Question

Nationwide, the average waiting time until a electric utility customer service representative answers a call is 200 seconds per call. The Gigantic Kilowatt Energy Company took a sample of 30 calls and found that, on the average, they answered in 120 seconds per call. Moreover, it is know that the standard deviation of the times for all such calls is 25 seconds. At the .05 significance level, is there evidence that this company's mean response time is lower than the average utility?

Answer 1

Answer:

$z=\frac{120-200}{\frac{25}{\sqrt{30}}}=-17.527$  

$p_v =P(Z<-17.527) \approx 0$  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v<<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis.  

We can say that at 5% of significance the mean average waiting time is significantly less than 200 seconds per call.

Step-by-step explanation:

Data given and notation  

$\bar X=120$ represent the sample mean  

$\sigma=25$ represent the population standard deviation  

$n=30$ sample size  

$\mu_o =200$ represent the value that we want to test  

$\alpha=0.05$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the population mean is less than 200, the system of hypothesis are :  

Null hypothesis:$\mu \geq 200$  

Alternative hypothesis:$\mu < 200$  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

$z=\frac{120-200}{\frac{25}{\sqrt{30}}}=-17.527$  

P-value  

Since is a one-side left tailed test the p value would given by:  

$p_v =P(Z<-17.527) \approx 0$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v<<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis.  

We can say that at 5% of significance the mean average waiting time is significantly less than 200 seconds per call.