Question

2. Coronary heart disease (CHD) begins in young adulthood, and is the fifth leading cause of death among adults, and one of the crucial factor is the serum cholesterol level. It is reported that the serum cholesterol level for women in US is 168 mg/dl 10 years ago. A researcher wants to see if the serum cholesterol level has changed ever since. He collected a random sample of 30 females, and their serum cholesterol level is measured and recorded in the Excel file provided.  

• (1) What is the sample of the study? What is the population of the study? (Please describe in context words for the specific study, don’t use general definition).  

• (2) Use Excel to calculate mean and standard deviation of the sample data. (refer to the practices in Unit 5: AVERAGE and STDEV functions from Excel)  

?̅ = ____________ s = _____________  

(3) Assume that we know the standard deviation of cholesterol level for all women in US is 40 mg/dl. Use appropriate hypothesis test (z test or t test) to address the research question. (Please follow the scheme from question #1 for the 5-step procedure of the hypothesis test).  

Hint: If σ is known, you will need to do a z test, and if σ is unknown, you will need to do a t test.  

(4) Based on your conclusion, what type of error could you possibly make? (type I or type II).  

(5) Do the calculation part in Excel, from the DATA ANALYSIS add-in, choose “z-test: Two Sample for Means”, Please copy and paste the output result here. (Don’t forget to make the dummy variable, and variance is the square of the standard deviation)  

(6) Keep in mind, the Excel will only do the calculation for you (Step 3 of the hypothesis test). You will still need to be able to interpret the output yourself. So please write the full 5-step procedure based on the output from part (5).  

Serum Cholesterol
138.7
186.7
129.9
169.5
126.9
210.9
230.4
208.8
168.2
212.1
174.1
134.8
176.2
202.8
197.1
170.4
190
163.6
153.3
155
131.4
201.1
135.9
254.3
144.1
144.6
172.1
234.8
220.1
143.1

Answer 1

2
Given that,
population mean(u)=168
standard deviation, σ =40
sample mean, x =176.03
number (n)=30
null, Ho: μ=168
alternate, H1: μ!=168
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 176.03-168/(40/sqrt(30)
zo = 1.1
| zo | = 1.1
critical value
the value of |z α| at los 5% is 1.96
we got |zo| =1.1 & | z α | = 1.96
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 1.1 ) = 0.272
hence value of p0.05 < 0.272, here we do not reject Ho
ANSWERS
---------------
1.
sample of study
The sample of a study is the group of subjects in the study. Sampling is the process whereby a researcher chooses his or her sample.
sample data is given in the data that is example of sample study
population of study
A study of a group of individuals taken from the general population who share a common characteristic, such as age, sex, or health condition.
population mean is 168mg/Dl serum cholesterol level
2.
sample mean = 176.03
sample standard deviation =34.527
3.
null, Ho: μ=168
alternate, H1: μ!=168
test statistic: 1.1
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.272
we have enough evidence to support the claim that if the serum cholesterol level has changed
4.
Type 2 error is possible in this context,
Because when its fails to reject the null hypothesis
5.
dummy variable is 151.2 then sample size is 31
Given that,
population mean(u)=168
standard deviation, σ =40
sample mean, x =175.229
number (n)=31
null, Ho: μ=168
alternate, H1: μ!=168
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 175.229-168/(40/sqrt(31)
zo = 1.006
| zo | = 1.006
critical value
the value of |z α| at los 5% is 1.96
we got |zo| =1.006 & | z α | = 1.96
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 1.006 ) = 0.314
hence value of p0.05 < 0.314, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=168
alternate, H1: μ!=168
6.
test statistic: 1.006
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.314
we have enough evidence to support the claim that if the serum cholesterol level has changed