CHAPTER 9 – HYPOTHESIS TESTING AND ESTIMATION FOR …



CHAPTER 9 – HYPOTHESIS TESTING AND ESTIMATION FOR TWO POPULATION VALUES

1.

a. If the calculated F > 4.405, reject Ho, otherwise do not reject Ho

b. F = 232/192 = 1.4654

Since 1.4654 < 4.405 do not reject Ho

2.

a. If the calculated F > 2.255, reject Ho, otherwise do not reject Ho

b. F = 2302/2102 = 1.1995

Since 1.1995 < 2.255 do not reject Ho

3. s1 = 2.8975 s2 = 2.7033

H0: σ12 = σ22

HA: σ12 ≠ σ22

If the calculated F > 3.179, reject Ho, otherwise do not reject Ho

F = 2.89752/2.70332 = 1.1488

Since 1.1488 < 3.179 do not reject Ho

4.

a. H0: σd2 = σw2

HA: σd2 ≠ σw2

b. If the calculated F > 3.284, reject Ho, otherwise do not reject Ho

F = 22/1.22 = 2.7778

Since 2.7778 < 3.284 do not reject Ho and conclude that there is no difference in the standard deviations

5.

a. H0: σm2 < σf2

HA: σm2 > σf2

b. If the calculated F > 1.984, reject Ho, otherwise do not reject Ho

F = 2.52/1.342 = 3.4807

Since 3.4807 > 1.984 reject Ho and conclude that there is more variability in male donations than in female donations.

6. sA = 7.1375 sB = 8.6929

H0: σA2 = σB2

HA: σA2 ≠ σB2

If the calculated F > 3.0274, reject Ho, otherwise do not reject Ho

F = 8.69292/7.13752 = 1.4833

Since 1.4833 < 3.0274 do not reject Ho and conclude that there is no difference in the standard deviation of dollars returned between the two brochures.

7.

| | | |

|Sales Plan |Data |Total |

|Basic Plan |Count of Cell Phone Minutes |52.000 |

|  |StdDev of Cell Phone Minutes |34.434 |

|Business Plan |Count of Cell Phone Minutes |148.0000 |

|  |StdDev of Cell Phone Minutes |29.9187 |

a. H0: σBasic2 < σBus.2

HA: σBasic2 > σBus2

b. If the calculated F > 1.4341, reject Ho, otherwise do not reject Ho

F = 34.4342/29.91872 = 1.3246

Since 1.3246 < 1.4341 do not reject Ho and conclude that the standard deviation in minutes used by the Business Plan is not less than the Basic Plan

8. df = 100 + 120 – 2 = 218

a. If t > 1.96 or t < -1.96 reject Ho, otherwise do not reject Ho

b. Since the degrees of freedom are not available in the F table you must use Excel’s FINV function to find the critical F. If the calculated F > 1.4658, reject Ho, otherwise do not reject Ho

F = 242/202 = 1.44

Since 1.44 < 1.4658 do not reject Ho and conclude that the variances are equal

c. sp = [pic] = 22.2727

t = (430 – 405)/(22.2727[pic]) = 8.2898

Since 8.2898 > 1.96 reject Ho

9.

a. If the calculated F > 2.108, reject Ho, otherwise do not reject Ho

b. H0: σ12 = σ22

HA: σ12 ≠ σ22

If the calculated F > 2.108 (using α = 0.10, critical value for α = 0.05 not available in tables), reject Ho, otherwise do not reject Ho

F = 322/302 = 1.1378

Since 1.1378 < 2.108 do not reject Ho and conclude that the variances are equal

10.

a. Ho: μd < 0

Ha: μd > 0

b.

|Sample 1 |Sample 2 |Difference (d) |(d - [pic])2 |

|50 |38 |12 |36 |

|47 |44 |3 |9 |

|44 |38 |6 |0 |

|48 |37 |11 |25 |

|40 |43 |-3 |81 |

|36 |44 |-8 |196 |

|43 |31 |12 |36 |

|46 |38 |8 |4 |

|72 |39 |33 |729 |

|40 |54 |-14 |400 |

|55 |41 |14 |64 |

|38 |40 |-2 |64 |

| | |72 |1644 |

[pic]= 72/12 = 6

sd = [pic] = 12.2252

t = (6 – 0)/(12.2252/[pic]) = 1.7001

Decision Rule:

If t > 2.7181 reject Ho otherwise do not reject Ho

Since t = 1.7001 < 2.7181 do not reject Ho and conclude that the mean of Population 1 is not greater than the mean of Population 2

c. Must first test for equal variances at alpha = 0.02 (must use Excel’s FINV function)

H0: σ12 = σ22

HA: σ12 ≠ σ22

If the calculated F > 4.462, reject Ho, otherwise do not reject Ho

F = 9.64332/5.53432 = 3.0362

Since 3.3062 < 4.462, fail to reject Ho and conclude equal variances so can use pooled standard deviation

Ho: μ1 – μ2 < 0

Ha: μ1 – μ2 > 0

df = 12 + 12 – 2 = 22

Using Excel’s average and stdev functions students can determine the sample mean and sample standard deviation of each of the samples

Sample 1: Sample 2:

Mean = 46.5833 Mean = 40.5833

St. Dev. = 9.6433 St. Dev. = 5.5343

If t > 2.5083 reject Ho, otherwise do not reject Ho

sp = [pic] = 7.8620

t = (46.5833 – 40.5833)/(7.8620[pic]) = 1.8694

Since 1.8694 < 2.5083 do not reject Ho and conclude that the mean of Population 1 is not greater than the mean of Population 2; same conclusion as reached in part b.

