252Y032app 3/25/03 Solutions for First Problem in Take-Home Part of Second Exam

252y032app 3/25/03 Solutions for first problem in Take-Home part of Second Exam

Variations on Solution: If your Social Security Number ended in 9. To summarize the information in the problem -

Defect Rate / <1 yr / 1-4 yr / 5-9yr / Total
High / 6 / 9 / 9 / 24
Average / 9 / 19 / 23 / 51
Low / 7 / 8 / 10 / 25
Total / 22 / 36 / 42 / 100

a) We are comparing , and .

Our we are testing . So the null hypothesis is

a) Let . So and our hypotheses become and . or and .

Note that and that and are between zero and one.

Use one of the following:

Confidence interval: Since the alternate hypothesis is , the confidence interval will be or . This does not contradict since any value of between 0 and .1412 satisfies both the null hypothesis and the confidence interval, so do not reject .

Test ratio: . Make a diagram of a Normal curve with zero in the middle. The ‘reject’ zone is the area below . Since the test ratio is not in this zone, do not reject .

Critical value: Because the alternate hypothesis is , we need a critical value below zero. Use Make a diagram of a Normal curve with zero in the middle. The ‘reject’ zone is the area below -.1574. Since is not in this zone, do not reject .

b) The p-value for this problem is Since this is not below do not reject .

c) or -.2031 to .1713

252y032app 3/25/03

d) . The proportions in rows, , are used with column totals to get the items in . is at the top of the page. Note that row and column sums in are the same as in except for a possible small rounding error. (Note that is computed two different ways here - only one way is needed.)

Row

1 6 5.28 -0.72 0.5184 0.098182 6.8182

2 9 11.22 2.22 4.9284 0.439251 7.2193

3 7 5.50 -1.50 2.2500 0.409091 8.9091

4 9 8.64 -0.36 0.1296 0.015000 9.3750

5 19 18.36 -0.64 0.4096 0.022309 19.6623

6 8 9.00 1.00 1.0000 0.111111 7.1111

7 9 10.08 1.08 1.1664 0.115714 8.0357

8 23 21.42 -1.58 2.4964 0.116545 24.6965

9 10 10.50 0.50 0.2500 0.023810 9.5238

Total 100 100 0.00 1.35101 101.3510

Since this is less than 9.4877, do not reject .

(Diagram!)

252y032app 3/25/03

Variations on Solution: If your Social Security Number ended in 8. To summarize the information in the problem -

Defect Rate / <1 yr / 1-4 yr / 5-9yr / Total
High / 6 / 9 / 8 / 23
Average / 9 / 19 / 23 / 51
Low / 7 / 8 / 10 / 25
Total / 22 / 36 / 41 / 99

a) We are comparing , and .

Our we are testing . So the null hypothesis is

a) Let . So and our hypotheses become and . or and .

Note that and that and are between zero and one.

Use one of the following:

Confidence interval: Since the alternate hypothesis is , the confidence interval will be or . This does not contradict since any value of between 0 and .1379 satisfies both the null hypothesis and the confidence interval, so do not reject .

Test ratio: . Make a diagram of a Normal curve with zero in the middle. The ‘reject’ zone is the area below . Since the test ratio is not in this zone, do not reject .

b) The p-value for this problem is Since this is not below do not reject .

c) or -.2048 to .1730

252y032app 3/25/03

d) The proportions in rows, , are used with column totals to get the items in . is at the top of the page. Note that row and column sums in are the same as in except for a possible small rounding error. (Note that is computed two different ways here - only one way is needed.)

Note slight rounding error.

Row

1 6 5.11 -0.89 0.7921 0.155010 7.0450

2 9 11.33 2.33 5.4289 0.479162 7.1492

3 7 5.56 -1.44 2.0736 0.372950 8.8129

4 9 8.36 -0.64 0.4096 0.048995 9.6890

5 19 18.55 -0.45 0.2025 0.010916 19.4609

6 8 9.09 1.09 1.1881 0.130704 7.0407

7 8 9.52 1.52 2.3104 0.242689 6.7227

8 23 21.12 -1.88 3.5344 0.167348 25.0473

9 10 10.35 0.35 0.1225 0.011836 9.6618

Total 99 98.99 0.00 1.61961 100.6296

or 1.619

Since this is less than 9.4877, do not reject .

(Diagram!)

252y032app 3/25/03

Variations on Solution: If your Social Security Number ended in 7. To summarize the information in the problem -

Defect Rate / <1 yr / 1-4 yr / 5-9yr / Total
High / 6 / 9 / 7 / 22
Average / 9 / 19 / 23 / 51
Low / 7 / 8 / 10 / 25
Total / 22 / 36 / 40 / 98

a) We are comparing , and .

Our we are testing . So the null hypothesis is

a) Let . So and our hypotheses become and . or and .

Note that and that and are between zero and one.

Use one of the following:

Confidence interval: Since the alternate hypothesis is , the confidence interval will be or . This does not contradict since any value of between 0 and .1324 satisfies both the null hypothesis and the confidence interval, so do not reject .

