In: Statistics and Probability
A study was undertaken to determine if there is a difference, between taking the tunnel and taking the bridge, in the average time it takes for Canadians entering the US to be interrogated by US customs and border security officials. A random sample of 6 times was selected, with one Canadian car during each of those times also being randomly selected. The time spent being interrogated for each of those cars (in minutes) was recorded as follows:
Time period tunnel time (min.) bridge time (min.)
A 1.3 1.6
B 2.7 3.1
C 3.7 4.8
D 2.3 2.1
E 0.6 1.1
F 3.0 2.7
At the .05 level of significance is there sufficient evidence to indicate that there is a difference, between taking the tunnel and taking the bridge, in the average time it takes for Canadians entering the US to be interrogated by US customs and border security officials? In answering this question manually, you must include the two hypotheses, the value of the test statistic, the decision rule in terms of the test statistic, the p-value (using the t-table) and a two-part conclusion.
Excel Assignment
After completing your solution you are then to use Excel to:
Answer the above question using Excel’s t-test: Paired Two Samples for Means (as illustrated in class)
Answer the above question using Excel’s t-test; Two Samples Assuming Equal Variances (as illustrated in class)
When answering each part of this question, you are to shade: in yellow the value of your test statistic; in blue the critical value of the test; and, in green the p-value of your test. And, when submitting your Excel solution make certain that each column is wide enough so that no words are cutoff.
Level of significance = = 0.05
We want to test "wether there is a difference, between taking the tunnel and taking the bridge, in the average time it takes for Canadians entering the US to be interrogated by US customs and border security officials."
Let's write null hypothesis (H0) and the alternative hypothesis (H1)
H0 : = 0
H1: 0
Let's make table in excel using the given data set:
The formula of t test statistic for pair t test isas follow:
Decision rule:
1) If | t | >= positive critical t tvalue then we reject null hypothesis.
2) If | t | < positive critical t tvalue then we fail to reject null hypothesis.
So first we need to find critical t value for two tailed pair t test
degrees of freedom = 5
= 0.05
The critical t value is 2.57
Since | t | = 1.44 < 2.57 = critical t value
We fail to reject null hypothesis.
p -value = 2* P( | t | > 1.44) = 0.2091
Decision rule:
1) If p-value < level of significance (alpha) then we reject null hypothesis
2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.
Here p value = 0.2091 > 0.05 so we used second rule.
That is we reject the null hypothesis
Conclusion: At 5% level of significance there are not sufficient evidence to say that the means are different.
Let's used excel :
Step 1) Enter the given data in excel columns.
Step 2) Click on Data >>>Data Analysis >>> t-Test: Pair Two sample for Means
then click on OK
In variable 1 Range select data of Ending SBP (Because here we test Before - After < 0)
In variable 1 Range select data of Starting SBP ?
Hypothesized mean difference : 0
Select Labels
Then click on Output range and select any empty cell
See the following box
Then click on OK
So we get the following output