11.

a. Ho: μd = 0

Ha: μd ≠ 0

b.

|Sample 1 |Sample 2 |Difference (d) |(d - [pic])2 |

|4.4 |3.7 |0.7 |4.6464 |

|2.7 |3.5 |-0.8 |0.4298 |

|1.0 |4.0 |-3.0 |2.3853 |

|3.5 |4.9 |-1.4 |0.0031 |

|2.8 |3.1 |-0.3 |1.3353 |

|2.6 |4.2 |-1.6 |0.0209 |

|2.4 |5.2 |-2.8 |1.8075 |

|2.0 |4.4 |-2.4 |0.8920 |

|2.8 |4.3 |-1.5 |0.0020 |

| | |-13.1 |11.5222 |

[pic]= -13.1/9 = -1.4556

sd = [pic] = 1.2001

t = (-1.4556 – 0)/(1.2001/[pic]) = -3.6387

Decision Rule:

If t > 1.8595 or t < -1.8595 reject Ho otherwise do not reject Ho

Since t = -3.6387 < -1.8595 reject Ho and conclude that the means of the populations are different

c. –1.4556 + 1.8595(1.2001/[pic]); -2.1995 ----- -0.7117; yes this is consistent with the answer in part b because the interval range does not include 0 which would mean that the mean of the differences is significantly different from 0.

12. You must first determine whether the population standard deviations are assumed to be equal. If tested at a significance level of 0.02 the results are as follows:

H0: σ12 = σ22

HA: σ12 ≠ σ22

If the calculated F > 2.6591, reject Ho, otherwise do not reject Ho

F = 0.082/0.062 = 1.7778

Since 1.7778 < 2.6591 do not reject Ho and conclude that the standard deviations are equal

sp = [pic] = 0.0707

a. (0.145 – 0.107) + 1.6772(0.0707) [pic]; 0.0045 ----- 0.0715; because 0 is not included in the confidence interval you would conclude that the means are different and since you are in the positive range you can conclude that the mean of Population 1 is greater than the mean of Population 2

b. (0.145 – 0.107) + 2.0106(0.0707) [pic]; -0.0022 ----- 0.0782; because 0 is in this range you would conclude that the means are the same.

c. sp = [pic] = 0.0707

90% confidence interval:

(0.145 – 0.107) + 1.6606(0.0707) [pic]; 0.0145 ----- 0.0615

95% confidence interval:

(0.145 – 0.107) + 1.9845(0.0707) [pic]; 0.0099 ----- 0.0661

The pooled standard deviation does not change because it is essentially a weighted average of the two standard deviations and since you doubled both sample sizes the weights were not changed. The critical t-value changes because the sample sizes change and the reciprocals of the sample sizes change.

13.

a. Ho: μF – μM < 1

Ha: μF – μM > 1

df = 60 + 60 – 2 = 118

If t > 1.6579 reject Ho, otherwise do not reject Ho

Must first determine if population standard deviations are equal. Test at alpha = 0.02

H0: σ12 = σ22

HA: σ12 ≠ σ22

If the calculated F > 1.8459, reject Ho, otherwise do not reject Ho

F = 1.56 2/1.22 = 1.69

Since 1.69 < 1.8459 do not reject Ho and conclude that the standard deviations are equal

sp = [pic] = 1.3917

t = ((14.65 – 13.24) – 1)/(1.3917[pic]) = 1.6136

Since 1.6136 < 1.6579 do not reject Ho and conclude that the difference is not greater than 1

b. P(t > 1.6136) = 0.0546; p-value

14.

a. Ho: μC – μR < 0.25

Ha: μC – μR > 0.25

df = 25 + 25 – 2 = 48

If t > 1.6772 reject Ho, otherwise do not reject Ho

Must first determine if population standard deviations are equal. Test at alpha = 0.02

H0: σ12 = σ22

HA: σ12 ≠ σ22

If the calculated F > 2.6591, reject Ho, otherwise do not reject Ho

F = 0.87 2/0.792 = 1.2128

Since 1.2128 < 2.6591 do not reject Ho and conclude that the standard deviations are equal

sp = [pic] = 0.8310

t = ((3.74 – 3.26) – 0.25)/(0.8310[pic]) = 0.9785

Since 0.9785 < 1.6772 do not reject Ho and conclude that the difference is not greater than $0.25

b. Since you accepted the null hypothesis the type of error that could occur is accepting a false null hypothesis that is a Type II error.

15.

a. Ho: μN – μO < 0

Ha: μN – μO > 0

df = 35 + 30 – 2 = 63

If t > 1.2951 reject Ho, otherwise do not reject Ho

Must first determine if population standard deviations are equal. Test at alpha = 0.02

H0: σ12 = σ22

HA: σ12 ≠ σ22

If the calculated F > 2.3716, reject Ho, otherwise do not reject Ho

F = 16.23 2/15.912 = 1.0406

Since 1.0406 < 2.3716 do not reject Ho and conclude that the standard deviations are equal

sp = [pic] = 16.0835

t = ((288 – 279) – 0)/(16.0835[pic]) = 2.2491

Since 2.2491 > 1.2951 reject Ho and conclude that the new cartridge will result in a longer lasting product.

b. 90% confidence interval:

(288 – 279) + 1.6694(16.0835) [pic]; 2.3196 ----- 15.6804; yes this is consistent with the results in part a.

16.

Must first determine if population standard deviations are equal. Test at alpha = 0.02. Because of the sample sizes you must use Excel’s FINV function.

H0: σ12 = σ22

HA: σ12 ≠ σ22

If the calculated F > 1.351, reject Ho, otherwise do not reject Ho

F = 6.18 2/2.332 = 7.035

Since 7.035 > 1.351 reject Ho and conclude that the standard deviations are not equal

a. 90% confidence interval:

(22.48 – 12.56) + 1.645[pic]; 9.2012 ----- 10.6388

95% confidence interval:

(22.48 – 12.56) + 1.96[pic]; 9.0635 ----- 10.7765

b. Yes, since both confidence intervals contain the value of $10 this means the fee could be $10 more for federally chartered banks.

17. Must first determine if population standard deviations are equal. Test at alpha = 0.02. Because of the sample sizes you must use Excel’s FINV function.

H0: σ12 = σ22

HA: σ12 ≠ σ22

If the calculated F > 1.3184, reject Ho, otherwise do not reject Ho

F = 32 2/112 = 8.4628

Since 8.4628 > 1.3184 reject Ho and conclude that the standard deviations are not equal

a.

