Question

Need help calculating t-statistic and P-value: Listed below are systolic blood pressure measurements​ (mm Hg) taken from the right and left arms of the same woman. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Use a 0.05 significance level to test for a difference between the measurements from the two arms. What can be​ concluded?

Answer 1

Solution:

Here, we have to use paired t test.

H₀: µ_d = 0 versus H₁: µ_d ≠ 0

This is a two tailed test.

Test statistic for paired t test is given as below:

t = (Dbar - µ_d)/[S_d/sqrt(n)]

From given data, we have

Dbar = -25.2

S_d = 25.0240

n = 5

df = n – 1 = 4

α = 0.05

Critical values = - 2.7764 and 2.7764

We reject H₀ when test statistic t < - 2.7764 or t > 2.7764.

Test statistic is given as below:

t = (Dbar - µ_d)/[S_d/sqrt(n)]

t = (-25.2 - 0)/[ 25.0240/sqrt(5)]

t = -2.2518

P-value = 0.0875

(by using t-table or excel)

P-value > α = 0.05

So, we do not reject the null hypothesis

At 5% level of significance, there is not sufficient evidence to conclude that there is a statistically significant difference between the measurements from the two arms.

We conclude that there is no significant difference in the measurements from two arms.