Question

A study was conducted with vegetarians to see whether the number of grams of protein each ate per day was related to diastolic blood pressure. The data are given here. Grams x Pressure y 4 6.5 5 5.5 8 10 9 8.2 73 79 83 82 84 92 88 86 10.5 95 1. Plot the data points on graph paper. 2. Compute and interpret the coefficient of correlation between the Grams (x) and Pressure (y). 3. Conduct a "4" test for the significant of coefficient of correlation at 0.05 level of significance. What assumptions are necessary for the validity of this procedure? 4. Find the least-squares line for the data. 5. Graph the least-squares line to see how well it fits the data. 6. Use the least-squares line to estimate the Pressure for protein with 7 grams. 7. Compute Standard error of estimate (S2). 8. Do the data present sufficient evidence to indicate that the slope differs from 0? Test using a = .05 and give bounds for the attained significance level. 9. Find a 90% confidence interval for E(Y) when x = 1.8. 10. Predict the particular value of Y with 1 - a=.90 when x = 1.8.

Answer 1

X	Y	XY	X²	Y²
4	73	292	16	5329
6.5	79	513.5	42.25	6241
5	83	415	25	6889
5.5	82	451	30.25	6724
8	84	672	64	7056
10	92	920	100	8464
9	88	792	81	7744
8.2	86	705.2	67.24	7396
10.5	95	997.5	110.25	9025

Ʃx =	66.7
Ʃy =	762
Ʃxy =	5758.2
Ʃx² =	535.99
Ʃy² =	64868
Sample size, n =	9
x̅ = Ʃx/n = 66.7/9 =	7.411111111
y̅ = Ʃy/n = 762/9 =	84.66666667
SSxx = Ʃx² - (Ʃx)²/n = 535.99 - (66.7)²/9 =	41.66888889
SSyy = Ʃy² - (Ʃy)²/n = 64868 - (762)²/9 =	352
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 5758.2 - (66.7)(762)/9 =	110.9333333

1) Scatter plot:

Scatterplot 0 2 4 8 10 12 6 X

2) Correlation coefficient, r = SSxy/√(SSxx*SSyy) = 110.93333/√(41.66889*352) =    0.9160

3) Null and alternative hypothesis:  

Ho: ρ = 0

Ha: ρ ≠ 0  

Test statistic :    

t = r*√(n-2)/√(1-r²) = 0.916 *√(9 - 2)/√(1 - 0.916²) =    6.0401

df = n-2 =    7

Critical value, t_c = T.INV.2T(0.05, 7) =    2.3646

p-value = T.DIST.2T(ABS(6.0401), 7) =    0.0005

Conclusion:  

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

4) Slope, b = SSxy/SSxx = 110.93333/41.66889 =    2.662258013

y-intercept, a = y̅ -b* x̅ = 84.66667 - (2.66226)*7.41111 =    64.93637673

Regression equation :   

ŷ = 64.9364 + (2.6623) x  

5)

Scatterplot 0 2 4 6 8 10 12

6) Predicted value of y at x =    7

ŷ = 64.9364 + (2.6623) * 7 = 83.5722  

7) Sum of Square error, SSE = SSyy -SSxy²/SSxx = 352 - (110.93333)²/41.66889 = 56.66684

Standard error, se = √(SSE/(n-2)) = √(56.66684/(9-2)) =    2.84522

Estimate of variance, S² = SSE/(n-2) = 56.66684/(9-2) =    8.0953

8) Null and alternative hypothesis:  

Ho: β₁ = 0  

Ha: β₁ ≠ 0  

Test statistic:  

t = b/(se/√SSxx) =    6.0401

df = n-2 =    7

p-value = T.DIST.2T(ABS(6.0401), 7) =    0.0005

Conclusion:  

p-value < α Reject the null hypothesis.  

9) Predicted value of y at x =    1.8

ŷ = 64.9364 + (2.6623) * 1.8 = 69.7284  

Critical value, t_c = T.INV.2T(0.1, 7) = 1.8946  

90% Confidence interval :  

Lower limit = ŷ - tc*se*√((1/n) + ((x-x̅)²/(SSxx)))

= 69.7284 - 1.8946*2.8452*√((1/9) + ((1.8 - 7.4111)²/(41.6689))) =    64.7101

Upper limit = ŷ + tc*se*√((1/n) + ((x-x̅)²/(SSxx)))

= 69.7284 + 1.8946*2.8452*√((1/9) + ((1.8 - 7.4111)²/(41.6689))) =    74.7468

10) 90% Prediction interval :  

Lower limit = ŷ - tc*se*√(1 + (1/n) + ((x-x̅)²/(SSxx)))

= 69.7284 - 1.8946*2.8452*√(1 + (1/9) + ((1.8 - 7.4111)²/(41.6689))) =    62.3636

Upper limit = ŷ + tc*se*√(1 + (1/n) + ((x-x̅)²/(SSxx)))

= 69.7284 + 1.8946*2.8452*√(1 + (1/9) + ((1.8 - 7.4111)²/(41.6689))) =    77.0933