90% confidence interval:

(53 – 24) + 1.645[pic]; 25.9624 ----- 32.0376; yes there is a difference in how long males and females spend in the store per visit because the confidence interval does not contain the value 0.

b. (53 – 24) + 1.96[pic]; 25.3808 ----- 32.6192; no because the difference was so large in part a that changing the confidence interval to 95% would not change the upper and lower limits enough to cause them to include the value 0

c. Based upon the confidence intervals calculated in parts a and b it would actually be more than 20 minutes more.

d. P(men > 53) = P(z > (53 – 24)/(11/[pic])) = P(z > 41.68) which is essentially 0

18. Must first determine if population standard deviations are equal. Test at alpha = 0.02. Because of the sample sizes you must use Excel’s FINV function.

H0: σ12 = σ22

HA: σ12 ≠ σ22

If the calculated F > 1.3923, reject Ho, otherwise do not reject Ho

F = 5 2/3.62 = 1.929

Since 1.929 > 1.3923 reject Ho and conclude that the standard deviations are not equal

a.

(41.5 – 39) + 1.96[pic]; 1.6461 ----- 3.3539; yes because the interval does not contain the value 0 which would indicated no difference.

b. Company A:

-1.645 = (x – 41.5)/(3.6/[pic]); x = 41.0813

Company B:

-1.645 = (x – 39)/(5/[pic]); x = 38.4184

19. Must first determine if population standard deviations are equal. Test at alpha = 0.02. Because of the sample sizes you must use Excel’s FINV function.

H0: σ12 = σ22

HA: σ12 ≠ σ22

If the calculated F > 1.9626, reject Ho, otherwise do not reject Ho

F = 1.05 2/0.892 = 1.392

Since 1.392 < 1.9626 do not reject Ho and conclude that the standard deviations are equal

a. 5.26 – 6.19 = -0.93; The advantage of using this point estimate is that it is easy to calculate. The disadvantage is that it uses only 2 values calculated from the samples and it does not consider the spread or the variation in the sample at all.

b. sp = [pic] = 0.9733

the standard error would be 0.9733[pic] = 0.1947

c. -0.93 + 1.9845(0.1947); -1.3164 ----- -0.5436; Yes there is a difference since the interval does not contain the value 0

d. He could increase the sample size or decrease the confidence level.

20. Must first determine if population standard deviations are equal. Test at alpha = 0.02.

H0: σ12 = σ22

HA: σ12 ≠ σ22

If the calculated F > 3.905, reject Ho, otherwise do not reject Ho

F = 2.5 2/1.82 = 1.929

Since 1.929 < 3.905 do not reject Ho and conclude that the standard deviations are equal

a. sp = [pic] = 2.1783

(17.2 – 15.9) + 1.7056(2.1783)[pic]; -0.1043 ----- 2.7043; No because the interval contains the value 0 you cannot say that there is a difference setup time for the two additives.

b. No because again you cannot say that there is a difference in the setup time for the two additives.

21. Must first determine if population standard deviations are equal. Test at alpha = 0.02. Because of the sample sizes you must use Excel’s FINV function.

H0: σ12 = σ22

HA: σ12 ≠ σ22

If the calculated F > 2.5069, reject Ho, otherwise do not reject Ho

F = 2.9 2/1.62 = 3.2852

Since 3.2852 > 2,5069 reject Ho and conclude that the standard deviations are not equal

a. d.f. = (8.41/28 + 2.56/28)2/(0.0033+0.0003) = 43

(18.4 – 15.2) + 1.6811[pic] = 2.1478 ----- 4.2522; Yes the new additive does help reduce setup time.

b. Any time you increase the sample size you will automatically reduce the standard error.

c. The C.I. indicates that µold > µnew. Therefore, if they are primarily interested in reducing setup time, then yes they should switch to the new additive.

22.

a. Ho: μB – μA < 0.35

Ha: μB – μA > 0.35

df = 20 + 20 – 2 = 38

If t > 2.4286 reject Ho, otherwise do not reject Ho

Must first determine if population standard deviations are equal. Test at alpha = 0.02. Students can use Excel’s test to determine whether the standard deviations are equal

H0: σ12 = σ22

HA: σ12 ≠ σ22

If the calculated F > 3.0274, reject Ho, otherwise do not reject Ho

|F-Test Two-Sample for Variances | |

| | | |

|  |Brochure B |Brochure A |

|Mean |11.881 |9.3225 |

|Variance |75.56685158 |50.943725 |

|Observations |20 |20 |

|df |19 |19 |

|F |1.483339736 | |

Since 1.4833 < 2.0274 do not reject Ho and conclude the variances are equal

Students can use Excel’s data analysis tool to conduct the t-test assuming equal variances

|t-Test: Two-Sample Assuming Equal Variances |

| | | |

|  |Brochure B |Brochure A |

|Mean |11.881 |9.3225 |

|Variance |75.56685158 |50.943725 |

|Observations |20 |20 |

|Pooled Variance |63.25528829 | |

|Hypothesized Mean Difference |0.35 | |

|df |38 | |

|t Stat |0.878110123 | |

|P(T 1.6909 reject Ho, otherwise do not reject Ho

Must first determine if population standard deviations are equal. Test at alpha = 0.02. Students can use Excel’s test to determine whether the standard deviations are equal

H0: σ12 = σ22

HA: σ12 ≠ σ22

If the calculated F > 3.1296, reject Ho, otherwise do not reject Ho

|F-Test Two-Sample for Variances | |

| | | |

|  |Stay Home |Day Care |

|Mean |13.02 |15.12380952 |

|Variance |4.181714286 |3.988904762 |

|Observations |15 |21 |

|df |14 |20 |

|F |1.048336457 | |

Since 1.0483 < 3.1296 do not reject Ho and conclude the variances are equal

Students can use Excel’s data analysis tool to conduct the t-test assuming equal variances

|t-Test: Two-Sample Assuming Equal Variances |

| | | |

|  |Day Care |Stay Home |

|Mean |15.12380952 |13.02 |

|Variance |3.988904762 |4.181714286 |

|Observations |21 |15 |

|Pooled Variance |4.068296919 | |

|Hypothesized Mean Difference |0 | |

|df |34 | |

|t Stat |3.085347764 | |

|P(T 1.96 or if z < -1.96 reject Ho, otherwise do not reject Ho

Since the sample sizes are so large you can use Excel’s z-test for two means and use the sample variances for the known population variances.

