Week #7 Homework Answers

Stats 333 NJITWeek #7 Homework Answers9-59 a) n 1 or 9 degrees of freedomb) StDev = * .296 = .936 t = 1.9255p = .089 at = .05 results are not significant.c) Two sided test.d) 12.564 2.262 * .296 = (11.89, 13.23)e) = 1.905, at = .05 the critical t value is 1.833 so still reject H0f) Yes because 11.5 is not in the range of the C.I.9-62b)

One-Sample T: 9-62 a and e

Test of = 22.5 vs 22.5

Variable N Mean StDev SE Mean 95% CI T P9-62 5 22.496 0.378 0.169 (22.026, 22.966) -0.02 0.982

Confidence interval includes the 22.5, so failure to reject the null hypothesis not equal Power 9-62 c

1-Sample t Test

Testing mean = null (versus null)Calculating power for mean = null + difference = 0.05 Assumed standard deviation = 0.378

SampleDifference Size Power 0.25 5 0.208794

Sample Size 9-62 d

1-Sample t Test


Sample TargetDifference Size Power Actual Power 0.25 27 0.9 0.911040

9-63 b)

One-Sample T: 9-63 a & e

Test of = 98.6 vs 98.6


e) Since 98.6 is outside the confidence interval the results reject the null hypothesis that the mean female temperature is 98.6

Power and Sample Size 9-63 c1-Sample t Test


SampleDifference Size Power -0.6 25 0.999968

Sample Size 9-63 d

1-Sample t Test


Sample TargetDifference Size Power Actual Power -0.4 18 0.9 0.9123479-64 b

Notice I used a 98% confidence interval since = .01 and is a one sided test.One-Sample T: 9-64 a & e

Test of = 25 vs > 25

Variable N Mean StDev SE Mean 99%LowerBound T P9-64 20 26.03 4.78 1.07 23.32 0.97 0.173

Since the hypothesized mean of 25 is greater than the lower bound, we fail to reject the null hypothesis of = 25.

Power with sample size 20 9-64 c

1-Sample t Test

Testing mean = null (versus > null)Calculating power for mean = null + difference = 0.01 Assumed standard deviation = 4.78

SampleDifference Size Power 2 20 0.278664

Power > .9 and Sample Size 9-64d

1-Sample t Test

Testing mean = null (versus > null)Calculating power for mean = null + difference = 0.01 Assumed standard deviation = 4.78

Sample TargetDifference Size Power Actual Power 2.5 51 0.9 0.9043259-65 b

One-Sample T: 9-65 a & e

Test of = 130 vs 130


Confidence interval includes the hypothesized mean, so results are not significant and the weight of the boxes conforms to the standards.

Power and Sample Size of 20 9-65c

1-Sample t Test


SampleDifference Size Power 0.5 20 0.678000

Power > .75 and required Sample Size 9-65d

1-Sample t Test


Sample TargetDifference Size Power Actual Power 0.1 535 0.75 0.750388

9-90a) & b) two sided. Sample p = .356z = -1.477P-value = .13964

c)Yes, n * p and n * (1 p) > 5

9-91a) one-sided testb) Normal approximation since n * p and n * (1 p) > 5c) Sample p = .574. Upper bound = .61037. z = -1.1867. p = .118

d) 2 * p-value = .2369-100 a) z = = 1.389. P-value = .08235 Type I error

b) first calculate the z value of .75 versus the critical value of .63 = 8.074 which gives you a of approx. 0.9-101 a) H0 = .10 H1 > .10Z = .542, p-value = .294. Fail to reject the null hypothesis at = .05b) Calculate the , if in fact the p = .15Power with given Sample Size of 85 for 9-101b

Test for One Proportion

Testing p = 0.1 (versus > 0.1) = 0.05

SampleComparison p Size Power 0.15 85 0.463763

Beta = 1 - power. Beta = (1 - .4637) = .5363c) Power > .90 required Sample Size 101-c

