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

Using the t-distribution, it is found that the p-value is of 0.086.

At the null hypothesis, it is tested if there is no improvement, that is, the subtraction of the mean of the grades on test 1 by the mean of the grades on test 2 is of at least 0, that is:

At the alternative hypothesis, it is tested if there is improvement, that is, the grades on the second test were greater than on the first, hence:

We can find the standard deviation for the samples, hence, the t-distribution is used.

The p-value is found using a left-tailed test, as we are testing if the variable is less than a value, with 5 + 5 - 2 = 8 df and t = -1.5, using a calculator, the p-value is of 0.086.

A similar problem is given at link

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

walter needs to make at least 5 more cookies but no more than 40. 5 ≥ x ≥ 40;

step-by-step explanation:

walter has to make a minimum of 20 cookies. he already has 15 cookies made, so he must make 5 more to meet the minimum.

the maximum amount of cookies he can make is 55. he already has 15, so he has to make no more than 40 more cookies to meet the maximum.

any options?

step-by-step explanation: