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

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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.

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