Question

Fifteen students from Poppy High School were accepted at Branch University. Of those students, six were offered academic scholarships and nine were not. Mrs. Bergen believes Branch University may be accepting students with lower ACT scores if they have an academic scholarship. The newly accepted student ACT scores are shown here.

Academic scholarship: 25, 24, 23, 21, 22, 20
No academic scholarship: 23, 25, 30, 32, 29, 26, 27, 29, 27

Part A: Do these data provide convincing evidence of a difference in ACT scores between students with and without an academic scholarship? Carry out an appropriate test at the α = 0.02 significance level. (5 points)

Part B: Create and interpret a 98% confidence interval for the difference in the ACT scores between students with and without an academic scholarship. (5 points)

Answer 1

For academic scholarship

= 22.5, s1 = 1.87, n1 = 6

For no academic scholarship

= 27.56, s2 = 2.74, n2 = 9

A) H​​​​​​₀:

H₁:

The test statistic

  

At alpha = 0.02, the critical values are +/- t_0.01 = +/- 2.681

Since the test statistic value is less than the lower critical value (-4.25 < -2.681), so we should reject the null hypothesis.

At 0.05 significance level, there is sufficient evidence to conclude that a difference in ACT scores between students with and without an academic scholarship.

B) At 98% confidence level, the critical value is t^* = 2.681

The 98% confidence interval is

() +/- t^* * sqrt(s1^2/n1 + s2^2/n2)

= (22.5 - 27.56) +/- 2.681 * sqrt((1.87)^2/6 + (2.74)^2/9)

= -5.06 +/- 3.19

= -8.25, -1.87

We are 98% confidence that the true difference in the ACT scores between students with and without an academic scholarship lies in the above confidence interval.