Question

A research report claims that mice with an average life span 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. Is there any reason to believe that muμless than<40 if 70 mice that are placed on this diet have an average life of 3939 months with a standard deviation of 8.8 ​months? Use a​ P-value in your conclusion.

Answer 1

Answer:

$t=\frac{39-40}{\frac{8.8}{\sqrt{70}}}=-0.9508$      

$p_v =P(t_{69}<-0.9508)=0.1725$  

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly less than 40 months.

Step-by-step explanation:

1) Data given and notation      

$\bar X=39$ represent the sample mean

$s=8.8$ represent the standard deviation for the sample      

$n=70$ sample size      

$\mu_o =40$ represent the value that we want to test    

$\alpha$ represent the significance level for the hypothesis test.    

t 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 determine if the mean  is less than 40 months, the system of hypothesis would be:      

Null hypothesis:$\mu \geq 40$      

Alternative hypothesis:$\mu < 40$      

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:      

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)      

t-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:      

$t=\frac{39-40}{\frac{8.8}{\sqrt{70}}}=-0.9508$      

Calculate the P-value      

First we need to calculate the degrees of freedom given by:  

$df=n-1=70-1=69$  

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

$p_v =P(t_{69}<-0.9508)=0.1725$  

Conclusion      

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly less than 40 months.