|z-Test: Two Sample for Means | |

| | | |

|  |Vanguard Visibility |Scorpian Visibility |

|Mean |0.36552897 |1.688964286 |

|Known Variance |0.1291 |3.5203 |

|Observations |280 |280 |

|Hypothesized Mean Difference |0 | |

|Z |-11.59233709 | |

|P(Z 1.3923, reject Ho, otherwise do not reject Ho

F = 33.92 2/30.312 = 1.2524

Since 1.2524 < 1.3923 do not reject Ho and conclude that the standard deviations are equal

sp = [pic] = 32.1657

t = ((142.76 – 133.19) – 0)/(32.1657[pic]) = 2.9752

Since 2.9752 > 1.6487 reject Ho and conclude that regular ATM users are more profitable to the bank that customers who do not regularly use their ATM card.

c. (142.76 – 133.19) + 1.9659(32.1657)[pic]; 3.2465 ----- 15.8935; In this case you might prefer the confidence interval because it gives you an estimate as to how much more profitable they are.

26.

a. The research hypothesis would be that the mean speed for California drivers exceeded the mean speed of out-of-state drivers by more than 5 mph.

Ho: μC – μO < 5

Ha: μC – μO > 5

b. df = 140 + 75 – 2 = 213

If t > 1.2855 reject Ho, otherwise do not reject Ho

Must first determine if population standard deviations are equal. Test at alpha = 0.02. Students can use Excel’s test to determine whether the standard deviations are equal

H0: σ12 = σ22

HA: σ1 ≠ σ22

If the calculated F > 1.6370, reject Ho, otherwise do not reject Ho

|F-Test Two-Sample for Variances | |

| | | |

|  |California Cars |Out-of-State Cars |

|Mean |64.45 |61.96 |

|Variance |64.29244604 |59.76864865 |

|Observations |140 |75 |

|df |139 |74 |

|F |1.075688467 | |

|P(F 2.7073 reject Ho and conclude that the standard deviations are not equal

Using Equation 9-7 degrees of freedom = 12

(48,768.0769 – 12,212.0789) + 2.1788[pic] =

21,013.6964 ---- 52,098.2996; yes this provides evidence that there is a difference because the interval does not include the value of 0.

30.

a. Must first determine if population standard deviations are equal. Test at alpha = 0.02. Because of the sample sizes you must use Excel’s FINV function.

H0: σ12 = σ22

HA: σ12 ≠ σ22

If the calculated F > 2.9079, reject Ho, otherwise do not reject Ho

F = 9083.6074 2/3377.39952 = 7.2464

Since 7.2464 > 2.9079 reject Ho and conclude that the standard deviations are not equal

| |Employees |

| |< 20,000 |> 20,000 |

|Average Revenues | 3,043.0625 | 11,932.2105 |

|St. Dev. Revenues | 3,377.3995 | 9,083.6074 |

|Count |32 |19 |

Using Equation 9-7 degrees of freedom = 20

(11,932.2105 – 3,043.0625) + 2.0860[pic];

4,521.9535 ----- 13,411.0924

b. Must first determine if population standard deviations are equal. Test at alpha = 0.02. Because of the sample sizes you must use Excel’s FINV function.

H0: σ12 = σ22

HA: σ12 ≠ σ22

If the calculated F > 2.8876, reject Ho, otherwise do not reject Ho

F = 9498.3599 2/3211.07242 = 8.7498

Since 8.7498 > 2.8876 reject Ho and conclude that the standard deviations are not equal

Using Equation 9-7 degrees of freedom = 9

| |Employees |

| |< 30,000 |> 30,000 |

|Average Revenues | 3,626.4390 | 17,540.6000 |

|St. Dev. Revenues | 3,211.0724 | 9,498.3599 |

|Count |41 |10 |

(17,540.6 – 3,626.439) + 2.2622[pic];

7,025.2618 ----- 20,803.06012

c. The difference in means is larger and has a larger standard error which the dividing line is 30,000 as opposed to 20,000

31.

| | | |

|Manufacturing Plant |Data |Total |

|Boise |Average of Dollar Claim Amount |268.4358974 |

|  |StdDev of Dollar Claim Amount |50.89551041 |

|  |Count of Dollar Claim Amount |78 |

|Salt Lake City |Average of Dollar Claim Amount |277.3333333 |

|  |StdDev of Dollar Claim Amount |62.64299584 |

|  |Count of Dollar Claim Amount |24 |

Must first determine if population standard deviations are equal. Test at alpha = 0.02. Because of the sample sizes you must use Excel’s FINV function.

H0: σ12 = σ22

HA: σ12 ≠ σ22

If the calculated F > 2.06, reject Ho, otherwise do not reject Ho

F = 62.643 2/50.89552 = 1.5149

Since 1.5149 < 2.06 do not reject Ho and conclude that the standard deviations are equal

sp = [pic] = 53.8249

(277.3333 – 268.4359) + 2.3642(53.8249)[pic]; -20.8069 ----- 38.6011

No, since the confidence interval includes the value of 0 you cannot conclude that there is a difference between the two plants.

32.

a.

|Credit Card Balances - Male |

| | |

|Mean |746.512931 |

|Standard Error |19.33632279 |

|Median |738.5 |

|Mode |1018 |

|Standard Deviation |294.5220941 |

|Count |232 |

|Credit Card Account Balance - Female |

| | |

|Mean |778.1323529 |

|Standard Error |35.80014705 |

|Median |737 |

|Mode |600 |

|Standard Deviation |295.2155754 |

|Count |68 |

Since the sample size for both male and female are greater than 30 use equation 9-8 to determine the confidence interval.

