Answer to Question #248082 in Statistics and Probability for lameck

Question

An XYZ business can currently produce 251 per hour of 5 amp fuses. The supplier claims that a new machine has been purchased and installed that will boost the manufacturing rate. The mean hourly production on the new machine was 259, with a sample standard deviation of 6 per hour, according to a sample of ten hours randomly selected from last month.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\leq251$

$H_1:\mu>251$

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 $\alpha = 0.05,$

$df=n-1=10-1=9$ degrees of freedom, and the critical value for a right-tailed test is $t_c =1.833113.$

The rejection region for this right-tailed test is$R = \{t: t > 1.833113\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{259-251}{6/\sqrt{10}}\approx4.21637$$

Since it is observed that $t = 4.21637 > 1.833113=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for right-tailed, $\alpha=0.05, df=9, t=4.21637$ is $p =0.001126,$ and since $p=0.001126<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than 251, at the $\alpha = 0.05$ significance level.

Therefore, there is enough evidence to claim that the new machine is faster, at the $\alpha = 0.05$ significance level.