Question

The table below gives the number of hours spent unsupervised each day as well as the overall grade averages for seven randomly selected middle school students. Using this data, consider the equation of the regression line, yˆ=b0+b1xy^=b0+b1x, for predicting the overall grade average for a middle school student based on the number of hours spent unsupervised each day. Keep in mind, the correlation coefficient may or may not be statistically significant for the data given. Remember, in practice, it would not be appropriate to use the regression line to make a prediction if the correlation coefficient is not statistically significant.

Hours Unsupervised	1.51.5	2.52.5	33	3.53.5	4.54.5	55	66
Overall Grades	8989	8686	7777	7474	7373	7272	7171

Step 1 of 6:

Find the estimated slope. Round your answer to three decimal places.

Step 2 of 6:

Find the estimated y-intercept. Round your answer to three decimal places.

Step 3 of 6:

According to the estimated linear model, if the value of the independent variable is increased by one unit, then the change in the dependent variable yˆy^ is given by

Step 5 of 6:

Find the error prediction when x=5x=5. Round your answer to three decimal places.

Step 6 of 6:

Find the value of the coefficient of determination. Round your answer to three decimal places.

Answer 1

X	Y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
1.5	89	4.9030612	133.898	-25.6224
2.5	86	1.4744898	73.46939	-10.4082
3	77	0.5102041	0.183673	0.306122
3.5	74	0.0459184	11.7551	0.734694
4.5	73	0.6173469	19.61224	-3.47959
5	72	1.6530612	29.46939	-6.97959
6	71	5.2244898	41.32653	-14.6939

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	26	542	14.42857	309.7143	-60.1429
mean	3.714285714	77.4285714	SSxx	SSyy	SSxy

sample size ,   n =   7      
here, x̅ =   3.714285714       ȳ =   77.42857143
              
SSxx =    Σ(x-x̅)² =    14.42857143      
SSxy=   Σ(x-x̅)(y-ȳ) =   -60.14285714      

correlation coefficient ,    r = Sxy/√(Sx.Sy) =   -0.8997
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correlation hypothesis test      
Ho:   ρ = 0  
Ha:   ρ < 0  
n=   7  

df=n-2 = 5

alpha,α =    0.05  
correlation , r=   -0.8997  
t-test statistic =    t = r*√(n-2)/√(1-r²) =    -4.6085
critical t-value = -2.0150
p-value =    0.0029  
since,p-value<α , reject Ho

hence, correlation is significant .

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1)

SSxx =    Σ(x-x̅)² =    14.42857143
SSxy=   Σ(x-x̅)(y-ȳ) =   -60.14285714

slope ,    ß1 = SSxy/SSxx =   -4.168

2)

intercept,   ß0 = y̅-ß1* x̄ =   92.911

3)

so, regression line is   Ŷ =   92.9109   +   -4.1683   *x

According to the estimated linear model, if the value of the independent variable is increased by one unit, then the change in the dependent variable yˆ is given by decrease by 4.1683

5) when x=5

then

so, regression line is   Ŷ =   92.9109   +   -4.1683*5 = 72.069

error= observed-predicted value= 72-72.069 = - 0.069

6)

value of the coefficient of determination = r² = -0.8997² = 0.809