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

The degrees of freedom are given by:  

and the p value would be given by:

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

represent the sample mean    

represent the sample standard deviation

sample size  

represent the value to verify

represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

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:  

Alternative hypothesis:  

The statistic is given by:

 (1)  

Replacing the info we got:

The degrees of freedom are given by:  

and the p value would be given by:

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:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497