Question

A hypertension trial is mounted and 12 participants are randomly assigned to receive either a new treatment or a placebo. Each participant takes the assigned medication and their systolic blood pressure (SBP) is recorded after 6 months on the assigned treatment. The data are as follows.

Placebo	New Treatment
134	114
143	117
148	121
142	124
150	122
160	128

Is there a difference in mean SBP between treatments? Assume equal variances. Run the test at a 5% level of significance. Give each of the following to receive full credit: 1) the appropriate null and alternative hypotheses; 2) the appropriate test; 3) the decision rule; 4) the calculation of the test statistic; and 5) your conclusion including a comparison to alpha or the critical value. You MUST show your work to receive full credit. Partial credit is available.

Answer 1

Solution:

Here, we have to use two sample t test for the difference between two population means by assuming equal population variances.

H₀: µ₁ = µ₂ versus H_a: µ₁ ≠ µ₂

This is a two tailed test.

Test statistic formula for pooled variance t test is given as below:

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

From given data, we have

X1bar = 146.1667

X2bar = 121

S1 = 8.773065

S2 = 4.97996

n1 = 6

n2 = 6

df = n1 + n2 – 2 = 6 + 6 – 2 = 10

α = 0.05

Critical value = - 2.2281 and 2.2281

(by using t-table)

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = [(6 – 1)* 8.773065^2 + (6 – 1)* 4.97996^2]/(6 + 6 – 2)

S_p² = 50.8833

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

t = (146.1667 – 121) / sqrt[50.8833*((1/6)+(1/6))]

t = 25.1667 / 4.1184

t = 6.1108

P-value = 0.0001

(by using t-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that there is a difference in mean SBP between treatments.