Question

A sports scientist collects strength data from two groups of students. The first group is a group of athletes who might be expected to be strong, and the second group is a control group of nonathletic students. The first group has strength ratings of:

c(121.2, 128.9, 126.8, 121.1, 122.2, 123.1, 107)

And the second group has strength ratings of:

c(116.3, 132.3, 122.7, 129.8, 101.4, 125.4, 121.3, 114.2)

Is there evidence that the first group is actually stronger than the second?

(a) State a sensible null hypothesis

(b) State the precise definition of p-value and explain what “more extreme” means in this context

(c) Is a one-sided or two-sided test needed? justify

(d) Perform a student t-test using R and interpret

Notes:
• Show detailed working, including appropriate mathematical notation for each question. For most questions this will involve showing your working from R, (e.g. cut-and-paste commands and output from an R Studio session).

• Any question involving regression will score 0 marks unless a scattergraph is produced.

• No additional information provided

Answer 1

(a) State a sensible null hypothesis

The null hypothesis is

Ho : There is no difference in the both groups.

H1 : The first group is actually stronger than the second.

(c) Is a one-sided or two-sided test needed? justify

Answer : It is one sided test. Since our claim is to check the first group is actually stronger than the second therfore it is one sided test.

(d) Perform a student t-test using R and interpret

> A=c(121.2, 128.9, 126.8, 121.1, 122.2, 123.1, 107)
> B=c(116.3, 132.3, 122.7, 129.8, 101.4, 125.4, 121.3, 114.2)

# we check assumption of equal variance for both the groups
> res.ftest <- var.test(A,B)
> res.ftest

   F test to compare two variances

data: A and B
F = 0.51054, num df = 6, denom df = 7, p-value =
0.4308
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.0997413 2.9077397
sample estimates:
ratio of variances
0.5105355

# Since p-value 0.4308 > 0.05 level of significance so conclude that both group have equal variance.

> t.test(A, B, alternative = "greater", var.equal = TRUE)

   Two Sample t-test

data: A and B
t = 0.23362, df = 13, p-value = 0.4095
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
-6.885829 Inf
sample estimates:
mean of x mean of y
121.4714 120.4250
# Here the p-value =0.4095 > 0.05 level of significance so conclude that there is not enough evidance to conclude that the first group is actually stronger than the second.