Test for One Proportion

Testing p = 0.1 (versus > 0.1) = 0.05

Sample TargetComparison p Size Power Actual Power 0.15 362 0.9 0.900124

10-6 a) Testing 1 = 2 versus 1 2

Z = = -6.324. p value approx. 0b) -6 1.96 * = (-7.859, -4.141). Since 0 is not included in the confidence interval, we say the propellants are from different populations.c) Using the formula for on page 378 - = =

d) = 3

10-7 a) z0 = 7.25 > 1.645 reject H0 b) 89.6 92.5 1.96 * = (-3.684, -2.116)Which does not include 0.So therefore there is a difference.c) so sample size of 11.

10-8 a) = 750.2 = 756.875Z0 = = -.3797Since -1.28 < -.3797 the populations are not significantly different. P-value = .352

b) Confidence interval: -750.2 + 756.875 1.645 * = c)(-7.73, 21.08), since the confidence interval does include 10. The difference in the batch means is not greater than 10.

10-9 a) 65.22 68.42 1.96 * = (-5.83, -.5704).Z0 = = -2.385. P-value = .0171b) Since the difference in the mean values of the catalysts are significant, the mean active concentrations depend on the choice of catalyst.

c) = } - } = = .0386So the power is 1 - .0386 = .9614d) Since the power of the test > .9, the samples size is sufficientNormality seems reasonable.

10-50 Using Minitab:Paired T-Test and CI: Table 10-50(1), Table 10-50(2)

Paired T for Table 10-50(1) - Table 10-50(2)

N Mean StDev SE MeanTable 10-50(1) 14 29.23 10.36 2.77Table 10-50(2) 14 28.01 10.84 2.90Difference 14 1.21 12.68 3.39

95% CI for mean difference: (-6.11, 8.54)T-Test of mean difference = 0 (vs 0): T-Value = 0.36 P-Value = 0.726

Since the confidence interval includes 0, supports the p- value of the paired t-test.

10-51 Graphing calculator:Differences:26073020740150-805560390285t = 1.9039p-value = .0986 Confidence Interval of 99% = (-727.7, 2464.5)Since Confidence interval includes 0, no preference of brands.

10-52 using graphing calculatorDifferences: Performing a t-test of the differences. -1223-53612-1-2t = .779p-value = .45295% C.I. = (-1.217, 25502) Since the confidence interval includes 0, there is no preferable design language.Paired T-Test and CI: 10-52

In Minitab: N Mean StDev SE MeanC12 12 17.92 3.63 1.05C13 12 17.25 4.59 1.33Difference 12 0.667 2.964 0.856

95% CI for mean difference: (-1.217, 2.550)T-Test of mean difference = 0 (vs 0): T-Value = 0.78 P-Value = 0.452

10-61a) Paired t-test not appropriate as each group is independent. You perform a 2 sample t-test.Using graphing calculator: t = 3.357p-value = .0035P-value of .0035 < .01. Results are significant and non-confined has higher brain wave activity than confined.

11-5 Using Minitab:

Regression Equation

Rating = 14.18 +10.09yds/att.

2 = = Error 30 819.50 27.32Variance is the ratio of the sum of squared differences from the prediction equation to n 2 degrees of freedom. /(n 2)b) 14.18 + 10.09 * 7.5 = 90.6c)-10.09d) = .99e) Fitted value of 7.21 = 14.18+10.09*7.21= 86.95Residuals = 8.05 and 3.25, since there were two values of 7.21

