Question

A political scientist believes that she can show that the average age of the voters in a presidential election is less than 40 years. To this end an exit poll of 100 voters is obtained. The data obtained in the poll has a sample average age of 39.2 years with a sample standard deviation of 8 years. We establish a 1% level of significance.
a. Referring to Political Scientist, give the alternative hypothesis.
b. Referring to Political Scientist, describe the rejection region. Be sure to label the endpoint(s).
c. Referring to Political Scientist, what is the conclusion?

Answer 1

Answer:

a) Null hypothesis:$\mu \geq 40$  

Alternative hypothesis:$\mu < 40$  

b) The rejection zone for this test would be:

$(-\infty, -2.36)$

c) If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly less than 40 years at 1% of signficance.  

Step-by-step explanation:

1) Data given and notation  

$\bar X=39.2$ represent the sample mean  

$s=8$ represent the sample standard deviation

$n=100$ sample size  

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

$\alpha=0.01$ 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)  

Part a

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is less than 40, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 40$  

Alternative hypothesis:$\mu < 40$  

If we analyze the size for the sample is > 30 but we don't know the population deviation so 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:  

$t=\frac{39.2-40}{\frac{8}{\sqrt{100}}}=-1$    

Part b: Rejection zone

On this case we are conducting a left tailed test so we need to find first the degrees of freedom for the distribution given by:

$df=n-1=100-1=99$

Now we need to find a critical value on th t distribution with 99 degrees of freedom that accumulates 0.01 of the area on the right tail and this value is given by $t_{crit}=-2.36$

And we can find it using the following excel code: "=T.INV(0.01,99)"

So then the rejection zone for this test would be:

$(-\infty, -2.36)$

Part c: P-value and conclusion

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

$p_v =P(t_{(99)}<-1)=0.160$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly less than 40 years at 1% of signficance.