Check for corrections and see whether am correct

Go through my Statistics assignment and tell me whether I have done correctly. It is important to me, so please double check.
statistics.xls

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Among Groups:
DfA = c – 1 = 4 – 1 = 3
SSA = 80 * 3 = 240
Within Groups:
DfW = n – c = 32 – 4 = 28
Total:
DfT = n – 1 = 31
MSA = 80
SSW = 560
SST = 240 + 560 = 800
MSW = 560/28 = 20
(a)
Source
Among
Within
Total
Df
5
166
171
SS
3172
21246
24418
MS
634.40
127.99
F
4.957
P- value
0.0003
(b) Since 0.0003 < 0.05, the result is significant. There is evidence of a difference in the mean computer anxiety experienced b (c) Number of groups, N = DfW = MSW = 166 127.99 Tukey's Pair-wise Studentized Range Statistics Table (q Table) … Marketing Management Other Marketing 0.0000 0.4816 0.8741 Management 0.4816 0.0000 0.3925 Other 0.8741 0.3925 0.0000 Finance 1.0400 0.5585 0.1659 Accountancy 2.7559 2.2743 1.8818 MIS 4.9210 4.4394 4.0469 Finance 1.0400 0.5585 0.1659 0.0000 1.7159 3.8809 Accountancy 2.7559 2.2743 1.8818 1.7159 0.0000 2.1651 MIS 4.9210 4.4394 4.0469 3.8809 2.1651 0.0000 Other Finance Accountancy MIS 0.1173 1.3306 2.8616 1.2133 2.7442 1.5309 Finance Accountancy 0.2267 0.0067 0.1277 Groups Marketing Management Other Finance Accounatncy MIS Sums = Count 19 11 14 45 36 47 172 6 Mean (M) 44.37 43.18 42.21 41.8 37.56 32.21 241.33 Tukey's Pair-wise Comparisons Table (t Table) … Marketing Management Marketing Management 0.3405 Other 0.6181 0.2776 Finance 0.7354 0.3949 Accountancy 1.9487 1.6082 MIS 3.4797 3.1391 Tukey's Pair-wise Comparisons Table (p- values Table) … Df = 166 Marketing Management Other Marketing Management 0.7339 Other 0.5374 0.7817 Finance 0.4631 0.6934 0.9067 Accountancy 0.0530 0.1097 0.1851 MIS 0.0006 0.0020 0.0048 MIS As seen from the above table, the MIS majors differ in the mean amxiety scores from the majors Marketing, Management, O (a) Main 120.08 81.9 78.79 63.83 79.77 47.94 79.88 48.63 55.43 64.06 64.99 53.82 62.43 65.07 81.02 Sat 1 30.75 61.83 26.4 53.84 72.3 53.09 27.67 52.46 10.64 53.5 37.28 34.31 66 8.99 29.75 Sat 2 75.86 37.88 68.73 51.08 50.21 58.47 86.29 62.9 44.84 64.17 50.68 47.97 60.57 58.37 30.4 Sat 3 54.05 38.82 36.85 32.83 52.94 34.13 69.37 78.52 55.95 49.61 66.4 76.06 11.37 83.51 39.17 One factor ANOVA ANOVA table Source Treatment Error Mean 69.843 41.254 56.561 51.972 54.908 n 15 15 15 15 60 Std. Dev 18.2022 19.3901 14.3302 20.2061 20.5045 SS 6,312.4439 18,493.0937 df 3 56 MS 2,104.14795 330.23382 Main Sat 1 Sat 2 Sat 3 Total F 6.37 p-value .0009 Total 24,805.5375 59 Since 0.0009 < 0.05, the result is significant. There is evidence of a difference in the mean waiting times in the four locations. (b) Post hoc analysis Tukey simultaneous comparison t-values (d.f. = 56) Sat 1 Sat 3 Sat 2 Main Sat 1 41.254 Sat 3 51.972 Sat 2 56.561 1.62 2.31 4.31 0.69 2.69 2.00 0.05 0.01 2.65 3.26 Sat 1 41.254 Sat 3 51.972 Sat 2 56.561 .1119 .0248 .0001 .4920 .0093 .0502 41.254 51.972 56.561 69.843 Main 69.843 critical values for experimentwise error rate: p-values for pairwise t-tests Sat 1 Sat 3 Sat 2 Main 41.254 51.972 56.561 69.843 Main 69.843 As seen from the above table, the locations that differ in mean waiting time are Main and Satellite 1, Main and Satellite 3, an F Stat = MSA/MSW = 80/20 = 4 1, Main and Satellite 3, and Satellite 1 and Satellite 2 10.61 The following data (stored in the file ) represent the nationwide highest yield of different