Question

A random selection of students at a school were chosen at random and how long they spent eating compared to how many calories they consumed was recorded as the following:

Time, minutes	21.4	30.8	37.7	33.5	32.8	39.5	22.8
Calories	472	498	465	456	423	437	508
Time, minutes	34.1	33.9	43.8	42.4	43.1	29.2	31.3
Calories	431	479	454	450	410	504	437
Time, minutes	28.6	32.9	30.6	35.1	33.0	43.7
Calories	489	436	480	439	444	408

1. Make a scatterplot for the data

2. Find the LSRL

3. Describe the slope and the y-intercept, do they make sense?

4. Create a Residual Plot for the data

5. Create a 95% confidence interval for the slope of the line

Answer 1

1)

2)

x	y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
21.4	472	159.0121	256.0000	-201.7600
30.8	498	10.3041	1764.0000	-134.8200
37.7	465	13.6161	81.0000	33.2100
33.5	456	0.2601	0.0000	0.0000
32.8	423	1.4641	1089.0000	39.9300
39.5	437	30.1401	361.0000	-104.3100
22.8	508	125.6641	2704.0000	-582.9200
34.1	431	0.0081	625.0000	-2.2500
33.9	479	0.0121	529.0000	-2.5300
43.8	454	95.8441	4.0000	-19.5800
42.4	450	70.39	36.00	-50.34
43.1	410	82.63	2116.00	-418.14
29.2	504	23.14	2304.00	-230.88
31.3	437	7.34	361.00	51.49
28.6	489	29.27	1089.00	-178.53
32.9	436	1.23	400.00	22.20
30.6	480	11.63	576.00	-81.84
35.1	439	1.19	289.00	-18.53
33	444	1.02	144.00	12.12
43.7	408	93.90	2304.00	-465.12

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	680.20	9120.00	758.06	17032.00	-2332.60
mean	34.01	456.00	SSxx	SSyy	SSxy

sample size ,   n =   20          
here, x̅ = Σx / n=   34.010   ,     ȳ = Σy/n =   456.000  
                  
SSxx =    Σ(x-x̅)² =    758.0580          
SSxy=   Σ(x-x̅)(y-ȳ) =   -2332.6          
                  
estimated slope , ß1 = SSxy/SSxx =   -2332.6   /   758.058   =   -3.07707
                  
intercept,   ß0 = y̅-ß1* x̄ =   560.65126          
                  
so, regression line is   Ŷ =   560.651   +   -3.077   *x

3) slope: for every minutes increase in time spent on eating, calorie consumed get decreased by 3.077

4)

5)

confidence interval for slope                  
α=   0.05              
t critical value=   t α/2 =    2.101   [excel function: =t.inv.2t(α/2,df) ]      
estimated std error of slope = Se/√Sxx =    23.39803   /√   758.06   =   0.850
                  
margin of error ,E= t*std error =    2.101   *   0.850   =   1.785
estimated slope , ß^ =    -3.0771              
                  
                  
lower confidence limit = estimated slope - margin of error =   -3.0771   -   1.785   =   -4.8625
upper confidence limit=estimated slope + margin of error =   -3.0771   +   1.785   =   -1.2917