Sample size in CI

[shigella123]

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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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95% CI = mean +/- [2 * standard_deviation/sqrt(sample size)] = 110 +/- (2*20/10)
=[106, 110]

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How is this person getting the sample size of 10?

 

[Brain Bucket]

HE IS USING THE POWER OF CAPS LOCK! And he's terrible at subtraction, but look at the formula more carefully.

 

[HatWobble]

How is this person getting the sample size of 10?

The square root of the sample size is 10 (i.e. 100). Though to me it seems it should be 86.

 

[Brain Bucket]

The square root of the sample size is 10 (i.e. 100). Though to me it seems it should be 86.

The question is poorly worded. It explicitly states that 100 completed the test, and then goes on to say that 20 had to be dropped because they couldn't complete two or more portions bla bla...

 

[shigella123]

Answer:

Confidence interval= Mean +/- Z score multiplied by SD/square root of sample number

For 95% confiedence interval,standard Z score is 1.95 rounding off to 2

So,

Confidence Interval= 110 +/- 2 multiplied by 20/10(square root of 100 is 10)
= 110 +/- 4
= 106 to 114

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F.

The formula for the confidence interval of the mean is:

Mean+/- Z(s/sqrtof N)

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 ± 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 ± 2S).
Choice B gives the given mean of 110 ± 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 ± 1S).
For choice D, see comment on choice B. In this case the result is for 110 ± 1S.
Choice E is simply the given mean (110) ± 5. If this seemed to the right answer to you, you made a calculation error.

 

[Brain Bucket]

If you had the answer, why'd you ask? That question was such a waste of time.

 

[shigella123]

If you had the answer, why'd you ask? That question was such a waste of time.

Mr. Brain Bucket, I found it online. I agree with you that part "dropping 20" got me confused too. It's a good practice q, don't you think? Thanks for your help. 🙂

I need more practice on biostat cal qs and the Hardy Weinberg as I'm barely getting them right.

 

[HatWobble]

That was mad dumb, yo.

 

[BlueArc]

lol at 'Mr.' Brain Bucket !

 

[CherryRedDracul]

For your pleasure.