Question

I plug this in my calculator L1 and L2 list, then stat-test-2 sampttest-data- enter data for L1 table and L2 table- freq1:1- freq2:1- u1/=u2- pooled no- calculate. but answer is always wrong. what am I doing wrong. can someone please explain. especially what freq means?

You wish to test the following claim ( H a ) at a significance level of α = 0.05 . For the context of this problem, μ d = P o s t T e s t − P r e T e s t where the first data set represents a pre-test and the second data set represents a post-test. (Each row represents the pre and post test scores for an individual. Be careful when you enter your data and specify what your μ 1 and μ 2 are so that the differences are computed correctly.) H o : μ d = 0 H a : μ d ≠ 0 You believe the population of difference scores is normally distributed, but you do not know the standard deviation. You obtain the following sample of data: pre-test post-test 71.9 74.5 49.9 47.2 57 46.5 29.3 20.9 44.6 34.1 31.5 27 39.3 40.8 50.3 43.7 48.8 37.6 39.3 15.4 76.2 80.8 52.4 63.1 63.6 44.7 54.7 48.4 55.3 33.2 46.8 39.9 73.4 57.9 64.4 70.8 64 68.6 71.9 72.3 68.9 36 42.5 52.5 What is the test statistic for this sample? test statistic = 1.3274 Incorrect (Report answer accurate to 4 decimal places.) What is the p-value for this sample? p-value = .9040 Incorrect (Report answer accurate to 4 decimal places.) The p-value is... less than (or equal to) α greater than α Correct This test statistic leads to a decision to... reject the null accept the null fail to reject the null Correct As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that the mean difference of post-test from pre-test is not equal to 0. There is not sufficient evidence to warrant rejection of the claim that the mean difference of post-test from pre-test is not equal to 0. The sample data support the claim that the mean difference of post-test from pre-test is not equal to 0. There is not sufficient sample evidence to support the claim that the mean difference of post-test from pre-test is not equal to 0.

Answer 1

using minitab>stat>basic stat>paired t test

we have

Paired T-Test and CI: post test, pre test

Paired T for post test - pre test

N Mean StDev SE Mean
post test 22 48.00 17.88 3.81
pre test 22 54.36 13.67 2.91
Difference 22 -6.37 11.40 2.43

95% CI for mean difference: (-11.42, -1.31)
T-Test of mean difference = 0 (vs ≠ 0): T-Value = -2.6201 P-Value = 0.0160

test statistic = -2.6201

p-value = .0160

The p-value is less than (or equal to) α.

This test statistic leads to a decision to reject the null

As such, the final conclusion is that. The sample data support the claim that the mean difference of post-test from pre-test is not equal to 0.