Respuesta :
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
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:
[tex]H_0: \mu_1 - \mu_2 \geq 0[/tex]
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:
[tex]H_1: \mu_1 - \mu_2 < 0[/tex]
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 https://brainly.com/question/13873630
