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The intelligence quotient (IQ) scores are obtained for a sample of 100 patients diagnosed with various types of schizophrenia who completed a standard IQ test battery. An additional 20 patients had to be dropped from the sample because they lacked the functional capacity to complete two or more portions of the test. Four other patients refused to take the test battery when offered the opportunity. Results for the patients who completed the test battery gave an average IQ of 110 and a standard deviation of 20.
Using this information, compute the 95% confidence interval for this estimate of the mean.
A. 70 to 130
B. 70 to 150
C. 85 to 115
D. 90 to 130
E. 105 to 115
F. 106 to 114
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i think AAAAAAA?I DNT KNOW FOR SURE..
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lala g is right .. good going.
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The correct answer is F.
The formula for the confidence interval of the mean is:
The mean is given as 110. To achieve a 95% confidence interval we use a Z-score =1.96 (or 2.0 to make the calculation easier). The standard deviation (S) is given as 20 and the sample size is given as 100. Inserting these values into the formula returns the result of 110 ¡À 4 or 106 to 114. Note that the information about those who either were unable or refused to participate in the study is not relevant to answering the question asked here.
Choice A is incorrect. Asking for a 95% confidence interval is not the same thing as asking for 95% of the cases in a normal distribution. The first is an inferential statistic, trying to decide what the true mean might be, while the second is a descriptive statistic, looking for 95% of the cases. This choice tells us the answer to the question: ¡°in the general population, 95% of the population has an IQ in what range? With a standard mean of 100 and a standard deviation of 15, 95% of the cases would fall between 70 and 130 (mean ¡À 2S).
Choice B gives the given mean of 110 ¡À 2S. But the question asked for the 95% confidence interval, not for the scores of 95% of the people who took the test.
See comment on choice A. Choice C is the result for 68% of the cases in the population (mean ¡À 1S).
For choice D, see comment on choice B. In this case the result is for 110 ¡À 1S.
Choice E is simply the given mean (110) ¡À 5. If this seemed to the right answer to you, you made a calculation erro
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The correct answer is F.
The formula for the confidence interval of the mean is:
The mean is given as 110. To achieve a 95% confidence interval we use a Z-score =1.96 (or 2.0 to make the calculation easier). The standard deviation (S) is given as 20 and the sample size is given as 100. Inserting these values into the formula returns the result of 110 ¡À 4 or 106 to 114. Note that the information about those who either were unable or refused to participate in the study is not relevant to answering the question asked here.
Choice A is incorrect. Asking for a 95% confidence interval is not the same thing as asking for 95% of the cases in a normal distribution. The first is an inferential statistic, trying to decide what the true mean might be, while the second is a descriptive statistic, looking for 95% of the cases. This choice tells us the answer to the question: ¡°in the general population, 95% of the population has an IQ in what range? With a standard mean of 100 and a standard deviation of 15, 95% of the cases would fall between 70 and 130 (mean ¡À 2S).
Choice B gives the given mean of 110 ¡À 2S. But the question asked for the 95% confidence interval, not for the scores of 95% of the people who took the test.
See comment on choice A. Choice C is the result for 68% of the cases in the population (mean ¡À 1S).
For choice D, see comment on choice B. In this case the result is for 110 ¡À 1S.
Choice E is simply the given mean (110) ¡À 5. If this seemed to the right answer to you, you made a calculation erro
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lala usmle .. plz tell what is the formula for determining the confindence interval here
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ok
the formula is:
confidence interval on the mean : Mean +/- 95%CI (Zscore)
In this qs : 110 +/- 2x (20/ square root of 100(n)) = 110 +/- 2(20/10) = 110 +/- 4
Answer: 106 and 114
Hope that help.
