Question

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

Answer 1

Answer:

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

Step-by-step explanation:

We have a matched-pair test, with a test statistic t=-14.9.

The degrees of freedom in a sample of 5 students is:

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

For a  t=-14.9 and 4 degrees of freedom, a left-tail test will have a P-value of:

$P-value=P(t_4$

The claim is that the new unit on taking square roots is helping students to learn. This test concludes that there is statistical evidence to support the claim that the new unit is helping students to learn.

Answer 2

Answer:

P-value < 0.001

Step-by-step explanation:

Test statistic = -14.9

Due to symmetry of t with df = n-2 = 5-2 = 3

Claim : There has been improvement

Consider,

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

P-value = P(t > 14.9)

P-value = 0.0003

P-value < 0.001

We observed the t-table