Question

A sample of 64 observations is selected from a normal population. The sample mean is 215, and the population standard deviation is 15. Conduct the following test of hypothesis using the .03 significance level. What is the p-value?H0 : μ ≥ 220H1 : μ < 220

Answer 1

Answer:

$z=\frac{215-220}{\frac{15}{\sqrt{64}}}=-2.67$      

$p_v =P(z<-2.67)=0.0038$  

If we compare the p value and the significance level given $\alpha=0.03$ we see that $p_v<\alpha$ then we have enough evidence to conclude that the true mean is significantly lower than 220 at 3% of significance.

Step-by-step explanation:

Data given

$\bar X=215$ represent the sample mean

$\sigma=15$ represent the population standard deviation

$n=64$ sample size      

$\mu_o =220$ represent the value that we want to test    

$\alpha=0.03$ 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 true mean is lower than 220, the system of hypothesis would be:      

Null hypothesis:$\mu \geq 220$      

Alternative hypothesis:$\mu < 220$      

The statistic is given by:

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

Calculate the statistic      

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

$z=\frac{215-220}{\frac{15}{\sqrt{64}}}=-2.67$      

P-value      

Since is a one-side lower test the p value would be:      

$p_v =P(z<-2.67)=0.0038$  

Conclusion      

If we compare the p value and the significance level given $\alpha=0.03$ we see that $p_v<\alpha$ then we have enough evidence to conclude that the true mean is significantly lower than 220 at 3% of significance.