types of accounts (extracted f Money Market Six Month CD One-Year CD 2.5 Year CD Five-Year CD 4 3.85 4.05 3.8 4.3 3.82 3.6 3.61 3.7 4.15 3.8 3.6 3.6 3.51 4.1 3.75 3.45 3.6 3.35 4 3.75 3.43 3.5 3.35 4 a. At the 0.05 level of significance, is there evidence of a difference in the mean yields of the different accounts? b. If appropriate, determine which accounts differ in mean yields c. At the 0.05 level of significance, is there evidence of a difference in the variation in yields among the different accounts? d. What effect does your result in (c) have on the validity of the results in (a) and (b)? (a) One factor ANOVA ANOVA table Source Treatment Error Total Mean 3.824 3.586 3.672 3.542 4.110 3.747 n 5 5 5 5 5 25 Std. Dev 0.1031 0.1680 0.2160 0.2039 0.1245 0.2603 SS 1.0563 0.5704 1.6267 df 4 20 24 MS 0.26409 0.02852 Money Market Six Month CD One-Year CD 2.5 Year CD Five-Year CD Total F 9.26 p-value .0002 Since 0.0002 < 0.05, the result is significant. There is evidence of a difference in the mean yields of the different accounts. (b) Post hoc analysis Tukey simultaneous comparison t-values (d.f. = 20) 2.5 Year CD 3.542 2.5 Year CD 3.542 Six Month CD 3.586 0.41 One-Year CD 3.672 1.22 Money Market 3.824 2.64 Five-Year CD 4.110 5.32 Six Month CD 3.586 One-Year CD 3.672 0.81 2.23 4.91 1.42 4.10 Money MarketFive-Year CD 3.824 4.110 2.68 critical values for experimentwise error rate: 0.05 0.01 2.99 3.74 p-values for pairwise t-tests 2.5 Year CD Six Month CD One-Year CD Money Market Five-Year CD 3.542 3.586 3.672 3.824 4.110 2.5 Year CD 3.542 Six Month CD 3.586 One-Year CD 3.672 .6848 .2377 .0157 3.32E-05 .4302 .0375 .0001 .1701 .0006 Money MarketFive-Year CD 3.824 4.110 .0145 From the above table, we see that the Five-year CD differs in mean yield from those of all the rest. Also, Money Market CD di
(c) Levene’s test for homogeneity of variance
Money Market Six Month CD One-Year CD 2.5 Year CD
Five-Year CD
4
3.85
4.05
3.8
4.3
3.82
3.6
3.61
3.7
4.15
3.8
3.6
3.6
3.51
4.1
3.75
3.45
3.6
3.35
4
3.75
3.43
3.5
3.35
4
Medians =
3.8
3.6
3.6
3.51
4.1
Table of absolute deviations from medians:
Money Market Six Month CD One-Year CD 2.5 Year CD
Five-Year CD
0.2
0.25
0.45
0.29
0.2
0.02
0
0.01
0.19
0.05
0
0
0
0
0
-0.05
-0.15
0
-0.16
-0.1
-0.05
-0.17
-0.1
-0.16
-0.1
One factor ANOVA
ANOVA table
Source
Treatment
Error
Total
Mean
0.024
-0.014
0.072
0.032
0.010
0.025
n
5
5
5
5
5
25
Std. Dev
0.1031
0.1680
0.2160
0.2039
0.1245
0.1568
SS
0.0200
0.5704
0.5904
df
4
20
24
MS
0.00501
0.02852
Money Market
Six Month CD
One-Year CD
2.5 Year CD
Five-Year CD
Total
F
0.18
p-value
.9484
Since 0.9484 > 0.05, the result is not significant. There is no evidence of a difference in the variation in yields among the diffe
(d) Our result in (c) means that the assumption of homogeneity of variance which was made in parts (a) and (b) is valid and it
11-25 Where people turn to for news is different for various age groups. A study indicated where different age groups primar
MEDIA Under 36 36-50 50+
Local TV 107 119 133
National TV 73 102 127
Radio 75 97 109
Local Newspaper 52 79 107
Internet 95 83 76
At the 0.05 level of significance, is there evidence of a significant relationship between the age group and where people prim
Ho: There is no relationship between age group amd media
Ha: There is a significant relationship between age group and media
Media
Local TV
National TV
Radio
Local Newspaper
Internet
Under 36
107
73
75
52
95
Age Group
36 – 50
119
