Question

NO SOFTWARE OR EXCEL

Blood pressure measurement consists of two numbers: the systolic pressure, which is the maximum pressure taken when the heart is contracting, and the diastolic pressure, which is the minimum pressure taken at the beginning of the heartbeat. Blood pressure values were measured for a sample of 8 adults; the data are given below.

SYSTOLIC	DIASTOLIC
134	87
115	83
113	77
123	77
119	69
118	88
130	76
116	70

h) Perform a one-tail hypothesis testing to test the hypothesis that the systolic and diastolic pressures are independent to each other (Use ? = 0.05)

(i) Compute the p-value for the computed sample statistic (? ) in (h)

(j) If the systolic pressures of two people differ by 10 units, by how much would you have predicted their diastolic pressures differed?

(Use point values of the mean predictions) (Note: You cannot use any software applications. All the calculations need to be shown on paper. Failure to do so will result in a zero score.)

Answer 1

Solution

[Note:

1. Solutions are based on Correlation-Regression Analysis.

2. Final answers are given first. Back-up Theory and Detailed Working follow at the end.]

Part (a)

The test on significance of correlation coefficient reveals that the null hypothesis of zero correlation is accepted at 5% significance level and hence we conclude that the systolic and diastolic pressures are independent to each other. Answer

Part (b)

p-value for the test = 0.2543 Answer

Part (c)

Treating systolic pressure as x, and diastolic pressure as y, the estimated regression equation is:

ycap = 46.5816 + 0.2628x

The regression coefficient, 0.2628, which represents the slope of the regression equation is interpreted as the change if y per unit change in x. So, in this scenario, when the systolic pressures of two people differ by 10 units implies x changes by 10 units and hence, diastolic pressures represented by y would differ by (10 x 0.2628) = 2.628 units Answer

Back-up Theory and Detailed Working

Correlation and Regression

The linear regression model Y = β₀ + β₁X + ε, ………………………………................................................………..(1)

where ε is the error term, which is assumed to be Normally distributed with mean 0 and variance σ².

Estimated Regression of Y on X is given by: Y = β_0cap + β_1capX, ………...........................................…………….(2)

where β_1cap = Sxy/Sxx = r.√(Syy/Sxx) = r.(SDy/SDx) and β_0cap = Ybar – β_1cap.Xbar..……………….………….…..(3)

Mean X = Xbar = (1/n) Σ(i = 1 to n)x_i ………………………………………….……….….(4)

Mean Y = Ybar = (1/n) Σ(i = 1 to n)y_i ………………………………………….……….….(5)

Sxx = Σ(i = 1 to n)(x_i – Xbar)² ………………………………………………..…………....(6)

Syy = Σ(i = 1 to n)(y_i – Ybar)² ……………………………………………..………………(7)

Sxy = Σ(i = 1 to n){(x_i – Xbar)(y_i – Ybar)} …………………………………….………….(8)

Correlation coefficient, r = Sxy/sqrt(Sxx. Syy) ……………………………….....…….. (10)

To test for significance of correlation coefficient

One-sided

Hypotheses:

Null: H₀: ρ = 0 Vs Alternative H₁: ρ > 0

Test Statistic:

t = r√{(n - 2)/(1 – r²)}

Distribution, Significance Level, α, Critical Value

t ~ t_{n – 2}

So, Critical Value = upper (α) % point of t_{n – 2}

Decision

H₀ is rejected/accdepted when t_cal > < t_crit or equivalently, when p-value < > α

Summary of Calculations

i	Xi (systolic)	Yi (diastolic)
1	134	87
2	115	83
3	113	77
4	123	77
5	119	69
6	118	88
7	130	76
8	116	70

n	8
Xbar	121.00000000
ybar	78.3750
Sxx	392
Syy	355.875
Sxy	103
β1cap	0.262755102
β0cap	46.58163265
r	0.275769004
r^2	0.076048544
Test for significance of r
Hypotheses: Null: H0: ρ = 0 Vs Alternative H1: ρ > 0 Test Statistic: t = r√{(n - 2)/(1 – r2)}
1- r^2		0.923951456
(n-2)/(1-r^2)		6.493847658
sqrt		2.548302898
tcal		0.702742953
tcrit		1.943180274
p-value		0.254274913

H₀ is accepted since t_cal < t_crit or equivalently, p-value > α.

DONE