8656 \ge 2.821433=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed test, $df=9$ degrees of freedom, $t=2.8656$ is $p = 0.009305,$ and since $p =0.009305 \le 0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is greater than 120, at the $\alpha = 0.01$ significance level.