9. A study is being planned to investigate if the mean serum cholesterol of men
ID: 3362140 • Letter: 9
Question
9. A study is being planned to investigate if the mean serum cholesterol of men who do not have heart disease (healthy men) is lower than the mean serum cholesterol of men with heart disease. From a study of men with heart disease, it is known that the population mean cholesterol for men with heart disease is 244 mg/100 ml and the population standard deviation (for all men) is 41 mg/100 ml. The researchers will sample 25 men without heart disease (healthy men) and test the null hypothesis of H 0: =244 versus Ha: < 244 at the one-sided = 0.05 level. The investigators believe that the true mean cholesterol level for healthy men is 219 mg/100 ml. If this mean is true, what will be the power of this test?
Explanation / Answer
Given that,
Standard deviation, =41
Sample Mean, X =244
Null, H0: =244
Alternate, H1: <244
Level of significance, = 0.05
From Standard normal table, Z /2 =1.6449
Since our test is left-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-244)/41/(n) < -1.6449 OR if (x-244)/41/(n) > 1.6449
Reject Ho if x < 244-67.4409/(n) OR if x > 244-67.4409/(n)
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Suppose the size of the sample is n = 25 then the critical region
becomes,
Reject Ho if x < 244-67.4409/(25) OR if x > 244+67.4409/(25)
Reject Ho if x < 230.51182 OR if x > 257.48818
Implies, don't reject Ho if 230.51182 x 257.48818
Suppose the true mean is 219
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(230.51182 x 257.48818 | 1 = 219)
= P(230.51182-219/41/(25) x - / /n 257.48818-219/41/(25)
= P(1.40388049 Z 4.69368049 )
= P( Z 4.69368049) - P( Z 1.40388049)
= 1 - 0.9198 [ Using Z Table ]
= 0.0802
For n =25 the probability of Type II error is 0.0802
power of the test = 1 - = 1-0.0802 = 0.9198
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