Test ratio: . Make a diagram of a Normal curve with zero in the middle. The ‘reject’ zone is the area below . Since the test ratio is not in this zone, do not reject .

b) The p-value for this problem is Since this is not below do not reject .

c) or -.2287 to .1631

252y032app 3/25/03

Row

1 6 4.94 -1.06 1.1236 0.227449 7.2874

2 9 11.45 2.45 6.0025 0.524236 7.0742

3 7 5.61 -1.39 1.9321 0.344403 8.7344

4 9 8.08 -0.92 0.8464 0.104752 10.0248

5 19 18.73 -0.27 0.0729 0.003892 19.2739

6 8 9.19 1.19 1.4161 0.154091 6.9641

7 7 8.98 1.98 3.9204 0.436570 5.4566

8 23 20.82 -2.18 4.7524 0.228261 25.4083

9 10 10.20 0.20 0.0400 0.003922 9.8039

Total 98 98 0.00 2.02758 102.0276

Since this is less than 9.4877, do not reject .

(Diagram!)

252y032app 3/25/03

Variations on Solution: If your Social Security Number ended in 6. To summarize the information in the problem -

Defect Rate / <1 yr / 1-4 yr / 5-9yr / Total
High / 6 / 9 / 6 / 21
Average / 9 / 19 / 23 / 51
Low / 7 / 8 / 10 / 25
Total / 22 / 36 / 39 / 97

a) We are comparing , and .

Our we are testing . So the null hypothesis is

a) Let . So and our hypotheses become and . or and .

Note that and that and are between zero and one.

Use one of the following:

Confidence interval: Since the alternate hypothesis is , the confidence interval will be or . This does not contradict since any value of between 0 and .1619 satisfies both the null hypothesis and the confidence interval, so do not reject .

Test ratio: . Make a diagram of a Normal curve with zero in the middle. The ‘reject’ zone is the area below . Since the test ratio is not in this zone, do not reject .

b) The p-value for this problem is Since this is not below do not reject .

c) or -.2271 to .1587

252y032app 3/25/03

Row

1 6 4.76 -1.24 1.5376 0.323025 7.5630

2 9 11.57 2.57 6.6049 0.570864 7.0009

3 7 5.67 -1.33 1.7689 0.311975 8.6420

4 9 7.79 -1.21 1.4641 0.187946 10.3979

5 19 18.93 -0.07 0.0049 0.000259 19.0703

6 8 9.28 1.28 1.6384 0.176552 6.8966

7 6 8.44 2.44 5.9536 0.705403 4.2654

8 23 20.51 -2.49 6.2001 0.302296 25.7923

9 10 10.05 0.05 0.0025 0.000249 9.9502

Total 97 97 0.00 2.57857 99.5786

Since this is less than 9.4877, do not reject .

(Diagram!)

252y032app 3/25/03

Variations on Solution: If your Social Security Number ended in 5. To summarize the information in the problem -

Defect Rate / <1 yr / 1-4 yr / 5-9yr / Total
High / 6 / 9 / 5 / 20
Average / 9 / 19 / 23 / 51
Low / 7 / 8 / 10 / 25
Total / 22 / 36 / 38 / 96

a) We are comparing , and .

Our we are testing . So the null hypothesis is

a) Let . So and our hypotheses become and . or and .

Note that and that and are between zero and one.

Use one of the following:

Confidence interval: Since the alternate hypothesis is , the confidence interval will be or . This does not contradict since any value of between 0 and .1412 satisfies both the null hypothesis and the confidence interval, so do not reject .

Test ratio: . Make a diagram of a Normal curve with zero in the middle. The ‘reject’ zone is the area below . Since the test ratio is not in this zone, do not reject .

Critical value: Because the alternate hypothesis is , we need a critical value below zero. Use Make a diagram of a Normal curve with zero in the middle. The ‘reject’ zone is the area below -.1642. Since is not in this zone, do not reject .

b) The p-value for this problem is Since this is not below do not reject .

c) or -.2365 to .1547

252y032app 3/25/03

d) . The proportions in rows, , are used with column totals to get the items in . is at the top of the page. Note that row and column sums in are the same as in except for a possible small rounding error. (Note that is computed two different ways here - only one way is needed.)

Note small rounding error.

Row

1 6 4.58 -1.42 2.0164 0.44026 7.8603

2 9 11.69 2.69 7.2361 0.61900 6.9290

3 7 5.73 -1.27 1.6129 0.28148 8.5515

4 9 7.50 -1.50 2.2500 0.30000 10.8000

5 19 19.13 0.13 0.0169 0.00088 18.8709

6 8 9.37 1.37 1.8769 0.20031 6.8303

7 5 7.92 2.92 8.5264 1.07657 3.1566

8 23 20.19 -2.81 7.8961 0.39109 26.2011

9 10 9.90 -0.10 0.0100 0.00101 10.1010

Total 96 96.01 0.01 3.30060 99.3006

or 3.311

Since this is less than 9.4877, do not reject .

(Diagram!)

252y032app 3/25/03

Variations on Solution: If your Social Security Number ended in 4. To summarize the information in the problem -

Defect Rate / <1 yr / 1-4 yr / 5-9yr / Total
High / 6 / 9 / 4 / 19
Average / 9 / 19 / 23 / 51
Low / 7 / 8 / 10 / 25
Total / 22 / 36 / 37 / 95

a) We are comparing , and .

Our we are testing . So the null hypothesis is

a) Let . So and our hypotheses become and . or and .

Note that and that and are between zero and one.

Use one of the following:

Confidence interval: Since the alternate hypothesis is , the confidence interval will be or . This does not contradict since any value of between 0 and .1176 satisfies both the null hypothesis and the confidence interval, so do not reject .