In: Math
Mondays |
Fridays |
43 |
24 |
27 |
16 |
25 |
33 |
45 |
28 |
32 |
26 |
41 |
17 |
38 |
32 |
24 |
23 |
Score
Group 1 (No Units Of Alcohol) |
Group 2 (Two Units Of Alcohol) |
23 |
21 |
22 |
16 |
28 |
17 |
25 |
14 |
25 |
20 |
26 |
19 |
23 |
18 |
26 |
16 |
24 |
17 |
25 |
16 |
a)
Xˉ1=34.375 and Xˉ2=24.875
Also, the provided sample standard deviations are:
s1 =8.45 and s2 =6.244
and the sample size is n = 8. For the score differences we have
Dˉ=9.5 and sD =10.433
(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
Ho: μD= 0
Ha:μD≠ 0
This corresponds to a two-tailed test, for which a t-test for two paired samples be used.
(2) Rejection Region
Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=7.
Hence, it is found that the critical value for this two-tailed test is tc =2.365, for α=0.05 and df=7.
The rejection region for this two-tailed test is R={t:∣t∣>2.365}.
(3) Test Statistics
The t-statistic is computed as shown in the following formula:
(4) Decision about the null hypothesis
Since it is observed that ∣t∣=2.575>tc =2.365, it is then concluded that the null hypothesis is rejected.
Using the P-value approach: The p-value is p=0.0367, and since p=0.0367<0.05, it is concluded that the null hypothesis is rejected.
(5) Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that mean number of products produced in Monday μ1 is different than Friday μ2, at the 0.05 significance level.
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