Question

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A study is conducted to examine the influence of ‘screen time’ on student performance on Statistics exams. A class of 12 students is observed over a period of time, with the independent variable being the average amount of time per day each student spends on TV/internet, and the dependent variable being their subsequent Statistics exam score, in %. The data is shown in the table below:

Student	Hrs/Day Watching TV or on Internet	Exam Score (%)
1	0	100
2	4.6	79
3	2.5	65
4	7.2	26
5	3.0	41
6	3.9	94
7	4.2	94
8	3.9	52
9	8.6	13
10	3.9	84
11	2.8	66
12	4.0	61

(a) Determine the equation of the line of best fit, relating Y = exam score (%) to X = hrs/day spent watching TV or on the internet.

(b) Use the line of best fit calculated in Part (a) to calculate an estimate for the exam score that a student would get, to the nearest %, if they spent an average of 2.5 hrs/day watching TV or on the internet. Repeat for an estimate of the exam score that would result after a student spent an average of 5.0 hrs/day watching TV or on the internet.

(c) Plot the raw data from the table on an x-y graph, and then draw the line of best fit showing at least 2 calculated points that are on that line (Hint: your answers for Parts (a) and (b) provide you with 4 such points).

(d) What is the predicted exam score, to the nearest %, for a student who completely avoids the TV or internet? Repeat for a predicted exam score for a student who spends an average of 12.0 hrs/day watching TV or on the internet. Comment briefly on your answers for these two estimates, and what it implies about the limitations of the linear regression model generated in Part (a).

(e) Calculate the covariance between X and Y.

(f) Calculate the correlation coefficient for this data set.

(g) Calculate the coefficient of determination for this data set, and explain what its value means with respect to the line of best fit calculated in Part (a).

(h) Conduct a hypothesis test on the significance of correlation between hrs/day spent watching TV or on the internet, and exam performance, using the critical-value method at LOC = 95%.

(i) Use the p-value method to determine the common values of LOC (if any) for which your decision in Part (h) would be that there is no significant correlation between X and Y, and the common LOC values (if any) for which the opposite decision would be made.

Answer 1

X	Y	XY	X²	Y²
0	100	0	0	10000
4.6	79	363.4	21.16	6241
2.5	65	162.5	6.25	4225
7.2	26	187.2	51.84	676
3	41	123	9	1681
3.9	94	366.6	15.21	8836
4.2	94	394.8	17.64	8836
3.9	52	202.8	15.21	2704
8.6	13	111.8	73.96	169
3.9	84	327.6	15.21	7056
2.8	66	184.8	7.84	4356
4	61	244	16	3721
Ʃx =	Ʃy =	Ʃxy =	Ʃx² =	Ʃy² =
48.6	775	2668.5	249.32	58501

Sample size, n =	12
x̅ = Ʃx/n = 48.6/12 =	4.05
y̅ = Ʃy/n = 775/12 =	64.5833333
SSxx = Ʃx² - (Ʃx)²/n = 249.32 - (48.6)²/12 =	52.49
SSyy = Ʃy² - (Ʃy)²/n = 58501 - (775)²/12 =	8448.91667
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 2668.5 - (48.6)(775)/12 =	-470.25

(a)

Slope, b = SSxy/SSxx = -470.25/52.49 = -8.958849

y-intercept, a = y̅ -b* x̅ = 64.58333 - (-8.95885)*4.05 = 100.86667

Regression equation :

ŷ = 100.8667 + (-8.9588) x

(b)

Predicted value of y at x = 2.5

ŷ = 100.8667 + (-8.9588) * 2.5 = 78.4695

Predicted value of y at x = 5

ŷ = 100.8667 + (-8.9588) * 5 = 56.0724

(c) Scatter plot:

Scatterplot Exam score 10 2 4 6 8 Hrs/Day Watching Tv or on Internet

(d)

Predicted value of y at x = 0

ŷ = 100.8667 + (-8.9588) * 0 = 100.8667

Predicted value of y at x = 12

ŷ = 100.8667 + (-8.9588) * 12 = -6.6395

(e)

Covariance between X and Y. = SSxy/(n-1) = -470.25/11 = -42.75

(f)

Correlation coefficient, r = SSxy/√(SSxx*SSyy) = -470.25/√(52.49*8448.91667) = -0.7061

(g)

Coefficient of determination, r² = (SSxy)²/(SSxx*SSyy) = (-470.25)²/(52.49*8448.91667) = 0.4986

49.86% variation in y is due to the linear relationship between y and x variables.   

(h)

Null and alternative hypothesis:

Ho: ρ = 0 ; Ha: ρ ≠ 0

α = 0.05

Correlation, r = -0.7061

Test statistic :  

t = r*√(n-2)/√(1-r²) = -0.7061 *√(12 - 2)/√(1 - -0.7061²) = -3.1536

df = n-2 = 10

Critical value, t_c = T.INV.2T(0.05, 10) = 2.2281

Conclusion:

p-value < α Reject the null hypothesis. There is a correlation between x and y.

(i)

if p-value > 0.05, then there is no significant correlation between X and Y