Question

All of number 3 please including letters d and e at the bottom

(a) Use the method of moments to find a point estimate for p. 3.(100] 6.5-3. The midterm and final exam scores of 10 students in a statistics course are tabulated as shown. (a) Calculate the least squares regression line for these data. (b) Plot the points and the least squares regression line on the same graph. (c) Find the value of σ2. Midterm Final Midterm Final 70 87 67 70 64 74 73 83 79 91 94 74 79 84 98 80 96 (d) Compute and plot residuals (e) What is the expected final exam score of a student whose midterm score was x = 68?

Answer 1

a) The following data are provided:

Midterm	Final
70	87
74	79
80	88
84	98
80	96
67	73
70	83
64	79
74	91
82	94

The independent variable is Midterm, and the dependent variable is Final. In order to compute the regression coefficients, the following table needs to be used:

	Midterm	Final	Midterm*Final	Midterm2	Final2
	70	87	6090	4900	7569
	74	79	5846	5476	6241
	80	88	7040	6400	7744
	84	98	8232	7056	9604
	80	96	7680	6400	9216
	67	73	4891	4489	5329
	70	83	5810	4900	6889
	64	79	5056	4096	6241
	74	91	6734	5476	8281
	82	94	7708	6724	8836
Sum =	745	868	65087	55917	75950

Based on the above table, the following is calculated:

$\bar X = \frac{1}{n} \sum_{i=1}^{n} X_i = \frac{ 745}{ 10} = 74.5$

$\bar Y = \frac{1}{n} \sum_{i=1}^{n} Y_i = \frac{ 868}{ 10} = 86.8$

$SS_{XX} = \sum_{i=1}^{n} X_i^2 - \displaystyle\frac{1}{n}\left(\sum_{i=1}^{n} X_i\right)^2 = 55917 - 745^2/10 = 414.5$

$SS_{YY} = \sum_{i=1}^{n} Y_i^2 - \displaystyle\frac{1}{n}\left(\sum_{i=1}^{n} Y_i\right)^2 = 75950 - 868^2/10 = 607.599$

$SS_{XY} = \sum_{i=1}^{n} X_i Y_i - \displaystyle\frac{1}{n}\left(\sum_{i=1}^{n} X_i\right) \left(\sum_{i=1}^{n} Y_i\right) = 65087 - 745 \times 868/10 = 421$

Therefore, based on the above calculations, the regression coefficients (the slope m, and the y-intercept n) are obtained as follows:

$m = \frac{SS_{XY}}{SS_{XX}} = \frac{ 421}{ 414.5} = 1.0157$

$n = \bar Y - \bar X \cdot m = 86.8 - 74.5 \times 1.0157 = 11.1317$

Therefore, we find that the regression equation is:

Final = 11.1317 + 1.0157 * Midterm

y = 11.1317 + 1.0157 * x

b)

Graphically:

Scatter Plot and Regression Line 103 98 93 LL 83 78 73 680 62 64 66 68 70 72 74 7678 80 82 84 8688 Midtern -Regression equation: Y = 11.1317 + 1.0157 * X

c)

$\hat \sigma^2 = \frac{S_{yy} - \frac{S_{xy}^2}{S_{xx}}}{n-2} =\frac{607.59 - \frac{421^2}{414.5}}{10-2} = 22.49850875$

d)

	Midterm(x)	Final(y)	y^ = 11.1317 + 1.0157 * x	residual = y - y^
	70	87	82.2307	4.7693
	74	79	86.2935	-7.2935
	80	88	92.3877	-4.3877
	84	98	96.4505	1.5495
	80	96	92.3877	3.6123
	67	73	79.1836	-6.1836
	70	83	82.2307	0.7693
	64	79	76.1365	2.8635
	74	91	86.2935	4.7065
	82	94	94.4191	-0.4191
Sum =	745	868	868.0135	-0.0135

e)

When Midterm = x = 68

Final = y = 11.1317 + 1.0157 * 68 = 80.1993

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