102
97
79
83
50+
133
127
109
107
76
Chi-square Contingency Table Test for Independence
Local TV
Observed
Expected
O-E
(O – E)² / E
National TV
Observed
Expected
O-E
(O – E)² / E
Radio
Observed
Expected
O-E
(O – E)² / E
Local Newspaper Observed
Expected
O-E
(O – E)² / E
Internet
Observed
Expected
O-E
(O – E)² / E
Under 36
107
100.64
6.36
0.40
73
84.66
-11.66
1.61
75
78.77
-3.77
0.18
52
66.72
-14.72
3.25
95
71.21
23.79
7.95
36 – 50
119
120.17
-1.17
0.01
102
101.09
0.91
0.01
97
94.06
2.94
0.09
79
79.67
-0.67
0.01
83
85.02
-2.02
0.05
50+
133
138.19
-5.19
0.20
127
116.25
10.75
0.99
109
108.17
0.83
0.01
107
91.62
15.38
2.58
76
97.77
-21.77
4.85
Total
359
359.00
0.00
0.61
302
302.00
0.00
2.61
281
281.00
0.00
0.28
238
238.00
0.00
5.84
254
254.00
0.00
12.85
Total
Observed
Expected
O-E
(O – E)² / E
402
402.00
0.00
13.39
480
480.00
0.00
0.17
552
552.00
0.00
8.63
1434
1434.00
0.00
22.18
22.18 chi-square
8 df
.0046 p-value
Since 0.0046 < 0.05, the result is significant. There is evidence of a significant relationship between the age group and where Contribution to ?2 is highest from Internet and Local Newspaper. It appears that it is these media that bear strong relationshi 12.1 Fitting a straight line to a set of data yields the following prediction line: Yi = 2 + 5Xi a. Interpret the meaning of the Y intercept, b0. b. Interpret the meaning of the slope, b1. c. Predict the value of Y for X = 3. (a) b0 = 2 is the value of Y when X = 0 (b) b1 = 5 is the increase in the value of Y for a unit increase in the value of X (c) When X = 3, Y = 2 + 5(3) = 17 t. Also, Money Market CD differs in yield from both 2.5 Years CD and Six Month CD 12.5 Circulation is the lifeblood of the publishing business. The larger the sales of a magazine, the more it can charge advertis Magazine Reported (X) Audited (Y) YM 621 299.6 CosmoGirl 359.7 207.7 Rosie 530 325 Playboy 492.1 336.3 Esquire 70.5 48.6 TeenPeople 567 400.3 More 125.5 91.2 Spin 50.6 39.1 Vogue 353.3 268.6 Elle 263.6 214.3 Magazine Reported (X) Audited (Y) YM 621 299.6 CosmoGirl 359.7 207.7 Rosie 530 325 Playboy 492.1 336.3 Esquire 70.5 48.6 TeenPeople 567 400.3 More 125.5 91.2 Spin 50.6 39.1 Vogue 353.3 268.6 Elle 263.6 214.3 a. Construct a scatter plot. For these data, b0 = 26.724 and b1 = 0.5719. b. Interpret the meaning of the slope, b1, in this problem. c. Predict the audited newsstand sales for a magazine that reports newsstand sales of 400,000. (a) 450 400 350 Audited 300 250 y = 0.5719x + 26.724 R² = 0.9015 200 150 100 50 0 0 100 200 300 400 500 600 700 Reported (b) b1 = 0.5719 means the audited figures increase by about 572 units for every 1000 units' increase in the reported sales (c) When X = 400, Y = 0.5719(400) + 26.724 = 255.484, that is 255,484 12.9 An agent for a residential real estate company in a large city would like to be able to predict the monthly rental cost for a. Construct a scatter plot. b. Use the least-squares method to find the regression coefficients b0 and b1. c. Interpret the meaning of b0 and b1 in this problem. d. Predict the monthly rent for an apartment that has 1,000 square feet. e. Why would it not be appropriate to use the model to predict the monthly rent for apartments that have 500 square feet? f. Your friends Jim and Jennifer are considering signing a lease for an apartment in this residential neighborhood. They are try Rent Size 950 850 1600 1450 1200 1085 1500 1232 950 718 1700 1485 1650 1136 935 72 6 875 700 1150 956 1400 1100 1650 