Question

Financial analysts know that January credit card charges will generally be much lower than those of the month before. What about the difference between January and the next​ month? Does the trend​ continue? The accompanying data set contains the monthly credit card charges of a random sample of 99 cardholders. Complete parts​ a) through​ e) below.

January	February
902.74	641.04
7212.18	4565.35
4235.42	2270.56
79.92	300.09
4045.57	1377.72
89.29	−120.74
3289.59	1928.85
2419.54	2609.97
83.81	144.83
6.42	392.85
0.00	40.46
564.69	295.63
2712.23	848.62
187.12	162.12
3265.86	2412.45
1523.59	956.31
1359.23	38.03
733.33	2656.79
75.09	64.94
70.29	−70.32
634.53	1862.61
1041.23	478.07
553.08	994.64
1016.27	774.54
1304.94	3368.08
249.39	5.52
48.78	96.93
872.34	890.89
485.94	485.21
616.52	1485.52
1574.18	890.46
422.34	391.43
770.85	323.19
56.53	0.00
1486.78	2253.73
495.28	390.19
1064.88	1065.85
510.65	131.33
5637.68	4942.63
5.49	5.51
871.63	591.44
1636.66	3364.19
92.13	85.99
669.34	1367.13
829.32	280.85
69.24	67.99
830.54	1057.56
2301.44	3317.76
270.67	14.13
210.42	160.52
1012.36	519.35
1044.96	2021.35
298.64	635.44
−29.99	0.00
1634.61	393.34
1731.93	1323.33
0.00	65.16
31.43	28.75
4.95	77.15
1088.69	892.78
26.88	29.03
120.31120.31	32.23
2007.48	815.63
291.31	779.47
104.02	0.00
53.01	66.25
2842.52	1530.91
675.47	293.45
221.86	171.92
37.79	4.78
533.25	880.96
1932.71	1063.55
692.17	915.55
6804.35	5941.41
393.36	466.47
1309.18	302.89
796.21	497.02
0.00	266.64
1040.29	59.45
565.12	206.62
339.14	412.34
5275.34	5324.54
40.09	72.58
43.39	38.45
653.63	480.25
1071.23	416.29
2337.04	1787.19
91.47	175.32
1433.01	1107.78
719.86	307.79
28.61	24.19
980.34	1216.35
1576.18	1810.23
0.00	468.24
161.96	147.68
494.32	1995.28
534.11	935.24
462.45	114.51
1478.23	2093.37

​a) Build a regression model to predict February charges from January charges.

Feb=____+____Jan ​(Round to two decimal places as​ needed.)

Check the conditions for this model. Select all of the true statements related to checking the conditions.

A. All of the conditions are definitely satisfied.

B. The Linearity Condition is not satisfied.

C. The Randomization Condition is not satisfied.

D. The Equal Spread Condition is not satisfied.

E. The Nearly Normal Condition is not satisfied. ​

b) How​ much, on​ average, will cardholders who charged ​$2000 in January charge in​ February? ​$____ ​(Round to the nearest cent as​ needed.)

​c) Give a​ 95% confidence interval for the average February charges of cardholders who charged ​$2000 in January.

($___,$___) ​(Round to the nearest cent as​ needed.) ​

d) From part​ c), give a​ 95% confidence interval for the average decrease in the charges of cardholders who charged ​$2000 in January.

​($___,$___) (Round to the nearest cent as​ needed.)

​e) What​ reservations, if​ any, would a researcher have about the confidence intervals made in parts​ c) and​ d)? Select all that apply.

A. The residuals show increasing​ spread, so the confidence intervals may not be valid.

B. The residuals show a curvilinear​ pattern, so the confidence intervals may not be valid.

C. The data are not​ linear, so the confidence intervals are not valid.

D. The data are not​ independent, so the confidence intervals are not valid.

E. A researcher would not have any reservations. The confidence intervals are valid. Click to select your answer(s).

Answer 1

Ʃx =	104470.25
Ʃy =	91867.94
Ʃxy =	236361665.1
Ʃx² =	302938229.1
Ʃy² =	228933590.9
Sample size, n =	99
x̅ = Ʃx/n = 104470.25/99 =	1055.255051
y̅ = Ʃy/n = 91867.94/99 =	927.9589899
SSxx = Ʃx² - (Ʃx)²/n = 302938229.1245 - (104470.25)²/99 =	192695470.18448
SSyy = Ʃy² - (Ʃy)²/n = 228933590.8566 - (91867.94)²/99 =	143683910.05010
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 236361665.059 - (104470.25)(91867.94)/99 =	139417557.39451

a)

Slope, b = SSxy/SSxx = 139417557.39451/192695470.18448 = 0.7235124

y-intercept, a = y̅ -b* x̅ = 927.95899 - (0.72351)*1055.25505 = 164.4689

Regression equation :

ŷ = 164.47 + (0.72) x

conditions.

D. The Equal Spread Condition is not satisfied.

E. The Nearly Normal Condition is not satisfied.

----

b) Predicted value of y at x = 2000

ŷ = 164.4689 + (0.7235) * 2000 = 1611.4937

----

c) Significance level, α = 0.05

Critical value, t_c = T.INV.2T(0.05, 97) = 1.9847

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 143683910.0501 - (139417557.39451)²/192695470.18448 = 42813581.9

Standard error, se = √(SSE/(n-2)) = √(42813581.90361/(99-2)) = 664.36220

95% Confidence interval :

Lower limit = ŷ - tc*se*√((1/n) + ((x-x̅)²/(SSxx))) = 1451.45

Upper limit = ŷ + tc*se*√( (1/n) + ((x-x̅)²/(SSxx))) = 1771.54

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d) 95% Prediction interval :

Lower limit = ŷ - tc*se*√(1 + (1/n) + ((x-x̅)²/(SSxx))) = 283.24

Upper limit = ŷ + tc*se*√(1 + (1/n) + ((x-x̅)²/(SSxx))) = 2939.75

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e) A. The residuals show increasing​ spread, so the confidence intervals may not be valid.