(778.1324 – 746.5129) + 1.96[pic]; -48.1297 ----- 111.3687

b. Student answers will vary but should include comments that based upon this confidence interval it cannot be concluded that there is a difference between males and females credit cards balances because the interval includes the value 0.

c. (778.1324 – 746.5129) + 2.575[pic]; -73.1531 ----- 136.3921

33.

a.

|  |Sex (M/F/U) |  |

|Data |F |M |

|Count of Total Chges | 70.00000 | 67.00000 |

|Average of Total Chges | 6,033.48286 | 6,859.00746 |

|StdDev of Total Chges | 3,694.54293 | 5,303.85551 |

Since both sample sizes are greater than 30 students should use Equation 9-8 to calculate this confidence interval.

(6859.0075 – 6033.4829) + 1.96[pic]; -711.369 ----- 2,362.418

Based upon this confidence interval it cannot be concluded that there is a difference between the way males and females are charged since the interval contains the value 0.

b.

|Principal Payer |CARE | |

| | | |

|  |Sex (M/F/U) |  |

|Data |F |M |

|Count of Total Chges | 59.00000 | 56.00000 |

|Average of Total Chges | 6,160.20576 | 6,683.20321 |

|StdDev of Total Chges | 3,833.22315 | 5,580.58925 |

Since both sample sizes are greater than 30 students should use Equation 9-8 to calculate this confidence interval.

(6683.2032 – 6160.2058) + 1.96[pic]; -1,235.73 ----- 2,281.73

You cannot conclude there is a difference since the interval contains the value 0.

34.

a.

|Cozine Garbage Truck Weights |

| | |

|Mean |42260.64 |

|Standard Error |278.2153265 |

|Median |42325.5 |

|Mode |40010 |

|Standard Deviation |3934.55888 |

|Count |200 |

Since both sample sizes are greater than 30 students should use Equation 9-8 to calculate this confidence interval.

(42,260.64 – 39,700) + 1.645[pic]; 1,914.601 ----- 3,206.679

b. Average trips for Toner: 500,000/39,700 = 12.5945

Average trips for Cozine: 500,000/42,260.64 = 11.8313

Average difference will be 12.5945 – 11.8313 = 0.7632 @ $20 = $15.264

35.

a.

|Difference |

| | |

|Mean |-1.3 |

|Standard Error |4.472195 |

|Median |-3.5 |

|Mode |-4 |

|Standard Deviation |20.00026 |

|Sample Variance |400.0105 |

|Kurtosis |-0.53901 |

|Skewness |0.03075 |

|Range |69 |

|Minimum |-34 |

|Maximum |35 |

|Sum |-26 |

|Count |20 |

|Confidence Level(95.0%) |9.360414 |

95% confidence interval:

-1.3 + 9.3604; -10.6604 ----- 8.0604

Based upon this confidence interval you cannot conclude that there is a difference in the two sign systems because the interval contains the value 0

b. Student answers will vary but one strength is this controls for individual shopper differences and familiarity with a store. A weakness is that you must run the tests sequentially so this increases the time required to perform the test. In this particular example they also had the expense of transporting customers from one city to another city.

36.

Decision Rule:

If p-value < 0.05 reject Ho, otherwise do not reject Ho

[pic]= (34+30)/(100+100) = 0.32

p1 = 30/100 = 0.30

p2 = 34/100 = 0.34

z = (0.34 – 0.3)/[pic] = 0.6063

p-value = 2(0.5 – 0.2291) = 0.5418

Since 0.5418 > 0.5 do not reject Ho and conclude that the population proportions are equal

37. Decision Rule:

If z > 1.645 or z < -1.645 reject Ho, otherwise do not reject Ho

[pic]= (87+80)/(200+150) = 0.4771

p1 = 87/200 = 0.435

p2 = 80/150 = 0.5333

z = [(0.435 – 0.5333)-.05]/[pic] = -2.7594

Since z = -2.7594 < -1.645 reject Ho, and conclude that the difference in the population proportions is not equal to 0.05.

38.

a. p1 = 88/300 = 0.2933; n1p1 = 0.2933(300) = 87.99> 5; n1(1-p1) = 300(1-0.2993) = 210.21>5

p2 = 136/400 = 0.34; n2p2 = 0.34(400) = 136> 5; n2(1-p2) = 400(1-0.34) = 264>5

b. (0.34 – 0.2933) + 1.28[pic] ; 0.0014 ----- 0.0920

39.

a. n1p1 = 0.62(745) = 462> 5; n1 (1-p1) = 745(1-0.62) = 283>5

n2p2 = 0.49(455) = 223> 5; n2(1-p2) = 455(1-0.49) = 232>5

b. Ho: π1 – π2 = 0

Ha: π1 – π2 ≠ 0

Decision Rule:

If z > 1.96 or z < -1.96 reject Ho, otherwise do not reject Ho

[pic]= (462+223)/(745+455) = 0.5708

z = [(0.62 – 0.49)-0]/[pic] = 4.414

Since z = 4.414 > 1.96 reject Ho, and conclude that there is a difference in the proportion of homes that watch a national news broadcast.

40.

a. Ho: πp – πg = 0

Ha: πp – πg ≠ 0

n1p1 = 22 > 5; n1(1-p1) = 48 > 5

n2p2 = 34 > 5; n2(1-p2) = 31 > 5

Decision Rule:

If z > 1.96 or z < -1.96 reject Ho, otherwise do not reject Ho

[pic]= (22+19)/(70+50) = 0.3417

pp = 22/70 = 0.3143

pg = 19/50 = 0.38

z = [(0.3143 – 0.38)-0]/[pic] = -0.7481

Since z = -0.7481 > -1.96 do not reject Ho, and conclude that there is no difference in the proportion of employees who give to United Way depending on whether the employer is a private business or a government agency.

b. (0.38 – 0.3143) + 1.96[pic] ; -0.1073 ----- 0.2387

Yes they give compatible results. The hypothesis test concluded that there was no difference and the confidence interval concludes the same thing because the interval contains the value 0.

41. p>=2 = 123/596 = 0.2064

p 1.96 or z < -1.96 reject Ho, otherwise do not reject Ho

pC = 19/105 = 0.1810

pR = 31/200 = 0.1550

[pic]= (19+31)/(105+200) = 0.1639

z = [(0.1810 – 0.1550)-0]/[pic] = 0.5828

Since 0.5282 < 1.96 do not reject Ho and conclude that there is no difference in the default rate between residential and commercial loans.

b. Student reports and graphs will vary.

c. Students will need to verify that the sample sizes are sufficient to assume the normal distribution.