1285 2300 1985 1800 1369 1400 1175 1450 1225 1100 1245 1700 1259 1200 1150 1150 896 1600 1361 1650 1040 1200 755 800 1000 1750 1200 Size (X) 850 1450 1085 1232 718 1485 1136 726 700 956 1100 1285 1985 1369 1175 1225 1245 1259 1150 896 1361 1040 755 1000 1200 (a) Rent (Y) 950 1600 1200 1500 950 1700 1650 935 875 1150 1400 1650 2300 1800 1400 1450 1100 1700 1200 1150 1600 1650 1200 800 1750 2500 y = 1.0651x + 177.12 R² = 0.7226 2000 Rent 1500 1000 500 0 0 500 1000 1500 Size (b) b0 = 177.12 and b1 = 1.07 2000 2500 (c) b0 = 177.12 has no meaning in this context since we can't have a house with 0 size. b1 = 1.07 is the dollar increase in rent (d) When X = 1000, Y = 177.12 + 1.07(1000) = $1247.12 (e) The data has houses of size in the range 700 sq ft to 2000 sq ft and therefore, the predicted rent for a 500 sq ft house will (f) For X = 1000, predicted Y = 177.12 + 1.07(1000) = $1247.12 and for X = 1200, predicted Y = 177.12 + 1.07(1200) = $1461.12 Since $1425 < the predicted value of $1461.12, leasing the 1200 sq ft apartment is a better deal. 12.17 In Problem 12.5 on page 417, you used reported magazine newsstand sales to predict audited sales (stored in the file C Magazine Reported Audited YM 621.0 299.6 CosmoGirl 359.7 207.7 Rosie 530.0 325.0 Playboy 492.1 336.3 Esquire 70.5 48.6 TeenPeople 567.0 400.3 More 125.5 91.2 Spin 50.6 39.1 Vogue 353.3 268.6 Elle 263.6 214.3 a. Determine the coefficient of determination, r2, and interpret its meaning. b. Determine the standard error of the estimate. c. How useful do you think this regression model is for predicting audited sales? (a) R^2 = SSR/SST = 130301.41/144538.64 = 0.9015 (b) DfE = 10 - 2 = 8, SSE = SST - SSR = 14237.23 MSE = SSE/DfE = 14237.23/8 = 1779.65 Standard error of estimate = vMSE = v1779.65 = 42.186 (c) R^2 value is quite high. This means the model is a good fit and can be used for predicting audited sales. more it can charge advertisers. Recently, a circulation gap has appeared between the publishers’ reports of magazines’ newsstand sales an he monthly rental cost for apartments, based on the size of the apartment, as defined by square footage. A sample of 25 apartments (sto neighborhood. They are trying to decide between two apartments, one with 1,000 square feet for a monthly rent of $1,275 and the other nt for a 500 sq ft house will have greater error. Therefore, it not be appropriate to use the model to predict the monthly rent for apartmen s of magazines’ newsstand sales and subsequent audits by the Audit Bureau of Circulations. The data in the file represent the reported and e. A sample of 25 apartments (stored in the file Rent) in a particular residential neighborhood was selected, and the information gathered nthly rent of $1,275 and the other with 1,200 square feet for a monthly rent of $1,425. Based on (a) through (d), which apartment do you he file represent the reported and audited newsstand yearly sales (in thousands) for the following 10 magazines: 12.21 In Problem 12.9 on page 418, an agent for a real estate company wanted to predict the monthly rent for apartments, b Rent Size 950 850 1600 1450 1200 1085 1500 1232 950 718 1700 1485 1650 1136 935 726 875 700 1150 956 1400 1100 1650 1285 2300 1985 1800 1369 1400 1175 1450 1225 1100 1245 1700 1259 1200 1150 1150 896 1600 1361 1650 1040 1200 755 800 1000 1750 1200 Size (X) 850 1450 1085 1232 718 1485 1136 726 700 956 1100 1285 1985 1369 1175 1225 1245 1259 1150 896 1361 1040 755 1000 