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# linear regression mean inference vs individual prediction

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• Assumptions: quantitative data from incr. normal popn as sample gets smaller, r. sampleParticular form of the test statisticTwo-tailed test: effect on the p-value (have to double tail prob) or rejection region (have to take given alpha &amp; divide it between two tails)
• ### linear regression mean inference vs individual prediction

1. 1. STA291Statistical MethodsLecture 29
2. 2. SE, and the confidence interval, becomes smallerwith increasing n.SE, and the confidence interval, are larger forsamples with more spread around the line (when seis larger).Standard Errors for Mean ValuesnsxxbSESE evv2212ˆvnv SEty ˆˆ *2Confidence Interval for the Mean ResponseLast time, we said we were modeling our line toinfer the “line of means”—the expected value ofour response variable for each given value of theexplanatory variable.The confidence interval for the meanresponse, v, at a value xv, is:where:
3. 3. Standard Errors for Mean ValuesnsxxbSESE evv2212ˆvnv SEty ˆˆ *2Confidence Interval for the Mean ResponseThe confidence interval for the meanresponse, v, at a value xv, is:where:SE becomes larger the furtherxν gets from . That is, theconfidence interval broadensas you move away from .(See figure at right.)xx
4. 4. Standard Errors for Predicted ValuesBecause of the extra term , theconfidence interval for individualvalues is broader that those for thepredicted mean value.2esPrediction Interval for an Individual ResponseNow, we tackle the more difficult (as far asadditional variability) of predicting a single value ata value xv. When conditions are met, that intervalis:where:ˆyv ±tn-2*´SE ˆyv( )SE ˆyv( )= SE2b1( )´ xv - x( )2+se2n+ se2
5. 5. Difference Between Confidence andPrediction IntervalsConfidence interval for a mean:The result at 95% means:“We are 95% confident that the mean valueof y is between 4.40 and 4.70 when x =10.1.”ˆ 10.1 4.55 0.15nsxxbSEty evnv2212*2ˆ
6. 6. Prediction interval for an individual value:The result at 95% means:“We are 95% confident that a singlemeasurement of y will be between 3.95and 5.15 when x = 10.1.”ˆ 10.1 4.55 0.60yDifference Between Confidence andPrediction Intervals22212*2ˆ eevnv snsxxbSEty
7. 7. Using Confidence and PredictionIntervalsExample : External Hard DisksA study of external disk drives reveals a linear relationshipbetween the Capacity (in GB) and the Price (in \$).Regression resulted in the following:Price = 18.64 + 0.104Capacityse = 17.95, and SE(b1) = 0.0051Find the predicted Price of a 1000 GB hard drive.Find the 95% confidence interval for the mean Price of all1000 GB hard drives.Find the 95% prediction interval for the Price of one 1000 GBhard drive.
8. 8. Example : External Hard DisksA study of external disk drives reveals a linearrelationship between the Capacity (in GB) and the Price(in \$). Regression resulted in the following:Price = 18.64 + 0.104Capacityse = 17.95, and SE(b1) = 0.0051Find the predicted Price of a 1000 GB hard drive.Using Confidence and PredictionIntervalsPrice = 18.64 + 0.104(1000) = 122.64
9. 9. Example : External Hard DisksA study of external disk drives reveals a linearrelationship between the Capacity (in GB) and the Price(in \$). Regression resulted in the following:Price = 18.64 + 0.104Capacityse = 17.95, and SE(b1) = 0.0051Find the 95% confidence interval for the mean Price ofall 1000 GB hard drives.Using Confidence and Prediction Intervals14.140,\$14.105\$50.17\$64.122\$795.17111010000051.0571.264.122\$ˆ2222212*2nsxxbSEty evnv
10. 10. Example : External Hard DisksA study of external disk drives reveals a linearrelationship between the Capacity (in GB) and the Price(in \$). Regression resulted in the following:Price = 18.64 + 0.104Capacityse = 17.95, and SE(b1) = 0.0051Find the 95% prediction interval for the price of one1000 GB hard drive.Using Confidence and Prediction Intervalsˆyv ±tn-2*SE2b1( )´ xv - x( )2+se2n+ se2= \$122.64±2.571 0.00512´ 1000 -1110( )2+17.9527+17.952= \$122.64±\$49.36 = 73.28,172.00[ ]. CI: \$105.14,\$140.14[ ]
11. 11. Looking backo Construct and interpret aconfidence interval for the meanvalueo Construct and interpret aprediction interval for an individualvalue
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