43. H0: σ12 = σ22

HA: σ12 ≠ σ22

a. If the calculated F > 2.5769, reject Ho, otherwise do not reject Ho

F = 300 2/2502 = 1.44

Since 1.44 < 2.5769 do not reject Ho and conclude that the standard deviations are equal

b. If the calculated F > 3.9052, reject Ho, otherwise do not reject Ho

The conclusion does not differ

c. Use Excel’s FDIST(1.44,13,13) which gives 0.26 which would be for a two-tailed test you would need an alpha of 0.26(2) = 0.52

44.

a. You must assume that the populations are normally distributed.

b. Ho: μA – μB < 0

Ha: μA – μB > 0

df = 7 + 7 – 2 = 12

Using Excel’s average and stdev functions students can determine the sample mean and sample standard deviation of each of the samples

Before: After:

Mean = 1991.571 Mean = 2137.286

St. Dev. = 620.5395 St. Dev. = 578.6161

If t > 1.7823 reject Ho, otherwise do not reject Ho

sp = [pic] = 599.9441

t = (2137.286 – 1991.571)/(599.9441[pic]) = 0.4544

Since 0.4544 < 1.7823 do not reject Ho and conclude that there is no difference in average sales before and after the ad.

c. H0: σB2 = σA2

HA: σB2 ≠ σA2

If the calculated F > 8.466, reject Ho, otherwise do not reject Ho

F = 620.5395 2/578.61612 = 1.150

Since 1.150 < 8.466 do not reject Ho and conclude that the variances are equal

45. H0: σn2 < σc2

HA: σn2 > σc2

If the calculated F > 2.2756, reject Ho, otherwise do not reject Ho

F = 3.452/2.872 = 1.445

Since 1.445 < 2.2756 do not reject Ho and conclude that the new model is not more consistent than the old model

46.

a. H0: σA2 < σT2

HA: σA2 > σT2

If the calculated F > 2.5342, reject Ho, otherwise do not reject Ho

F = 0.2022/0.142 = 2.0818

Since 2.0818 < 2.5342 do not reject Ho and conclude that the Trenton plant is not less variable than the Atlanta plant.

b. You would have rejected a true null hypothesis which is a Type I error. You could decrease the alpha level to decrease the probability of a Type I error or you could increase the sample size.

47.

a. H0: σB2 = σA2

HA: σB2 ≠ σA2

sA = 134.9919 sB = 218.2298

If the calculated F > 4.8491, reject Ho, otherwise do not reject Ho

F = 218.2298 2/134.99192 = 2.6134

Since 2.6134 < 4.8491 do not reject Ho and conclude that the variation in customer orders for the two locations is the same.

b. Develop a confidence interval for the difference in the means.

c. A 90% confidence interval would be:

Mean of A = 362.7273 Mean of B = 375.3636

sp = [pic] = 181.4484

90% confidence interval:

(375.3636 – 362.7273) + 1.7247(181.4484) [pic]; -120.8035 ----- 146.0761

48.

a. H0: σD2 = σW2

HA: σD2 ≠ σW2

Decision Rule:

If the calculated F > 3.2839, reject Ho, otherwise do not reject Ho

F = 2.0 2/1.22 = 2.7778

Since 2.7778 < 3.2839 do not reject Ho and conclude that the variation in service times for customers who use the drive-through versus those who go inside is the same.

b. Ho: μD – μW = 0

Ha: μD – μW ≠ 0

df = 13 + 9 – 2 = 20

If t > 2.0860 or t < -2.0860 reject Ho, otherwise do not reject Ho

sp = [pic] = 1.7251

t = (8.5 – 8)/(1.7251[pic]) = 0.6684

Since 0.6684 < 2.0860 do not reject Ho and conclude that there is no difference in the mean service times.

This test is valid because you did test that the variances were equal.

49.

a. H0: σM2 > σF2

HA: σM2 > σF2

Decision Rule:

If the calculated F > 1.9838, reject Ho, otherwise do not reject Ho

F = 2.5 2/1.342 = 3.4807

Since 3.4807 > 1.9838 reject Ho and conclude that the meals ordered by males does have greater variability than those ordered by females.

b. Ho: μM – μF = 1

Ha: μM – μF ≠ 1

df = 36 using Formula 9-5

If t > 2.0281 or t < -2.0281 reject Ho, otherwise do not reject Ho

t = (12.4 – 8.92) – [1/[pic]] = 4.3716

Since 4.3716 > 2.0281 reject Ho and conclude that there is at least $1.00 difference in the mean price of meals ordered by male and female customers.

50.

a. Ho: πU – πG < 0

Ha: πU – πG > 0

Decision Rule:

If z > 1.645 reject Ho, otherwise do not reject Ho

PG = 59/100 = 0.59

PU = 138/200 = 0.69

[pic]= (59+138)/(100+200) = 0.6567

z = [(0.69 – 0.59)-0]/[pic] = 1.7196

Since 1.7196 > 1.645 reject Ho and conclude that the percentage of undergraduates who purchase used textbooks is higher than the percentage of graduate students who purchase used textbooks.

b. They should not do this. You cannot necessarily assume that the characteristics of students at all universities reflect the same population characteristics of students at ASU.

51.

a. Ho: π > 0.75

Ha: π < 0.75

Decision Rule:

If z < -1.645 reject Ho, otherwise do not reject Ho

p = 386/420 = 0.9190

z = (0.9190 – 0.75)/ [pic] = 7.9985

Since 7.9985 > -1.645 do not reject Ho and conclude that the percentage of credit hours taken by the student body as a whole that can be counted toward the degree program is not less than 75%

b. 0.9190 + 3([pic]) = 0.8556 ----- 0.9824

52.

a. Ho: πA – πN = 0

Ha: πA – πN ≠ 0

Decision Rule:

If z > 1.96 or z < -1.96 reject Ho, otherwise do not reject Ho

PA = 144/200 = 0.72

PN = 402/500 = 0.804

[pic]= (144+402)/(200+500) = 0.78

z = [(0.804 – 0.72)-0]/[pic] = 2.4237

Since 2.4237 > 1.96 reject Ho and conclude that there is a difference between athletes and non-athletes.

b. Student letters will vary, but could address the issue that currently athletes are not taking as many degree credit courses as non-athletes and that currently the non-athletes appear to be exceeding the 75% proposed standard.

