Question

Use Excel to test. For each paired difference, compute After – Before. In Data Analysis, t-Test: Paired Two Sample for means, select the After data for Variable 1 Range. Note that the critical value output by Data Analysis for this test is always positive. In this problem, the sign of the critical value is negative corresponding to 1-tailed test with lower reject region and negative lower critical value.

           
   Person   Before   After
   1   176   164
   2   192   191
   3   185   176
   4   177   176
   5   196   185
   6   178   169
   7   196   196
   8   181   172
   9   158   158
   10   201   193
   11   191   185
   12   193   189
   13   176   175
   14   212   210
   15   177   173
   16   183   180
   17   210   204
   18   198   192
   19   157   152
   20   213   200
   21   161   161
   22   177   166
   23   210   203
   24   192   186
   25   178   170

What is your conclusion?

A. Do not reject the NULL Hypothesis because the actual value is greater than critical value

B. Reject the NULL Hypothesis because the actual value is less than the critical value

C. Reject the NULL Hypothesis because the actual value is greater than critical value

D. Do not reject the NULL Hypothesis because the actual value is less than the critical value

Answer 1

Solution:

We have to use excel to test paired difference.

Thus null and alternative hypothesis are:

Vs

We have to make conclusion.

Use following steps:

Copy this data and paste it in Excel

Click on Data tab

Select Data Analysis

Select t-Test: Paired Two Sample for Means

Click in Variable 1 range and select After column with its label

Click in Variable 2 range and select before column with its label

Hypothesized difference = 0

Check box for Label

Click on OK

We will get Output:

t-Test: Paired Two Sample for Means
	After	Before
Mean	181.04	186.72
Variance	237.79	254.7933333
Observations	25	25
Pearson Correlation	0.967973241
Hypothesized Mean Difference	0
df	24
t Stat	-7.086724825
P(T<=t) one-tail	1.25776E-07
t Critical one-tail	1.71088208
P(T<=t) two-tail	2.51552E-07
t Critical two-tail	2.063898562

t test statistic value = -7.0867 = -7.087

t critical value = -1.71088 = -1.711

Decision Rule:

Reject null hypothesis H0 if t test statistic value < t critical value = -1.711, otherwise we fail to reject H0.

Since t test statistic value = -7.087 < t critical value = -1.711, we reject null hypothesis H0.

Thus correct option is:

B. Reject the NULL Hypothesis because the actual value is less than the critical value