euphoric
Forum Junior
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A standard nutrition srvey of 100 adults living in a northestern city found that they consumed an average of 2650 calories a day , with a SD of 200.Twenty-one additional ppl who were approached refused to participate in the survey.Using this info ,which of the following is the 95% confidence interval for this estimate of the mean?
A)2250 to 3050
B)2646 to2656
C)2610 to 2690
D)2450 to 2850
E)2630 to 2670
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| babydoc4usmle
Forum Guru
Topics: 18
Posts: 634
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C
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| robin082006
Forum Hero
Topics: 471
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C
___________________
The Key to Succeed is Patience.
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| euphoric
Forum Junior
Topics: 15
Posts: 79
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You guys rock! Can u explain me i'm very poor in calculation.
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| babydoc4usmle
Forum Guru
Topics: 18
Posts: 634
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it's pretty simple
you have to use formula for confidence interval+/- (2 (because we need 95% CI, if it was 99, we would take 2.5)*(standard deviation/square root from number people in study)
at the end we have: +/- 2*(200/sq.r from 100=10)
2650+/- 2*20
2650+/-40 = from 2610 to 2690
do not count those who declined to participate in survey, it's just a distration
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