Question

A theory predicts that the mean age of stars within a particular type of star cluster is 3.3 billion years, with a standard deviation of .4 billion years. (Their ages are approximately normally distributed.) You think the mean age is actually greater, and that this would lend support to an alternative theory about how the clusters were formed. You use a computer to randomly select the coordinates of 50 stars from the catalog of known stars of the type you're studying and you estimate their ages. You find that the mean age of stars in your sample is 3.4 billion years.

To test whether the population mean is greater than 3.3 billion years, what would your null and alternative hypothesis be?

Answer 1

Answer:

Null hypothesis:$\mu \leq 3.3$  

Alternative hypothesis:$\mu > 3.3$  

$z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768$  

Since is a one right tailed test the p value would be:  

$p_v =P(z>1.768)=0.039$  

Step-by-step explanation:

Data given and notation

$\bar X=3.4$ represent the sample mean  

$\sigma=0.4$ represent the population deviation for the sample

$n=50$ sample size  

$\mu_o =3.3$ 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 to be tested  

We need to conduct a hypothesis in order to determine if the mean is higher than 3.3, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 3.3$  

Alternative hypothesis:$\mu > 3.3$  

Compute the test statistic  

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:  

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (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".  

We can replace in formula (1) the info given like this:  

$z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768$  

P value

Since is a one right tailed test the p value would be:  

$p_v =P(z>1.768)=0.039$