Question

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)

x = ____________          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.

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).

B C D je on А 1 Serum Cholesterol 2 138.7 3 186.7 4 129.9 5 169.5 6 126.9 7 210.9 230.4 9 208.8 10 168.2 11 212.1 12 174.1 13 134.8 14 176.2 15 202.8 16 197.1 17 170.4 18 190 19 163.6 20 153.3 21 155 22 131.4 23 201.1 24 135.9 25 254.3 144.1 27 144.6 28 172.1 29 234.8 30 220.1 31 143.1 32 26 33

Answer 1

i.
Given that,
population mean(u)=168
sample mean, x =176.03
standard deviation, s =34.52
number (n)=30
null, Ho: μ=168
alternate, H1: μ!=168
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.045
since our test is two-tailed
reject Ho, if to < -2.045 OR if to > 2.045
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =176.03-168/(34.52/sqrt(30))
to =1.274
| to | =1.274
critical value
the value of |t α| with n-1 = 29 d.f is 2.045
we got |to| =1.274 & | t α | =2.045
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 1.2741 ) = 0.2127
hence value of p0.05 < 0.2127,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=168
alternate, H1: μ!=168
test statistic: 1.274
critical value: -2.045 , 2.045
decision: do not reject Ho
p-value: 0.2127
we do not have enough evidence to support the claim that the serum cholesterol level for women in US is 168 mg/dl 10 years ago.
Type 2 error is possible because it fails to reject the null hypothesis.
ii.
if population standard deviation is known then Z test for single mean
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
---------------
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 do not have enough evidence to support the claim that the serum cholesterol level for women in US is 168 mg/dl 10 years ago.
Type 2 error is possible because it fails to reject the null hypothesis.