Answer to Question #334273 in Statistics and Probability for Dahz

Question

 the mean monthly Internet bill is $32.79 per household. A sample of 50 households in the Belize District showed a sample mean of $30.63. Use a population standard deviation $5.60.

  

 What is the value of the test statistic? 

c) What is the p-value?

d) At α = .05, what is your conclusion?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=32.79$

$H_1:\mu\not=32.79$

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

b) The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}$$

c) The p-value is $p=2P(Z<z), z\le0$ or $p=2P(Z>z), z\ge0.$

d) 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 $R = \{z: |z| > 1.96\}.$

$$z=\dfrac{30.63-32.79}{5.60/\sqrt{50}}\approx-2.7274$$

Since it is observed that $|z| = 2.7274 >1.96= z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(Z<-2.7274)=0.006384,$ and since $p=0.006481<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 32.79, at the $\alpha = 0.05$ significance level.