Example (which tire lasts longer?)

• To determine whether a new steel-belted radial tire

lasts longer than a current model, the manufacturer

designs the following experiment.

– A pair of newly designed tires are installed on the rear

wheels of 20 randomly selected cars.

– A pair of currently used tires are installed on the rear

wheels of another 20 cars.

– Drivers drive in their usual way until the tires wear out.

– The number of miles driven by each driver were recorded.

See data next.

Matched Pairs ExperimentMatched Pairs Experiment

Solution• Compare two populations of

quantitative data.

• The parameter is 1 - 2

New-Design Exstng-Dsn70 4783 6578 5946 6174 7556 6574 7352 8599 9757 8477 7284 3972 7298 9181 6463 6388 7969 7454 7697 43

1

2

The hypotheses are:H0: (1 - 2) = 0Ha: (1 - 2) > 0

Mean distance driven before worn out occurs for the new design tires

Mean distance driven before worn out occurs for the existing design tires

• The hypotheses areH0: 1 - 2 = 0

H1: 1 - 2 > 0

The test statistic is

We run the t test, and

obtain the following

Excel results.

)n1

n1

(s

)(xxt

11

2p

2121

t-Test: Two-Sample Assuming Equal Variances

New Dsgn Exstng dsgnMean 73.6 69.2Variance 243.4105263 226.8Observations 20 20Pooled Variance 235.1052632Hypothesized Mean Difference0df 38t Stat 0.907447484P(T<=t) one-tail 0.184944575t Critical one-tail 1.685953066P(T<=t) two-tail 0.36988915t Critical two-tail 2.024394234

t-Test: Two-Sample Assuming Equal Variances

New Dsgn Exstng dsgnMean 73.6 69.2Variance 243.4105263 226.8Observations 20 20Pooled Variance 235.1052632Hypothesized Mean Difference0df 38t Stat 0.907447484P(T<=t) one-tail 0.184944575t Critical one-tail 1.685953066P(T<=t) two-tail 0.36988915t Critical two-tail 2.024394234

We conclude that there is insufficient evidence to reject H0 in favor of H1.

• Example continued (using the matched pairs approach)

• to eliminate variability among observations within each sample the experiment was redone.– One tire of each type was installed

on the rear wheel of 20 randomly selected cars (each car was sampled twice, thus creating a pair of observations).

– The number of miles until wear-out was recorded

Car New-Dsn Exst-Dsn1 57 482 64 503 102 894 62 565 81 786 87 757 61 508 62 499 74 7010 62 6611 100 9812 90 8613 83 7814 84 9015 86 9816 62 5817 67 5818 40 4119 71 6120 77 82

t-Test: Paired Two Sample for Means

New-Dsn Exst-DsnMean 73.6 69.05Variance 242.779 316.366Observations 20 20Pearson Correlation 0.91468Hypothesized Mean Difference 0df 19t Stat 2.81759P(T<=t) one-tail 0.0055t Critical one-tail 1.72913P(T<=t) two-tail 0.01099t Critical two-tail 2.09302

t-Test: Paired Two Sample for Means

New-Dsn Exst-DsnMean 73.6 69.05Variance 242.779 316.366Observations 20 20Pearson Correlation 0.91468Hypothesized Mean Difference 0df 19t Stat 2.81759P(T<=t) one-tail 0.0055t Critical one-tail 1.72913P(T<=t) two-tail 0.01099t Critical two-tail 2.09302

Matched Pairs ExperimentMatched Pairs Experiment

• Solving by hand– Calculate the difference for each pair.– Calculate the average differences and the

standard deviation of the differences.– Build the statistics as follows:

– Run the hypothesis test using t distribution with nD - 1 degrees of freedom.

ns

xt

D

DD

ns

xt

D

DD

– The hypotheses test for this problem is

H0: D = 0

H1: D > 0

The statistic is

817.2

2022186.7

055.4

ns

μxt

DD

DD

New-Dsn Exst-Dsn Difference57 48 964 50 14

102 89 1362 56 681 78 387 75 1261 50 1162 49 1374 70 462 66 -4

100 98 290 86 483 78 584 90 -686 98 -1262 58 467 58 940 41 -171 61 1077 82 -5

Average = 4.55Standard Deviation = 7.22186

The rejection region is:t > t with d.f. = 20-1 = 19.If = .05, t.05,19 = 1.729.

Since 2.817 > 1.729, thereis sufficient evidence in the datato reject the null hypothesis infavor of the alternative hypothesis.

Conclusion: At 5% significancelevel the new type tires last longer than the current type.

See file Tires.xls

3.384.5520

7.222.0934.55is12.4examplein

differencemeantheofintervalconfidence95%The

ns

tx

μofEstimatorintervalConfidence

D

D1nα/2,D

D

Estimating the mean difference

Checking the required conditionsfor the paired observations case

• The validity of the results depends on the normality of the differences.

0

2

4

6

8

-12 -6 0 6 12 More

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