Question

A teacher wants to see if a new unit on factoring 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 12 14 11 12 13 Post- Test 15 17 11 13 12 The test statistic equals -1.50 What would be the p-value? Group of answer choices p-value is between 0.01 and 0.025 p-value > 0.20 p-value < 0.002 p-value > 0.10

Answer 1

Answer:

Step-by-step explanation:

Given the sample data

Pre-test... 12 14 11 12 13

Post-Test 15 17 11 13 12

The mean of pre-test

x = ΣX / n

x = (12+14+11+12+13) / 5

x = 12.4

The standard deviation of pre-test

S.D = √Σ(X-x)² / n

S.D = √[(12-12.4)²+(14-12.4)²+(11-12.4)²+(12-12.4)²+(13-12.4)² / 5]

S.D = √(5.2 / 5)

S.D = 1.02.

The mean of post-test

x' = ΣX / n

x' = (15+17+11+13+12) / 5

x' = 13.6

The standard deviation of post-test

S.D' = √Σ(X-x)² / n

S.D' = √[(15-13.6)²+(17-13.6)²+(11-13.6)²+(13-13.6)²+(12-13.6)² / 5]

S.D = √(23.2 / 5)

S.D = 2.15

Test value

t = (sample difference − hypothesized difference) / standard error of the difference

t = [(x-x') - (μ- μ')] / (S.D / n — S.D'/n)

t = (12.4-13.6) - (μ-μ')/ (1.02/5 - 2.15/5)

-1.5 = -1.2 - (μ-μ') / -0.226

-1.5 × -0.226 = -1.2 -(μ-μ')

​0.339 = -1.2 - (μ-μ')

(μ-μ') = -1.2 -0.339

μ-μ' = -1.539

Then, μ ≠ μ'

We can calculate our P-value using table.

This is a two-sided test, so the P-value is the combined area in both scores.

The p-value is 0.172

The p value > 0.1