Question

Researchers compared two groups of competitive rowers: a group of skilled rowers and a group of novices. The researchers measured the angular velocity of each subject’s right knee, which describes the rate at which the knee joint opens as the legs push the body back on the sliding seat. The sample size n, the sample means, and the sample standard deviations for the two groups are given below.

Group	n	Mean	Standard Deviation
Skilled	16	4.2	0.6
Novice	16	3.2	0.8

The researchers wished to test the hypotheses

H₀: the mean knee velocities for skilled and novice rowers are the same

H_{a: the mean knee velocity for skilled rowers is larger than for novice rowers.}

The data showed no strong outliers or strong skewness so the researchers decided to use the two-sample t test. The value of the t test statistic is:

a. 1.00

b. 1.25

c. 2.0

d. 4.0

Answer 1

d. 4.0

Solution:

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

H₀: µ₁ = µ₂ versus H_a: µ₁ > µ₂

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)

We are given

X1bar = 4.2

X2bar = 3.2

S1 = 0.6

S2 = 0.8

n1 = 16

n2 = 16

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

S_p² = [(16 – 1)*0.6^2 + (16 – 1)*0.8^2]/(16 + 16 – 2)

S_p² = 0.5

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

t = (4.2 – 3.2) / sqrt[0.5*((1/16)+(1/16))]

t = 1/ 0.25

t = 4.0

d. 4.0