Answer to Question #340588 in Statistics and Probability for Ofy

Question

a printer manufacturing company claims that its new ink-efficient printer can print an average of 1600 pages of word documents with standard deviation of 62. thirty-five of those printers showed a mean of 1585 pages. does this support the company’s claim? use 95% confidence level.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=1600$

$H_a:\mu\not=1600$

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:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{1585-1600}{62/\sqrt{35}}\approx-1.4313$$

Since it is observed that $|z|= 1.4313<1.96= z_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=2P(Z<-1.4313)=0.152344,$ and since $p= 0.152344>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is different than 1600, at the $\alpha = 0.05$ significance level.