In: Math
Listed below are brain volumes (cm3 ) of twins.
First Born | 1005 | 1035 | 1281 | 1051 | 1034 | 1079 | 1104 | 1439 | 1029 | 1160 |
Second Born | 963 | 1027 | 1272 | 1079 | 1070 | 1173 | 1067 | 1347 | 1100 | 1204 |
Test the claim at the 5% significance level that the brain volume for the first born is different from the second-born twin.
(a) State the null and alternative hypotheses.
(b) Find the critical value and the test statistic.
(c) Should H0 be rejected at the 5% significance level? Make a conclusion.
(d) Construct a 95% confidence interval for the paired difference of the population means
Solution:
We have to test the claim at the 5% significance level that the brain volume for the first born is different from the second-born twin.
Part a) State the null and alternative hypotheses.
Part b) Find the critical value and the test statistic.
n = number of pairs = 10
df= n - 1 = 10 - 1 = 9
Level of significance = 0.05
From t table for df = 9 and Two tail area = Level of significance = 0.05, t critical value = 2.262
Since this is two tailed test, t critical values are : -2.262 and 2.262
Test statistic:
where
Thus we need to make following table:
First Born | Second Born | di = Firs Born-Second Born | di^2 |
1005 | 963 | 42 | 1764 |
1035 | 1027 | 8 | 64 |
1281 | 1272 | 9 | 81 |
1051 | 1079 | -28 | 784 |
1034 | 1070 | -36 | 1296 |
1079 | 1173 | -94 | 8836 |
1104 | 1067 | 37 | 1369 |
1439 | 1347 | 92 | 8464 |
1029 | 1100 | -71 | 5041 |
1160 | 1204 | -44 | 1936 |
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Thus
Thus t test statistic:
Part c) Should H0 be rejected at the 5% significance level? Make a conclusion
No, since t test statistic value is neither less than -2.262 , nor greater than 2.262.
Thus we conclude that: there is not sufficient evidence to support the claim that: the brain volume for the first born is different from the second-born twin.
Part d) Construct a 95% confidence interval for the paired difference of the population means
Formula:
where
tc is t critical value for c = 95% confidence level , that is for two tail area = 1 - c= 1 - 0.95 = 0.05 area.
df = n - 1 = 10 - 1 = 9
Thus tc = 2.262
Thus
Thus
Thus a 95% confidence interval for the paired difference of the
population means is inbetween :