The U.S. Department of Health has suggested that a healthy total cholesterol mea
ID: 3313897 • Letter: T
Question
The U.S. Department of Health has suggested that a healthy total cholesterol measurement should be 200 mg/dL or less. Records for 28 randomly selected individuals are presented in StatCrunch (the data set is called “Cholesterol”). Test the hypothesis that the mean cholesterol level is more than 200 using a significance level of 0.05.
Cholesterol
200.78
199.46
203.64
205.66
204.19
205.97
192.95
196.44
191.34
201.47
202.35
195.88
214.08
200.59
205.38
200.06
201.36
211.4
206.97
202.28
209.3
201.37
204.99
196.07
202.84
207.41
195.81
195.94
a) Conduct a full hypothesis test by following the steps below. Enter an answer for each of these steps in your document.
i) State the null and alternative hypotheses using correct notation.
ii) State the significance level for this problem.
iii) Calculate the test statistic “by-hand.” Show the work necessary to obtain the value by typing your work and provide the resulting test statistic.
iv) Calculate the p-value using StatCrunch.
v) State whether you reject or do not reject the null hypothesis and your reason for your answer in one sentence.
vi) State your conclusion in context of the problem (i.e. interpret your results and/or answer the question being posed) in one or two complete sentences.
vii) Use StatCrunch (Stat > T-Stats > ...) to verify your test statistic and p-value. Copy and paste this box into your document.
b) Construct a 95% confidence interval using the above data. Please do this “by hand” using the formula and showing your work (please type your work, no images accepted here). Then, verify your result using StatCrunch. Copy and paste your StatCrunch result in your document as well. Interpret the confidence interval as we learned in class.
Cholesterol
200.78
199.46
203.64
205.66
204.19
205.97
192.95
196.44
191.34
201.47
202.35
195.88
214.08
200.59
205.38
200.06
201.36
211.4
206.97
202.28
209.3
201.37
204.99
196.07
202.84
207.41
195.81
195.94
Explanation / Answer
Given that,
population mean(u)=200
sample mean, x =201.9992
standard deviation, s =5.4035
number (n)=28
null, Ho: =200
alternate, H1: >200
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.703
since our test is right-tailed
reject Ho, if to > 1.703
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =201.9992-200/(5.4035/sqrt(28))
to =1.9578
| to | =1.9578
critical value
the value of |t | with n-1 = 27 d.f is 1.703
we got |to| =1.9578 & | t | =1.703
make decision
hence value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 1.9578 ) = 0.03033
hence value of p0.05 > 0.03033,here we reject Ho
ANSWERS
---------------
null, Ho: =200
alternate, H1: >200
test statistic: 1.9578
critical value: 1.703
decision: reject Ho
p-value: 0.03033
evidence that the mean cholesterol level is more than 200
given that,
sample mean, x =201.9992
standard deviation, s =5.4035
sample size, n =28
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 27 d.f is 2.052
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 201.9992 ± t a/2 ( 5.4035/ Sqrt ( 28) ]
= [ 201.9992-(2.052 * 1.021) , 201.9992+(2.052 * 1.021) ]
= [ 199.904 , 204.095 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 199.904 , 204.095 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
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