Homework SourseQuestion 4 options A 95 confidence interval for the time2 t
Question 4 options:
A 95% confidence interval for the (time2 - time1) differences would contain the range (3.55 to 6.45).
The null hypothesis is H0: There is no difference in the average anxiety level at time 1 (10 min) and time 2 (1 min).
The observed t value is 6.9.
We would fail to reject the null hypothesis.
For a specific value of mu, looking at a 95% confidence interval and conducting a t-test with an alpha level of .05 will lead to the same conclusion about the null hypothesis.
The critical t value is +/- 1.98.
To test the hypothesis, a t-test is needed.
| Subject | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| Anxiety 10 minutes prior | 14 | 12 | 10 | 12 | 8 | 10 | 10 | 8 | 12 | 16 | 14 | 12 |
| Anxiety 1 minutes prior | 18 | 19 | 14 | 16 | 12 | 18 | 16 | 18 | 16 | 19 | 16 | 16 |
Solution
Set Up Hypothesis
Null Hypothesis , There Is No-Significance between them Ho: u1 = u2
Alternate Hypothesis, There Is Significance between them - H1: u1 != u2
Test Statistic
X(Mean)=11.5
Standard Deviation(s.d1)=2.4308 ; Number(n1)=12
Y(Mean)=16.5
Standard Deviation(s.d2)=2.067; Number(n2)=12
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =11.5-16.5/Sqrt((5.90879/12)+(4.27249/12))
to =-5.43
| to | =5.43
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 11 d.f is 2.201
We got |to| = 5.42825 & | t | = 2.201
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -5.4282 ) = 0
Hence Value of P0.05 > 0,Here we Reject Ho
