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Confidence Intervals OBJ: • Determine confidence intervals for statistical data Determine the area under the standard normal curve that lies between -1 and 1 z -3 -2 -1 0 1 2 3 Determine the area under the standard normal curve that lies between -2 and 2 z -3 -2 -1 0 1 2 3 Determine the area under the standard normal curve that lies between -3 and 3 z -3 -2 -1 0 1 2 3 The 68% - 95% - 99% Rule: Every normal distribution has three important properties: • About 68% of the distribution is within 1 standard deviation of the mean. • About 95% of the distribution is within 2 standard deviations of the mean. • About 99% of the distribution is within 3 standard deviations of the mean. p (1 – p) p – 3 s < p < p + 3s √ n EX: Of 400 households surveyed, 144 reported watching the Channel 7 news at least twice a week. Find a 99% confidence interval for p, the unknown proportion of households that watch thee Channel 7 news at least twice a week. p = 144 400 = .36 p (1 – p) = .36 .64 √ n 400 =.024 .36 – 3( ) < p < .36 + 3( ) .36 – 3(.024) < p < .36 + 3(.024) .288 < p < .432 28.8% < p < 43.2% p (1 – p) p – 2 s < p < p + 2s √ n EX: Suppose that the general manager of Channel 5 claims that less than 30% of the households watch the Channel 7 news at least twice a week. Use a 95% confidence interval to show that it is unlikely that the manager is correct. p = 144 400 = .36 p (1 – p) = .36 .64 √ n 400 =.024 .36 – 2( ) < p < .36 + 2( ) .36 – 2(.024) < p < .36 + 2(.024) . 312 < p < .408 31.2% < p < 40.8% EX: A new, medically approved wheelchair was judged to be an improvement over a standard wheelchair by 320 out of 400 patients in a survey. p = 320 400 = .8 p (1 – p) √ n Find a 99% confidence interval for p, the proportion of the wheelchair using population for whom the new wheelchair would be an improvement. p (1 – p) = .8 .2 √ n 400 =.02 p – 3 s < p < p + 3s .8 – 3( ) < p < .8 + 3( ) .8 – 3(.02)< p <.8 +3(.02) .74 < p < .86 74% < p < 86% If 72% of the wheel chair-using population has found that the standard wheelchair to be acceptable, does it appear that manufacturing the new wheelchair is a good idea? p (1 – p) √ n EX: a. If 50 out of 100 people surveyed favor brand X, find p and the standard deviation of p. p = 50 100 = .5 p (1 – p) = .5 .5 √ n 100 = .05 p – 2 s < p < p + 2s Find a 95% confidence interval of p, the proportion of the population favoring brand X. p – 2 s < p < p + 2s .5 – 2( ) < p < .5 + 2( ) .5 – 2(.05) < p < .5 + 2(.05) .4 < p < .6 40% < p < 60%