• View
14

0

Embed Size (px)

DESCRIPTION

Stats 333 NJIT

### Text of 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

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

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

Variable N Mean StDev SE Mean 95% CI T P9-63 25 98.2640 0.4821 0.0964 (98.0650, 98.4630) -3.48 0.002

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

Testing mean = null (versus null)Calculating power for mean = null + difference = 0.05 Assumed standard deviation = 0.4821

SampleDifference Size Power -0.6 25 0.999968

Sample Size 9-63 d

1-Sample t Test

Testing mean = null (versus null)Calculating power for mean = null + difference = 0.05 Assumed standard deviation = 0.4821

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

Variable N Mean StDev SE Mean 95% CI T P9-65 20 129.747 0.876 0.196 (129.337, 130.157) -1.29 0.212

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

Testing mean = null (versus null)Calculating power for mean = null + difference = 0.05 Assumed standard deviation = 0.876

SampleDifference Size Power 0.5 20 0.678000

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

1-Sample t Test

Testing mean = null (versus null)Calculating power for mean = null + difference = 0.05 Assumed standard deviation = 0.876

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

Recommended ##### Physics 590 Homework week 6 - Penn Arts & kennethp/ Physics 590 Homework, Week 6 Week 6, Homework
Documents ##### December Homework- Week 1 - Santee School ... December Homework- Week 1 Page 3: Cover Page for Homework
Documents ##### Web view Year 7 Poetry. Homework Booklet. Homework week 1 â€“ Similes worksheet page 3. Homework week
Documents ##### Homework Package Answers - .Strathaven Academy 4 April, 2009 Formal Homework Answers 2 of 20 HOMEWORK MARKING SCHEME Induction - Uncertainties & Significant Figures Homework 2 -
Documents Documents