The average cholesterol level is generally accepted to be 265 mg/dl. Due to incr
ID: 3155425 • Letter: T
Question
The average cholesterol level is generally accepted to be 265 mg/dl. Due to increasing awareness regarding the importance of diet, a cardiologist believes that the average cholesterol level is different from 265. The cardiologist collects a random sample of 625 obese patients. The following sample statistics are computed: = 270 mg/Dl, s = 50 mg/DI
a) Conduct the appropriate hypothesis test at level = 0.05
b) Compute the p-value for this test statistic.
c) Compute a 95% confidence interval for the mean cholesterol level.
Explanation / Answer
a)
Formulating the null and alternative hypotheses,
Ho: u = 265
Ha: u =/ 265
As we can see, this is a two tailed test.
Thus, getting the critical z, as alpha = 0.05 ,
alpha/2 = 0.025
zcrit = +/- 1.959963985
Getting the test statistic, as
X = sample mean = 270
uo = hypothesized mean = 265
n = sample size = 625
s = standard deviation = 50
Thus, z = (X - uo) * sqrt(n) / s = 2.5
As |z| > 1.96, we REJECT THE NULL HYPOTHESIS.
Hence, there is significant evidence that the average cholesterol level is different from 265. [CONCLUSION]
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b)
Also, the p value is, as this is two tailed,
p = 0.012419331 [ANSWER, P VALUE]
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c)
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 270
z(alpha/2) = critical z for the confidence interval = 1.959963985
s = sample standard deviation = 50
n = sample size = 625
Thus,
Margin of Error E = 3.919927969
Lower bound = 266.080072
Upper bound = 273.919928
Thus, the confidence interval is
( 266.080072 , 273.919928 ) [ANSWER]
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