Question

Consider the following matched samples representing observations before and after an experiment. Assume that the sample differences are normally distributed. Use Table 2.

Before	2.5	1.8	1.4	-2.9	1.2	-1.9	-3.1	2.5
After	2.9	3.1	3.9	-1.8	0.2	0.6	-2.5	2.9

Let the difference be defined as Before – After.

a.	Construct the competing hypotheses to determine if the experiment increases the magnitude of the observations.
	H0: μD = 0; HA: μD ≠ 0 H0: μD ≥ 0; HA: μD < 0 H0: μD ≤ 0; HA: μD > 0

b-1.	Implement the test at a 5% significance level. (Negative value should be indicated by a minus sign. Round all intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.)

Test statistic

b-2.	What is the p-value?
	0.005 < p-value < 0.01 0.01 < p-value < 0.025 0.025 < p-value < 0.05 0.05 < p-value < 0.10 0.1 < p-value < 0.2

b-3.	What is the conclusion to the hypothesis test?

We (Click to select)rejectdo not reject H0. At the 5% significance level, We (Click to select)cancannot conclude that the experiment increases the magnitude of the observations.

c.	Do the results change if we implement the test at a 1% significance level?
	Yes No

Answer 1

### By using Excel

Before	After	Difference(D)
2.5	2.9	-0.4
1.8	3.1	-1.3
1.4	3.9	-2.5
-2.9	-1.8	-1.1
1.2	0.2	1
-1.9	0.6	-2.5
-3.1	-2.5	-0.6
2.5	2.9	-0.4
	Mean(D)	-0.98
	SD(D)=	1.16

Mean of the difference is obtained by using function: =AVERAGE(C2:C9)

Standard deviation of the difference is obtained by using function:=STDEV(C2:C9)

a) We want to test the hypothesis that whether the experiment increases the magnitude of the observations.

The hypothesis testing problem is:

Vs

The test statistics is:

The P value is:

### By using P value table.

Therefore the pvalue is:

0.01 < p-value < 0.025

Therefore we reject the null hypothesis. At 5% level of significance we conclude that the experiment increases the magnitude of the observations.

c) If we implement the test at a 1% significance level the results will be change.

Since pvalue is 0.024 we are unable to reject the null hypothesis.