1200 Rent (Y) 950 1600 1200 1500 950 1700 1650 935 875 1150 1400 1650 2300 1800 1400 1450 1100 1700 1200 1150 1600 1650 1200 800 1750 a. Determine the coefficient of determination, r2, and interpret its meaning. b. determine the standard error of the estimate, and interpret its meaning. c. How useful do you think this regression model is for predicting the monthly rent? d. Can you think of other variables that might explain the variation in monthly rent? Regression Analysis r² 0.723 r 0.850 Std. Error 194.595 n 25 k 1 Dep. Var. Rent (Y) ANOVA table Source Regression Residual Total Regression output SS 2,268,776.5453 870,949.4547 3,139,726.0000 df 1 23 24 MS 2,268,776.5453 37,867.3676 F 59.91 p-value 7.52E-08 confidence interval variables Intercept Size (X) coefficients std. error 177.1208 1.0651 0.1376 t (df=23) 7.740 p-value 95% lower 7.52E-08 0.7805 (a) r^2 = 0.723. This means about 72.3% of the variation in the monthly rent is explained by the variation in size using this mo (b) SEE = 194.595. This means on average, the predicted rents using this model are about $194.60 higher than the actual rent (c) r^2 = 0.723 is a low value. The standard error of estimate is also quite high. This means the model is a poor fit, and may no (d) Monthly rents may also be affected by other factors such as location of the house (distance from city center), number of b 12.43 The data in the file Coffeedrink represent the calories and fat (in grams) of 16-ounce iced coffee drinks at Dunkin’ Donu Product Calories (X) Fat (Y) DD Iced Mocha Latte 240 8.0 Starbucks Frap. 260 3.5 DD Coolatta 350 22.0 Starbucks Mocha Expresso 350 20.0 Starbucks Mocha Frap. 420 16.0 Starbucks Chocolate Brownie Frap. 510 22.0 Starbucks Chocolate Frap. 530 19.0 a. Compute and interpret the coefficient of correlation, r. b. At the 0.05 level of significance, is there a significant linear relationship between calories and fat? Product DD Iced Mocha Latte Starbucks Frap. DD Coolatta Starbucks Mocha Expresso Starbucks Mocha Frap. Starbucks Chocolate Brownie Starbucks Chocolate Frap. Calories (X) 240 260 350 350 420 510 530 Fat (Y) 8 3.5 22 20 16 22 19 Regression Analysis r² 0.518 r 0.720 Std. Error 5.529 n 7 k 1 Dep. Var. Fat (Y) ANOVA table Source Regression Residual Total SS 164.0951 152.8335 316.9286 df 1 5 6 MS 164.0951 30.5667 F 5.37 p-value .0683 Regression output variables Intercept Calories (X) coefficients std. error -1.7794 0.0462 0.0200 t (df=5) 2.317 confidence interval p-value 95% lower .0683 -0.0051 (a) The correlation coefficient is r = 0.72 (b) P- value for the regression hypothesis test is 0.0683. Since this is > 0.05, the result is not significant. There is no sufficient e
he monthly rent for apartments, based on the size of the apartment (stored in the file Rent). Using the results of that problem
confidence interval
95% upper
1.3498
std. coeff.
0.000
0.850
y the variation in size using this model
194.60 higher than the actual rent
the model is a poor fit, and may not be reliable. It may not be appropriate to use the model to predict monthly rents.
ance from city center), number of bedrooms, number of bathrooms, whether the house has a pool etc.
iced coffee drinks at Dunkin’ Donuts and Starbucks:
confidence interval
95% upper
0.0975
std. coeff.
0.000
0.720
t significant. There is no sufficient evidence of a linear relationship between calories and fat.
results of that problem
…
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