1 pts A teacher wants to see if a new unit on fractions 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? The test statistic is -0.15. What would be the p-value? Student 1 2 3 4 5 Pre-test 15 14 15 19 14 Post- Test 19 13 12 20 18 Group of answer choices p-value less than 0.002 p-value less than 0.001 p-value greater than 0.10 p-value greater than 0.05

The required value of p is greater than 0.10.

Given that,

She has five randomly selected students take a pre-test and a post test on the material.

The scores are out of 20.

We have to determine,

The test statistic is -0.15. What would be the p-value.

According to the question,

The teacher is performing a paired t-test to check whether there is any improvement in the scores after using the new unit on fractions.

The hypothesis can be defined as:

H₀: There is no improvement in the scores after using the new unit on fractions, i.e. μₓ = 0.

Hₐ: There is an improvement in the scores after using the new unit on fractions, i.e., μₓ > 0.

The variable, μₓ is defined as the difference between the the scores on the post-test and pre-test given as;

μₓ = post-test - pre-test

This is provided that the test statistic value is, t = -0.15.

A sample of n = 5 students were selected.

The p-value of the test is given by,

By using t- table Thus, the p-value is greater than 0.10 .

Hence, The required value of p is greater than 0.10.

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The p-value is greater than 0.10.

Step-by-step explanation:

The dependent t-test (also known as the paired t-test or paired-samples t-test) compares the two means associated groups to conclude if there is a statistically significant difference amid these two means.

In this case, the teacher is performing a paired t-test to check whether there is any improvement in the scores after using the new unit on fractions.

The hypothesis can be defined as:

H₀: There is no improvement in the scores after using the new unit on fractions, i,e, μₓ = 0.

Hₐ: There is an improvement in the scores after using the new unit on fractions, i,e, μₓ > 0.

Here the variable, μₓ is defined as the difference between the the scores on the post-test and pre-test, i.e.

μₓ = post-test - pre-test

It is provided that the test statistic value is, t = -0.15.

A sample of n = 5 students were selected.

Compute the p-value of the test as follows:

*Use a t-table.

Thus, the p-value is greater than 0.10.

a)(x-2)^2+(y-1)^2=9

step-by-step explanation:

(picture explains)

p = 6s

step-by-step explanation: