A teacher wants to see if a new unit on taking square roots is helping students learn. She has five randomly selected students t

Question

2 months ago

5

A teacher wants to see if a new unit on taking square roots is helping students learn. She has five randomly selected students t

ake a pre-test and a post test on the material. The scores are out of 20. Has there been improvement? (pre-post) Student 1 2 3 4 5 Pre-test 11 9 10 14 10 Post- Test 18 17 19 20 18 The test statistic is -14.9. What is the p-value?

Mathematics

2 answers:

PIT_PIT [12.4K] · Answer 1 · 2026-03-11T23:06:46+03:00

Answer:

The P-value for this analysis is P=0.00006.

Step-by-step explanation:

We are conducting a matched-pair test, with a calculated test statistic of t=-14.9.

The degree of freedom for a sample size of 5 students is:

$df=n-1=5-1=4$

For t=-14.9 with 4 degrees of freedom, a left-tail test yields a P-value of:

$P-value=P(t_4$

This test aims to verify the claim that the new unit regarding square roots aids student learning. The data suggests statistical support that the new curriculum improves learning outcomes for students.

Inessa [12.5K] · Answer 2 · 2026-03-11T23:04:29+03:00

Answer:

P-value < 0.001

Step-by-step explanation:

Test statistic is -14.9

Thanks to the symmetry of t when n-2 degrees of freedom is applied, where n=5

Claim: There is significant improvement

Thus,

P-value = P(t < -14.9) (this is a left-tailed test)

P-value = P(t > 14.9)

P-value = 0.0003

P-value < 0.001

Information was gathered from the t-table.