53.

a. Ho: πU – πG < 0

Ha: πU – πG > 0

Decision Rule:

If z > 1.645 reject Ho, otherwise do not reject Ho

PG = 26/60 = 0.4333

PU = 46/80 = 0.575

[pic]= (26+46)/(60+80) = 0.5143

z = [(0.575 – 0.4333)-0]/[pic] = 1.66

Since 1.66 > 1.645 reject Ho and conclude that the percentage of undergraduates who will attend graduation does exceed the percentage of graduate students who will attend graduation.

b. Using the sample as a point estimate you could recommend that they reserve 0.4333(500) = 216.65 or 217 seats for the graduates and 0.575(2000) = 1,150 seats for the undergraduates.

54.

a. H0: σA2 = σB2

HA: σA2 ≠ σB2

b. Students can use Excel’s STDEV function to determine the standard deviations of each brochure.

Standard dev. of A = 7.1375 Standard Dev. of B = 8.6929

Decision Rule:

If the calculated F > 3.0274, reject Ho, otherwise do not reject Ho

F = 8.6929 2/7.13752 = 1.4833

Since 1.4833 < 3.0274 do not reject Ho and conclude that the variation in collections from the brochures are the same.

c. You must assume the populations are normally distributed. Yes, since you are assuming a normal distribution and you have determined that the variances of the populations are equal you would be able to use a two-sample t-test procedure to test if the mean sales amounts is different between those receiving Brochure A and those receiving Brochure B.

55.

a. Ho: πT – πB = 0

Ha: πT – πB ≠ 0

Decision Rule:

If p-value is < 0.05 reject Ho, otherwise to not reject Ho

PT = 11/90 = 0.1222

PB = 15/90 = 0.1667

[pic]= (11+15)/(90+90) = 0.1444

z = [(0.167 – 0.1222)-0]/[pic] = 0.8550

p-value = (0.5 – 0.3051)2 = 0.3898 > 0.05 so do not reject Ho and conclude that there will be no difference in the proportion of people returning the tune-up coupon and those returning the brake work coupon.

b. A Type I error would be if there is no difference but we concluded there was a difference. A Type II error would be if there is a difference but we concluded there was not a difference. The relative cost of a Type I error would be that you might send out more of one coupon than the other and would lose business because of not enough people receiving the coupon. The cost of a Type II error would be that you sent out equal numbers of each coupon when one coupon would be redeemed more frequently. A type II error could have been committed in this problem.

56.

a. nOpO = (81/280)(280) = 81> 5; nO(1-pO) = 280(1-0.2893) = 199>5

nApA = (74/280)(280) = 74> 5; nA(1-pA) = 280(1-0.2643) = 206>5

b. Ho: πO – πA = 0

Ha: πO – πA ≠ 0

Decision Rule:

If z > 1.645 or z < -1.645 reject Ho, otherwise to not reject Ho

PO = 81/280 = 0.2893

PA = 74/280 = 0.2643

[pic]= (81+74)/(280+280) = 0.2768

z = [(0.2893 – 0.2643)-0]/[pic] = 0.6611

Since 0.6611 < 1.645 do not reject Ho and conclude that there is no difference in the regions.

57. Ho: πci – πco < 0

Ha: πci – πco > 0

Decision Rule:

If z > 1.28 reject Ho, otherwise to not reject Ho

Pci = 62/100 = 0.62

Pco = 36/75 = 0.48

[pic]= (62+36)/(100+75) = 0.56

z = [(0.62 – 0.48)-0]/[pic] = 1.8464

Since 1.8464 > 1.28 reject Ho and conclude that there the proportion of city residents who favor this proposal is greater than the proportion of county residents who favor the proposal.

58. Ho: πUS – πF < 0

Ha: πUS – πF > 0

a. A Type I error would mean that the US cars would have better quality than foreign cars but the test would conclude that they had worse quality. The cost to this would be loss of car sales and additional advertising expenditures trying to convince the public that the US has good quality. A Type II error would mean that the US actually has lower quality but the test would conclude that they had better quality. This would mean individuals would purchase US cars only to have major repairs which would cost the US automakers money fixing cars under warranty and then having a poor reputation for poor quality.

b. Decision Rule:

If z > 2.05 reject Ho, otherwise to not reject Ho

PUS = 12/60 = 0.20

PF = 13/70 = 0.1857

[pic]= (12+13)/(60+70) = 0.1923

z = [(0.20 – 0.1857)-0]/[pic] = 0.2062

Since 0.2062 < 2.05 do not reject Ho and conclude that the proportion of US cars needing major repair within 2 years in not higher than foreign cars needing major repairs in 2 years.

59.

a. n1p1 = (19/300)(300) = 19 > 5; n1(1-p1) = 300(1-0.0633) = 281>5

n2p2 = (12/250)(250) = 12> 5; n2p2(1-p2) = 250(1-0.048) = 238>5

b. Ho: π1 – π2 = 0

Ha: π1 – π2 ≠ 0

Decision Rule:

If z > 2.17 of z < -2.17 reject Ho, otherwise to not reject Ho

P1 = 19/300 = 0.0633

PF = 12/250 = 0.048

[pic]= (19+12)/(300+250) = 0.0564

z = [(0.0633 – 0.048)-0]/[pic] = 0.7745

Since 0.7745 < 2.17 do not reject Ho and conclude that there is no difference in the packing materials.

