Question

Professor Gill has taught General Psychology for many years. During the semester, she gives three multiple-choice exams, each worth 100 points. At the end of the course, Dr Gil gives a comprehensive alworth 200 poset, and represent students scores on exams 1, 2, and respectively. Let X represent the student's score on the final exam. Last semester De Gil had 25 students in her cess. The student exam scores are shows below 73 75 152 93 185 91 90 180 99 100 196 23 66 70 142 53 55 101 69 74 77 149 47 56 115 175 770 70 73 70 65 14 93 95 91 14 73 BD 153 70 7370 93 89 193 78 75 16 81 9093 183 03 80 17 783 32 36 90 127 1 175 783 35 135 76 78 96 93 Since Professor Glas not changed the course much front semester to the present to the preceding date should be for constructing a repression model is meer as well () Generale sunmayistics, including the mean and standard deviation of each vor compute the coffen of vertente enth we (we decimale) Belative to sman, would you say that each eam had but the same spread of scores Most professors do not to get exam that is extremely cay or extremely hard. Would you say that all of the cams were about same level of my consider to means and spread of best scores) Yes the read is what the same, thout the same level of my No, the spread the best love of mult

Answer 1

(a) Generating the summary statistics: Using excel,

File Home Insert Page Layout View 21 21 UU Formulas Data Review Connections Properties Refresh All- Edit Links Connections Clear Reapply Advanced From Access Filter From Web At Sort From From Other Text Sources Get External Data Existing Connections Text to Remove Columns Duplicates Valid Da Sort & Filter N18 A B С D E ? X 1 x1 X2 X3 X4 73 80 75 152 2. 3 OK 93 88 93 185 4 89 91 90 180 Cancel 5 96 98 100 196 Data Analysis Analysis Tools Anova: Single Factor Anova: Two-Factor With Replication Anova: Two Factor Without Replication Correlation Covariance Descriptive Statistics Exponential Smoothing F-Test Two-Sample for Variances Fourier Analysis Histogram Help 6 73 66 70 142 7 53 55 101 46 74 8 69 77 149 9 47 56 60 115 10 87 79 90 175 11 79 70 88 164 12 69 70 73 141

А B с D E F G H J K 1 x1 X2 x3 X4 2 73 75 152 80 88 3 93 93 185 ? 4 x 89 91 90 Descriptive Statistics 180 5 96 98 100 196 142 OK 6 73 66 70 Input Input Range: Grouped By: 7 53 46 55 101 Cancel $A$1:$D$26 Columns Rows 8 69 74 77 149 Help 9 47 56 60 115 Labels in first row 10 87 79 90 175 11 79 70 88 164 12 69 70 73 141 13 70 65 74 141 14 93 95 91 184 Output options Qutput Range: SF $2 New Worksheet Ply: New Workbook Summary statistics Confidence Level for Mean: Kth Largest: 1 Kth Smallest: 1 15 79 80 73 152 16 70 73 78 148 95 17 93 89 96 192 18 78 75 68 147 19 81 90 93 183 20 88 92 86 177 21 78 83 77 159

We get the output:

A B C D E F G H K 1 x1 x2 x3 X4 2 73 80 75 152 x1 x2 x3 X4 3 93 88 93 185 4 89 91 90 180 Mean 79.04 81.48 162.04 79.48 2.50 5 96 100 196 2.46 2.35 4.81 98 66 6 73 70 142 79 82 85 164 7 53 46 55 101 93 83 175 8 69 74 77 149 12.28 12.50 90 11.77 138.43 24.04 9 47 56 60 115 150.79 156.34 578.12 10 87 79 90 0.82 Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count 0.95 -0.83 -0.50 -0.48 88 -0.91 11 12 79 69 70 70 73 49 175 164 141 141 184 152 45 52 46 0.34 -0.74 95 101 196 4051 70 65 74 47 93 95 91 98 13 14 15 16 17 96 1976 55 100 2037 1987 79 70 80 73 73 78 148 25 25 25 25 93 89 96 192

Substituting the mean and standard deviations to compute the Coefficient of Variation (in %):

CV 100 T

	T	s	CV(%)
x1	79.04	12.08	12.08 / 79.04 = 15.54
x2	79.48	12.50	15.73
x3	81.48	11.77	14.44
x4	162.04	24.04	14.84

Comparing the spread relative to the mean would be the same as comparing the CVs. We find the CV for all the four variables are approximately equal i.e. they do not appear to be significantly different. We may say that:

Yes, the spread is about the same.Yes, the tests are about the same level of difficulty.

(b) Computing the correlation coefficient:

SUM - Xfx =CORREL(A2:A26,B2:B26 A B C D E F G H - 1 x1 X2 x3 X4 r 2 73 80 75 152 r^2 =CORREL(A2:A26,B2:326 CORREL(array1, array2) 3 93 88 93 185 4 89 91 90 180 x1,x2 x1,x3 x1,x4 X2,3 x2,4 X3,4 5 96 98 100 196 6 73 66 70 142 7 53 46 55 101 8 69 74 77 149 47 56 60 115 9 10 90 87 79 79 70 70 175 164 11 88 12 69 73 141

	r	r2
x1,x2	0.90	0.81
x1,x3	0.89	0.80
x1,x4	0.95	0.90
x2,x3	0.85	0.72
x2,x4	0.93	0.86
x3,x4	0.97	0.95

We find the highest correlation coefficient obtained is 0.97 observed between 3 and 4. Hence, we may say that:

3. Beacause it has the highest correlation with 4. Yes, the other two still have a lot of influence because of their high correlations with 4.

