Determine whether there is any difference on the average for the two methods

# Determine whether there is any difference on the

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Determine whether there is any difference (on the average) for the two methods. Girder Karlsruhe Method Lehigh Method Difference d j S1/1 1.186 1.061 0.125 S2/1 1.151 0.992 0.159 S3/1 1.322 1.063 0.259 S4/1 1.339 1.062 0.277 S5/1 1.2 1.065 0.135 S2/1 1.402 1.178 0.224 S2/2 1.365 1.037 0.328 S2/3 1.537 1.086 0.451 S2/4 1.559 1.052 0.507 Sec 10-4 Paired t -Test
Example 10-11 Shear Strength of Steel Girder 30 The seven-step procedure is: 1. Parameter of interest: The parameter of interest is the difference in mean shear strength for the two methods. 2. Null hypothesis: H 0 : μ D = 0 3. Alternative hypothesis: H 1 : μ D 0 4. Test statistic: The test statistic is 5. Reject H 0 if: Reject H 0 if the P -value is <0.05. 6. Computations: The sample average and standard deviation of the differences d j are and s d = 0.1350, and so the test statistic is 7. Conclusions: Because t 0.0005.8 = 5.041 and the value of the test statistic t 0 = 6.15 exceeds this value, the P- value is less than 2(0.0005) = 0.001. Therefore, we conclude that the strength prediction methods yield different results. Interpretation: The data indicate that the Karlsruhe method produces, on the average, higher strength predictions than does the Lehigh method. n s d t d / 0 = 2769 . 0 = d 15 . 6 9 / 1350 . 0 2769 . 0 / 0 = = = n s d t d Sec 10-4 Paired t -Test
A Confidence Interval for μ D from Paired Samples 31 If and s D are the sample mean and standard deviation of the difference of n random pairs of normally distributed measurements, a 100(1 - α )% confidence interval on the difference in means 31 μ D = μ 1 - μ 2 is where t α /2, n -1 is the upper α /2% point of the t distribution with n - 1 degrees of freedom. d n s t d n s t d D n D D n / / 1 /2, 1 , /2 - α - α + μ - (10-25) Sec 10-4 Paired t -Test
Example 10-12 Parallel Park Cars 32 The journal Human Factors (1962, pp. 375–380) reported a study in which n = 14 subjects were asked to parallel park two cars having very different wheel bases and turning radii. The time in seconds for each subject was recorded and is given in Table 10-4. From the column of observed differences, we calculate d = 1.21 and s D = 12.68. Find the 90% confidence interval for μ D = μ 1 - μ 2. ( ) ( ) 21 . 7 79 . 4 14 / 68 . 12 771 . 1 21 . 1 14 / 68 . 12 771 . 1 21 . 1 / / μ - + μ - + μ - 0.05,13 0.05,13 D D D D D n s t d n s t d Notice that the confidence interval on μ D includes zero. This implies that, at the 90% level of confidence, the data do not support the claim that the two cars have different mean parking times μ 1 and μ 2 . That is, the value μ D = μ 1 - μ 2 = 0 is not inconsistent with the observed data. Subject 1( x 1 j ) 2( x 2 j ) ( d j ) 1 37.0 17.8 19.2 2 25.8 20.2 5.6 3 16.2 16.8 -0.6 4 24.2 41.4 -17.2 5 22.0 21.4 0.6 6 33.4 38.4 -5.0 7 23.8 16.8 7.0 8 58.2 32.2 26.0 9 33.6 27.8 5.8 10 24.4 23.2 1.2 11 23.4 29.6 -6.2 12 21.2 20.6 0.6 13 36.2 32.2 4.0 14 29.8 53.8 -24.0 Table 10-4 Sec 10-4 Paired t -Test
10-5.1 The F Distribution 10-5 Inferences on the Variances of Two Normal Populations We wish to test the hypotheses: 33 2 2 2 1 1 2 2 2 1 0 : : σ σ σ = σ H H Let W and Y be independent chi-square random variables with u and v degrees of freedom respectively. Then the ratio (10-28) has the probability density function (10-29) and is said to follow the distribution with u degrees of freedom in the numerator and v degrees of freedom in the denominator. It is usually abbreviated as F u,v .

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