Question

a normal distribution with a mean of 25 minutes with a population standard deviation of 5 minutes. The store manager sampled 100 customers and found that the mean waiting time was 24.25 minutes. At the 0.01 significance level, test if the mean waiting time is less than 25 minutes.

Answer 1

Answer:

$z=\frac{24.25-25}{\frac{5}{\sqrt{100}}}=-1.5$      

The p value for this case would be given with this probability:

$p_v =P(z$  

For this case since the p value is higher than the significance level of 0.01 we have enough evidence to fail to reject the null hypothesis and we don't have enough evidence to conclude that the true mean is significantly less than 25 min.

Step-by-step explanation:

Information given

$\bar X=242.5$ represent the sample mean      

$\sigma=5$ represent the population deviation

$n=100$ sample size      

$\mu_o =25$ represent the value to check

$\alpha=0.01$ represent the significance level

z would represent the statistic  

$p_v$ represent the p value

Hypotheis to test

We want to verify if the true mean is less than 25 min, the system of hypothesis would be:      

Null hypothesis:$\mu \geq 25$      

Alternative hypothesis:$\mu < 25$      

The statistic would be given by:

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

And replacing the info given we got:

$z=\frac{24.25-25}{\frac{5}{\sqrt{100}}}=-1.5$      

The p value for this case would be given with this probability:

$p_v =P(z$  

For this case since the p value is higher than the significance level of 0.01 we have enough evidence to fail to reject the null hypothesis and we don't have enough evidence to conclude that the true mean is significantly less than 25 min.