In a research report, Richard H. Weindruch of the UCLA medical school claims that mice with an average lifespan of 32 months will live to be about 40 months old when 40% of the calories in their diet are replaced by vitamins and protein (take the null hypothesis to be µ = 40 months). Is there any reason to believe that mice live, on average, less than 40 months if 64 mice that are placed on this diet have an average life of 38 months, assuming a population standard deviation of 5.8 months?

If we compare the p value and a significance level assumed for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we can reject the null hypothesis, and the the actual true mean for the average life is significant less than 40 months.

Step-by-step explanation:

Data given and notation

[tex]\bar X=38[/tex] represent the average life

[tex]\sigma=5.8[/tex] represent the standard deviation for the population

[tex]n=64[/tex] sample size

[tex]\mu_o =40[/tex] represent the value that we want to test

[tex]\alpha[/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 determine if the mean is less than 40 months, the system of hypothesis would be:

Null hypothesis:[tex]\mu \geq 40[/tex]

Alternative hypothesis:[tex]\mu < 40[/tex]

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (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:

[tex]z=\frac{38-40}{\frac{5.8}{\sqrt{64}}}=-2.759[/tex]

Calculate the P-value

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

[tex]p_v =P(Z<-2.759)=0.0029[/tex]

Conclusion

If we compare the p value and a significance level assumed for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we reject the null hypothesis, and the the actual true mean for the average life is significant lss than 40 months.