Question

Suppose we want to see if American children have higher levels of cholesterol than the average child (i.e., in the entire world - the total population). We find that the population average for cholesterol for children all over the world is 190. We test 25 US children and find an average of 201 with a standard deviation of 10. Conduct a hypothesis with a significance level of 0.05.

Answer 1

Answer:

$t=\frac{201-190}{\frac{10}{\sqrt{25}}}=5.5$    

$p_v =P(t_{(24)}>5.5)=0.00000589$  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 190

Step-by-step explanation:

Data given and notation  

$\bar X=201$ represent the mean

$s=10$ represent the sample standard deviation for the sample  

$n=25$ sample size  

$\mu_o =190$ represent the value that we want to test

$\alpha=0.05$ 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 check if the mean is higher than 190, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 190$  

Alternative hypothesis:$\mu > 190$  

If we analyze the size for the sample is < 30 and 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:  

$t=\frac{201-190}{\frac{10}{\sqrt{25}}}=5.5$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=25-1=24$  

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

$p_v =P(t_{(24)}>5.5)=0.00000589$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 190