Question

As part of the Honolulu Heart Study, a sample group of 100 individuals is selected and a study conducted to see if following a 6-month low-carbohydrate diet their collective cholesterol levels have changed. You know the sample mean for cholesterol was 203.8 with standard deviation (s) of 21.6. After the study, further research indicated the populations mean (µ) cholesterol is 221.4. Based on this, are cholesterol levels of persons eating a low-carbohydrate diet different from the average cholesterol level? Please test at α.=.01

Answer 1

Answer:

$t=\frac{203.8-221.4}{\frac{21.6}{\sqrt{100}}}=-8.148$    

The degrees of freedom are given by:  

$df=n-1=100-1=99$  

and the p value would be given by:

$p_v =P(t_{(99)}$  

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 221.4 the value of study.

Step-by-step explanation:

Information given

$\bar X=203.8$ represent the sample mean    

$s=21.6$ represent the sample standard deviation

$n=100$ sample size  

$\mu_o =221.4$ represent the value to verify

$\alpha=0.01$ 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

Hypothesis to test

We want to verify if the true mean is lower than 221.4, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 221.4$  

Alternative hypothesis:$\mu < 221.4$  

The statistic is given by:

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

Replacing the info we got:

$t=\frac{203.8-221.4}{\frac{21.6}{\sqrt{100}}}=-8.148$    

The degrees of freedom are given by:  

$df=n-1=100-1=99$  

and the p value would be given by:

$p_v =P(t_{(99)}$  

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 221.4 the value of study.