Answer to Question #262311 in Statistics and Probability for Veena

Question

The internet provider claims that their router gives a speed of 25 Mbps with a standard deviation of 4.8 Mbps. The computer engineer wants to check whether the speed is different than what the provider claims. For this study, the engineer collects the speed of 40 routers and found that the average speed is 23.2 Mbps for those 40 routers. Plot the acceptance and rejection region for a two-sided test to test the claim of engineer with 90% confidence.

Answer 1

The following null and alternative hypotheses need to be tested:

$\mu=25$

$\mu\not=25$

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 $\alpha = 0.1,$ and the critical value for a two-tailed test is $z_c = 1.6449.$

The rejection region for this two-tailed test is R = $\{z: |z| > 1.6449\}$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{23.2-25}{4.8/\sqrt{40}}=-2.3717$$

Since it is observed that $|z| = 2.3717 >1.6449= z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=2P(Z<-2.3717)=0.017106,$ it is concluded that the null hypothesis is rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than $25,$ at the $\alpha = 0.1$ significance level.