Question

You work in the HR department at a large franchise. you want to test whether you have set your employee monthly allowances correctly. the population standard deviation is 150. You want to test if the monthly allowances should be increased. A random sample of 40 employees yielded a mean monthly claim of $640.
1- at the 1% significance level, test if the average monthly allowances should be greater than 500(use the 5 steps hypothesis testing procedure)

2- Confirm your answer by the p value approach.

Answer 1

Answer:

1) Null hypothesis:$\mu \leq 500$  

Alternative hypothesis:$\mu > 500$  

$z=\frac{640-500}{\frac{150}{\sqrt{40}}}=5.90$  

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.01 of the area in the right and we got:

$z_{crit}= 2.33$

For this case we see that the calculated value is higher than the critical value

Since the calculated value is higher than the critical value we have enugh evidence to reject the null hypothesis at 1% of significance level

2) Since is a right tailed test the p value would be:  

$p_v =P(z>5.90)=1.82x10^{-9}$  

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, same conclusion for part 1

Step-by-step explanation:

Part 1

Data given

$\bar X=640$ represent the sample mean

$\sigma=150$ represent the population standard deviation

$n=40$ sample size  

$\mu_o =500$ represent the value that we want to test  

$\alpha=0.01$ 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)  

Step1:State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is higher than 500, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 500$  

Alternative hypothesis:$\mu > 500$  

Step 2: Calculate the statistic

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

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

$z=\frac{640-500}{\frac{150}{\sqrt{40}}}=5.90$  

Step 3: Calculate the critical value

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.01 of the area in the right and we got:

$z_{crit}= 2.33$

Step 4: Compare the statistic with the critical value

For this case we see that the calculated value is higher than the critical value

Step 5: Decision

Since the calculated value is higher than the critical value we have enugh evidence to reject the null hypothesis at 1% of significance level

Part 2

P-value  

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

$p_v =P(z>5.90)=1.82x10^{-9}$  

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, same conclusion for part 1