Fits and Diagnostics for All Observations

Obs Rating Fit Resid Std Resid 1 105.50 98.86 6.64 1.38 2 97.40 91.59 5.81 1.15 3 96.90 91.49 5.41 1.07 4 96.20 94.72 1.48 0.30 5 95.00 86.95 8.05 1.57 6 93.80 90.18 3.62 0.71 7 92.70 95.02 -2.32 -0.47 8 91.40 91.49 -0.09 -0.02 9 90.20 86.95 3.25 0.63 10 89.40 86.44 2.96 0.58 11 87.70 94.21 -6.51 -1.30 12 87.50 85.84 1.66 0.32 13 87.00 78.07 8.93 1.76 14 86.40 82.41 3.99 0.78 15 86.40 83.42 2.98 0.58 16 86.00 88.36 -2.36 -0.46 17 85.40 87.05 -1.65 -0.32 18 84.70 94.32 -9.62 -1.92 19 84.30 78.87 5.43 1.07 20 81.70 82.51 -0.81 -0.16 21 81.00 81.30 -0.30 -0.06 22 80.00 84.22 -4.22 -0.82 23 80.20 79.28 0.92 0.18 24 80.10 85.23 -5.13 -1.00 25 79.60 78.67 0.93 0.18 26 77.10 80.59 -3.49 -0.68 27 76.00 76.86 -0.86 -0.17 28 73.70 86.54 -12.84 -2.50 R 29 72.60 78.17 -5.57 -1.10 30 71.40 76.55 -5.15 -1.02 31 70.00 65.85 4.15 0.91 X 32 66.50 71.81 -5.31 -1.09

R Large residualX Unusual X

11-6 Regression Analysis: selling price/1000 versus taxes pd/1000 11-6

Analysis of Variance

Source DF Adj SS Adj MS F-Value P-ValueRegression 1 636.2 636.156 72.56 0.000 taxes pd/1000 1 636.2 636.156 72.56 0.000Error 22 192.9 8.768 = 2Total 23 829.0

Model Summary

S R-sq R-sq(adj) R-sq(pred)2.96104 76.73% 75.68% 72.91%

Coefficients

Term Coef SE Coef T-Value P-Value VIFConstant 13.32 2.57 5.18 0.000taxes pd/1000 3.324 0.390 8.52 0.000 1.00

Regression Equation

Selling price/1000 = 13.32 +3.324taxespd/1000


sellingObs price/1000 Fit Resid Std Resid 1 25.900 29.668 -3.768 -1.33 2 29.500 30.011 -0.511 -0.18 3 27.900 28.422 -0.522 -0.19 4 25.900 28.470 -2.570 -0.92 5 29.900 30.141 -0.241 -0.08 6 29.900 26.255 3.645 1.34 7 30.900 32.927 -2.027 -0.70 8 28.900 31.950 -3.050 -1.06 9 35.900 32.695 3.205 1.11 10 31.500 30.940 0.560 0.20 11 31.000 34.168 -3.168 -1.09 12 30.900 33.131 -2.231 -0.77 13 30.000 30.108 -0.108 -0.04 14 36.900 40.734 -3.834 -1.37 15 41.900 35.583 6.317 2.18 R 16 40.500 39.197 1.303 0.46 17 43.900 43.367 0.533 0.20 18 37.500 33.231 4.269 1.47 19 37.900 38.393 -0.493 -0.17 20 44.500 42.558 1.942 0.71 21 37.900 33.543 4.357 1.50 22 38.900 41.114 -2.214 -0.79 23 36.900 40.381 -3.481 -1.23 24 45.800 43.710 2.090 0.78 Y R Large residual

b) 13.32 +3.324* 7.50 = 38.25 thousandsc) fixed selling price is 30.9 * 1000 with a residual of -2.231 * 1000 see highlighted #12 above.d) See graph above. Obviously taxes paid is not the only factor of sale price of a house. There are market conditions such as interest rates and the general economy as other factors, but taxes paid is a good predictor of house sale price.

