Answer to Question #349357 in Statistics and Probability for Cecel

Question

A printer manufacturing company claims that it's new ink-efficient printer can print an average of 1500 pages of word documents with standard deviation of 60. Thirty-five of these printers showed a mean of 1475 pages. Does this support the company's claim? Use 95% confidence level.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=1500$

$H_1:\mu\not=1500$

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a two-tailed test is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z:|z|>1.96\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{1475-1500}{60/\sqrt{35}}=-2.465$$

Since it is observed that $|z|=2.465>1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(z<-2.465)=0.013701,$ and since $p=0.013701<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is different than 1500, at the $\alpha = 0.05$ significance level.