Answer:
[tex]z=\frac{215-220}{\frac{15}{\sqrt{64}}}=-2.67[/tex]
[tex]p_v =P(z<-2.67)=0.0038[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.03[/tex] we see that [tex]p_v<\alpha[/tex] 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
[tex]\bar X=215[/tex] represent the sample mean
[tex]\sigma=15[/tex] represent the population standard deviation
[tex]n=64[/tex] sample size
[tex]\mu_o =220[/tex] represent the value that we want to test
[tex]\alpha=0.03[/tex] represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
[tex]p_v[/tex] 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:[tex]\mu \geq 220[/tex]
Alternative hypothesis:[tex]\mu < 220[/tex]
The statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]z=\frac{215-220}{\frac{15}{\sqrt{64}}}=-2.67[/tex]
P-value
Since is a one-side lower test the p value would be:
[tex]p_v =P(z<-2.67)=0.0038[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.03[/tex] we see that [tex]p_v<\alpha[/tex] then we have enough evidence to conclude that the true mean is significantly lower than 220 at 3% of significance.