Question

A company decided to test the hypothesis that the average time a company’s employees are spending to check their private e-mails at work is more than 6 minutes. A random sample of 40 employees were selected and they averaged 6.6 minutes. It is believed that the population standard deviation is 1.7 minutes. The α is set to 0.05. The p-value for this hypothesis test would be ______.

Answer 1

Answer:

$z=\frac{6.6-6}{\frac{1.7}{\sqrt{40}}}=2.232$    

Since is a one sided right tailed test the p value would be:  

$p_v =P(z>2.232)=0.0128$  

Step-by-step explanation:

Data given and notation  

$\bar X=6.6$ represent the sample mean

$\sigma=1.7$ represent the population standard deviation

$n=40$ sample size  

$\mu_o =6$ represent the value that we want to test

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

t 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 true mean is higher than 6minutes or no, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 6$  

Alternative hypothesis:$\mu > 6$  

If we analyze the size for the sample is > 30 and we know the population deviation so 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:  

$z=\frac{6.6-6}{\frac{1.7}{\sqrt{40}}}=2.232$    

P-value

Since is a one sided right tailed test the p value would be:  

$p_v =P(z>2.232)=0.0128$