11-12 a) Yes

b) Regression Analysis: y versus x 11-12


Source DF Adj SS Adj MS F-Value P-ValueRegression 1 1273.54 1273.54 92.22 0.000 x 1 1273.54 1273.54 92.22 0.000Error 16 220.95 13.81 Lack-of-Fit 14 216.44 15.46 6.86 0.134 Pure Error 2 4.51 2.25Total 17 1494.49

Model Summary


Coefficients

Term Coef SE Coef T-Value P-Value VIFConstant 0.47 1.94 0.24 0.811x 20.57 2.14 9.60 0.000 1.00

Regression Equation

y = 0.47 +20.57x

Error 16 220.95 13.81 = 2


Obs y Fit Resid Std Resid 1 4.40 4.38 0.02 0.01 2 6.60 3.56 3.04 0.92 3 9.70 12.19 -2.49 -0.70 4 10.60 14.87 -4.27 -1.18 5 10.80 14.25 -3.45 -0.96 6 10.90 13.43 -2.53 -0.70 7 11.80 10.14 1.66 0.47 8 12.10 14.87 -2.77 -0.77 9 14.30 12.81 1.49 0.42 10 14.70 16.51 -1.81 -0.50 11 15.00 17.13 -2.13 -0.59 12 17.30 16.51 0.79 0.22 13 19.20 14.66 4.54 1.26 14 23.10 27.21 -4.11 -1.19 15 27.40 22.07 5.33 1.49 16 27.70 22.27 5.43 1.52 17 31.80 36.26 -4.46 -1.48 X 18 39.50 33.79 5.71 1.81

X Unusual X

c) Using the regression equation the fit for watershed that has a .01 roadway area is:0.47 + 20.57 * 1% = 21.04

d) The fitted value is in obs #7. Y = 11.8 and the residual is 1.66.47 + 20.57 * .47 = 10.14. The residual is 11.8 10.14 = 1.66

11-13 a) b) Regression Analysis: Strength versus Age 11-13


Source DF Adj SS Adj MS F-Value P-ValueRegression 1 1522819 1522819 155.21 0.000 Age 1 1522819 1522819 155.21 0.000Error 18 176602 9811Total 19 1699421

Model Summary


Coefficients

Term Coef SE Coef T-Value P-Value VIFConstant 2625.4 45.3 57.90 0.000Age -36.96 2.97 -12.46 0.000 1.00

Regression Equation

Strength = 2625.4 -36.96Age


Obs Strength Fit Resid Std Resid 1 2158.7 2052.5 106.2 1.10 2 1678.2 1747.5 -69.4 -0.76 3 2316.0 2329.7 -13.7 -0.14 4 2061.3 1997.0 64.3 0.67 5 2207.5 2440.6 -233.1 -2.50 R 6 1708.3 1923.1 -214.8 -2.26 R 7 1784.7 1738.3 46.4 0.51 8 2575.0 2533.0 42.0 0.46 9 2357.9 2348.2 9.7 0.10 10 2277.7 2218.8 58.9 0.61 11 2165.2 2144.9 20.3 0.21 12 2399.6 2486.8 -87.2 -0.95 13 1779.8 1701.3 78.5 0.87 14 2336.8 2265.0 71.7 0.75 15 1765.3 1812.2 -46.9 -0.50 16 2053.5 1960.1 93.4 0.98 17 2414.4 2403.6 10.8 0.11 18 2200.5 2163.4 37.1 0.38 19 2654.2 2551.5 102.7 1.14 20 1753.7 1830.7 -77.0 -0.82

R Large residual

Regression EquationStrength = 2625.4 -36.96Age2 = Error 18 176602 9811. Variance is a measure of error: The squared difference of the actual value predicted value / n 2 degrees of freedom.176602/18 = 9811, the variancec) 2625.4 36.96 * 20 = 1886.2d)

11-17 c) There are two potential points of interest.

a) Error 25 0.63329 0.02533. Variance is the ratio of the sum of squared differences of the actual values of y to the regression fit values of y divided by n 2 degrees of freedom = .02533Regression Equation

y = 2.0198 +0.02872x b) The predicted value age = 11:2.0198 + .02872 * 11 = 2.34

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Week #7 Homework Answers