60.

a. Ho: μN – μC = 0.4

Ha: μN – μC ≠ 0.4

df = 40 + 52 – 2 = 90

If t > 1.9867 reject Ho, otherwise do not reject Ho

Must first determine if population standard deviations are equal. Test at alpha = 0.02

H0: σ12 = σ22

HA: σ1 ≠ σ22

If the calculated F > 2.0055, reject Ho, otherwise do not reject Ho

F = 0.8 2/0.72 = 1.306

Since 1.306 < 2.0055 do not reject Ho and conclude that the standard deviations are equal

sp = [pic] = 0.745

t = ((5.2 – 4.3) – 0.4)/(0.745[pic]) = 3.191

Since 3.191 > 1.9867 reject Ho and conclude that the difference is not 0.4 tons more per acre

b. 0.002

61.

a. H0: σO2 < σN2

HA: σO2 > σN2

If the calculated F > 1.6333, reject Ho, otherwise do not reject Ho

F = 0.8 2/0.72 = 1.306

Since 1.306 < 1.6333 do not reject Ho and conclude that the standard deviation of the new is not less than the current leading brand.

b. You need to conclude the populations are normally distributed.

62.

a. Ho: μW – μO = 0

Ha: μW – μO ≠ 0

b. df = 65 + 85 – 2 = 148

If p-value is less than 0.05 reject Ho, otherwise do not reject Ho

Must first determine if population standard deviations are equal. Test at alpha = 0.02

H0: σ12 = σ22

HA: σ1 ≠ σ22

If the calculated F > 1.7517, reject Ho, otherwise do not reject Ho

F = 27920 2/248002 = 1.267

Since 1.267 < 1.7517 do not reject Ho and conclude that the standard deviations are equal

sp = [pic] = 26,615.7312

t = ((58740 – 54900) – 0)/(26615.7312[pic]) = 0.8756

p-value = 0.3827 > 0.05 so do not reject Ho and conclude that there is no difference in the average whole life insurance coverage for clients in Wisconsin and Ohio.

63.

a. H0: σ12 = σ22

HA: σ12 ≠ σ22

b. If the calculated F > 1.4836, reject Ho, otherwise do not reject Ho

F = 27920 2/248002 = 1.267

Since 1.267 < 1.4836 do not reject Ho and conclude that the standard deviations are equal

64.

a. Ho: μBB – μS < 0

Ha: μBB – μS > 0

b. df = 50 + 50 – 2 = 98

If t > 2.3650 reject Ho, otherwise do not reject Ho

Must first determine if population standard deviations are equal. Test at alpha = 0.02

H0: σ12 = σ22

HA: σ22 ≠ σ22

If the calculated F > 1.9626, reject Ho, otherwise do not reject Ho

F = 112 2/1052 = 1.1378

Since 1.1378 < 1.9626 do not reject Ho and conclude that the standard deviations are equal

sp = [pic] = 108.5564

t = ((653 – 691) – 0)/(108.5564[pic]) = -1.7502

Since –1.7502 < 2.3650 do not reject Ho and conclude that the Bounce Back backboards is not as durable, on the average, as the Swoosh Company backboards.

c. In this case it would not change the data since we are testing the alternative of greater than but based on this sample our calculated value is negative so our decision will not changed.

65.

a. H0: σs2 < σBB2

HA: σS2 > σBB2

b. If the calculated F > 1.6073, reject Ho, otherwise do not reject Ho

F = 112 2/1052 = 1.1378

Since 1.1378 < 1.6073 do not reject Ho and conclude that the standard deviation of the Bounce Back Company is not less than the standard deviation of the Swoosh Company

66.

a. Ho: μwhite – μwheat < 0

Ha: μwhite – μwheat > 0

b.

| |White |Wheat |

|Average |599.77273 |530.409091 |

|Standard Dev. |149.0551 |107.023655 |

df = 22 + 22 – 2 = 42

If t > 1.682 reject Ho, otherwise do not reject Ho

Must first determine if population standard deviations are equal. Test at alpha = 0.02

H0: σ12 = σ22

HA: σ12 ≠ σ22

If the calculated F > 2.8574, reject Ho, otherwise do not reject Ho

F = 149.0551 2/107.02372 = 1.9397

Since 1.9397 < 2.8574 do not reject Ho and conclude that the standard deviations are equal

sp = [pic] = 129.7526

t = ((599.7727 – 530.4091) – 0)/(129.7526[pic]) = 1.7730

Since 1.7730 > 1.682 reject Ho and conclude that white bread does outsell wheat bread

67. H0: σwhite2 = σwheat2

HA: σwhite2 ≠ σwheat2

If the calculated F > 2.0842, reject Ho, otherwise do not reject Ho

F = 149.0551 2/107.02372 = 1.9397

Since 1.9397 < 2.0842 do not reject Ho and conclude that the standard deviations are equal

68.

a. Ho: μM – μF = 0

Ha: μM – μF ≠ 0

|Credit Card Account Balance - Female |

| | |

|Mean |778.1323529 |

|Standard Error |35.80014705 |

|Median |737 |

|Mode |600 |

|Standard Deviation |295.2155754 |

|Count |68 |

|Credit Card Balances - Male |

| | |

|Mean |746.512931 |

|Standard Error |19.33632279 |

|Median |738.5 |

|Mode |1018 |

|Standard Deviation |294.5220941 |

|Count |232 |

df = 68 + 232 – 2 = 298

If t > 1.96 or t < -1.96 reject Ho, otherwise do not reject Ho

Must first determine if population standard deviations are equal. Test at alpha = 0.02

H0: σ12 = σ22

HA: σ22 ≠ σ22

If the calculated F > 1.543, reject Ho, otherwise do not reject Ho

F = 295.2156 2/294.52212 = 1.0047

Since 1.0047 < 1.543 do not reject Ho and conclude that the standard deviations are equal

sp = [pic] = 294.6782

t = ((778.1324 – 746.5129) – 0)/(294.6782[pic]) = 0.7781

Since 0.7781 < 1.645 do not reject Ho and conclude that there is no difference between the mean credit card balances between female and male customers.

b. H0: σ12 = σ22

HA: σ22 ≠ σ22

If the calculated F > 1.543, reject Ho, otherwise do not reject Ho

F = 295.2156 2/294.52212 = 1.0047

Since 1.0047 < 1.543 do not reject Ho and conclude that the standard deviations are equal.

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