(c) Running a multiple regression by regressing x₄ on the predictors x₁,x₂ and x₃, we get the output:

File Home Insert Page Layout Formulas Data Review View 21 z ZA From Access From Web From From Other Text Sources Get External Data Existing Connections Dil Connections Properties Refresh All Edit Links Connections Sort * Clear Reapply Filter Advanced Sort & Filter Text to Remove Columns Duplicates 15 A B с D E F G H J K 1 x1 x2 X3 X4 2 73 80 75 152 ? 3 93 88 185 4 89 OK 91 98 5 96 73 Cancel 66 180 196 142 101 149 115 93 90 100 70 55 77 60 90 Data Analysis Analysis Tools Exponential Smoothing F-Test Two-Sample for Variances Fourier Analysis Histogram Moving Average Random Number Generation Rank and Percentile Regression Sampling t-Test: Paired Two Sample for Means Help 53 69 6 7 8 9 10 11 12 13 46 74 56 47 79 175 87 79 69 70 164 70 141 88 73 74 91 70 65 141 14 93 95 184 15 79 8n 73 152

F2 for А B C D 1 x1 х x2 x3 X4 Regression 2 75 73 93 80 88 OK 3 93 Input Input Y Range: Input X Range: $D$1:$D$26 89 90 Cancel 91 98 $A$1:$C$26 96 152 185 180 196 142 101 149 Help 6 73 100 70 55 66 Labels Confidence Level: Constant is Zero 95 % 46 7 8 53 69 47 77 74 56 9 115 60 90 SF$2 10 87 79 175 11 79 70 88 164 Output options Output Range: New Worksheet Ply: O New Workbook Residuals Residuals Standardized Residuals 12 69 70 73 141 13 70 65 74 141 93 91 184 14 15 95 80 Residual Plots Line Fit Plots 79 73 152 16 70 73 78 148 Normal Probability Normal Probability Plots 17 93 89 96 192 18 78 75 68 147 19 81 90 93 183 20 88 92 86 177 21 78 83 77 159 22 82 86 90 177

A B с D E F G H 1 K L M 1 x1 x2 X3 X4 73 80 N 75 152 SUMMARY OUTPUT 93 88 93 185 89 91 90 180 3 4 5 5 96 98 100 196 73 66 70 Regression Statistics Multiple R 0.995 R Square 0.990 Adjusted R Square 0.988 Standard Error 2.614 Observations 25 53 46 142 101 149 55 3 69 74 77 m 47 56 60 115 LO 87 79 90 175 1 79 70 88 164 ANOVA 2 69 70 73 141 df SS F MS 4577.17 Significance F 0.000 3 70 65 74 141 3 13731.51 670.09 4 93 95 91 184 Regression Residual Total 21 6.83 143.45 13874.96 5 79 80 73 152 24 -6 70 73 78 148 -7 93 89 96 192 P-value Coefficients Standard Error -4.34 3.76 -8 78 75 68 147 t Stat -1.15 2.93 Intercept x1 0.262 81 90 93 183 0.36 9 20 0.608 0.12 0.10 Lower 95% Upper 95% - 12.164 3.492 0.103 0.333 0.752 0.953 1.382 88 92 86 177 X2 0.54 0.008 0.000 0.000 5.38 11.33 1 78 83 77 159 X3 1.17 0.10 22 82 86 90 177 175 23 86 82 89

(d) The fitted regression equation can be expressed as:

y = -4.34+ 0.36.21 +0.54r2 + 1.1773

where the intercept is estimated to be -4.34 and the slope of x₁ is estimated to be 0.36, other predictors in the model being constant; similarly for x₂ and x₃. Hence, we may say that:

If we hold all other explanatory variables as fixed constant, then we can look at one coefficient as slope.

Here, the slope of x₃ can be interpreted as: the mean score of x₄ increases by 1.17 units for a unit increase in x₃. Hence, if marks in x₃ increases by 14, x₄ marks is expected to increase by (14)(1.17) = 16.38 points

(e) We test the significance of the slope coefficients by testing the hypothesis:

Ho: =0,2 = 1,2,3
Vs
3; # , i = 1, 2, E

	t	P-value
$\beta _{1}$	2.93	0.008
$\beta _{2}$	5.38	0.000
$\beta _{3}$	11.33	0.000

Since, the p-value of the t test of all three predictors are significant at 5% level (since, 0.008,0.000 < 0.05), we may say that we reject all null hypothesis; there is sufficient evidence that differ from zero.

If a coefficient is different from zero, then it contributes to the regression equation.

(f) The 90% CI for slope can be constructed using the formula:

B 77,-2SE(3

where the critical value of t is obtained as:

ETINV(0.10,25-2 TINV(probability, deg_freedom)

= 1.714

	Lower Limit	Upper Limit
$\beta _{1}$	0.36 - 1.714 x 0.12 = 0.15	0.36 + 1.714 x 0.12 = 0.56
$\beta _{2}$	0.37	0.72
$\beta _{3}$	0.99	1.34

For x₁ = 68, x₂ = 72,x₃ = 75

y = -4.34 + 0.36(68) + 0.54(72) + 1.17(75)

= 146.49 = 146 (approx.)