Question

1. using the following data set, write a null hypothesis. Record both a generic version (through the use symbols) and an English version (using words) – for the generic version.

2. Compose an alternative hypothesis to accompany the test. Record both a generic version (through the use of symbols) and an English version (using words) – for the generic version.

3. What type of test should be used?

4. Interpret the results.

id	Pre	Post
1	2	4.00
2	2	4.00
3	4	6.00
4	1	0.00
5	4	6.00
6	3	5.00
7	0	2.00
8	2	3.00
9	7	6.00
10	5	4.00

Answer 1

Solution:-

1)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: There is no difference in Pre and Post scores.

Alternative hypothesis: There is difference in Pre and Post scores.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_d = 0

Alternative hypothesis: u_d ≠ 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

Id	Pre	Post	d	(d - dbar)^2
1	2	4	-2	1
2	2	4	-2	1
3	4	6	-2	1
4	1	0	1	4
5	4	6	-2	1
6	3	5	-2	1
7	0	2	-2	1
8	2	3	-1	0
9	7	6	1	4
10	5	4	1	4
Sum	30	40	-10	18
Mean	3	4	-1	1.8

s = sqrt [ (\sum (d_i - d)² / (n - 1) ]

s = 1.4142

SE = s / sqrt(n)

S.E = 0.4472

DF = n - 1 = 10 -1

D.F = 9

t = [ (x₁ - x₂) - D ] / SE

t = - 2.24

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 9 degrees of freedom is more extreme than -2.24; that is, less than - 2.24 or greater than 2.24

Thus, the P-value = 0.052

Interpret results. Since the P-value (0.052) is greater than the significance level (0.05), hence we failed to reject the null hypothesis.

Do not Reject H0. From the above test we do not have sufficient evidence in the favor of the claim that there is difference in Pre and Post scores.