Question

The average high-density lipoprotein (HDL) cholesterol level of adult females is 53 milligrams per deciliter. You want to determine whether the true mean female HDL cholesterol level is more than the average. You decide to pull a random sample of 12 adult females and measure their HDL cholesterol level. You find the sample mean is 54.1 milligrams per deciliter and the standard deviation is 1.78 milligrams per deciliter. What are the test statistic and the p-value?

Answer 1

Answer:

$t=\frac{54.1-53}{\frac{1.78}{\sqrt{12}}}=2.14$  

$df = n-1= 12-1=11$

$p_v = P(t_{11}>2.14) =0.0278$

Step-by-step explanation:

Data provided

$\bar X=54.1$ represent the sample mean in mg per deciliter of cholesterol level

$s=1.78$ represent the sample standard deviation

$n=12$ sample size  

$\mu_o =53$ represent the value that we want to test  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

System of hypothesis

We need to conduct a hypothesis in order to check if the mean is higher than 53 mg per deciliter, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 53$  

Alternative hypothesis:$\mu > 53$  

The statistic is given by:

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

Calculate the statistic  

Replacing into the formula we got:

$t=\frac{54.1-53}{\frac{1.78}{\sqrt{12}}}=2.14$  

P-value  

We need to find first the degrees of freedom:

$df = n-1= 12-1=11$

The p value for this case since we have a right tailed test is:

$p_v = P(t_{11}>2.14) =0.0278$