# Listed below are brain volumes, measured in cubic centimeters, of twins at birth. Use a 5% significance level to test the claim that there is no difference in the brain volume of the first born twin versus the second born twin. Assume that the distribution of brain volumes for newborn is normal.First Born   1005   1035   1281    1051   1034   1079   1104  1439   1029  11602nd Born    963     1027   1272    1079   1070   1173   1067  1347    1100  1204

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Listed below are brain volumes, measured in cubic centimeters, of twins at birth. Use a 5% significance level to test the claim that there is no difference in the brain volume of the first born twin versus the second born twin. Assume that the distribution of brain volumes for newborn is normal.

First Born   1005   1035   1281    1051   1034   1079   1104  1439   1029  1160

2nd Born    963     1027   1272    1079   1070   1173   1067  1347    1100  1204

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Step 1

Solution:

Null and alternative hypotheses:

Null hypothesis: µ1 − µ2 = 0 (equivalently µ1 = µ2)

Alternative hypothesis: µ1 − µ2 ≠ 0 (equivalently µ1 ≠ µ2)

Test statistic:

The appropriate to test the true mean of brain volumes of first born and 2nd born is independent sample t-test.

Step 2

Computation of test statistic value:

For first born: x1-bar = 1121.7, s1 = 138.2743, n1 = 10

For 2nd born: x2-bar = 1130.2, s2 = 117.4477, n2 = 10

Step 3

Degrees of freedom:

df = n1 + n2 – 2

= 10 + 10 − 2

= 18

Rejection region for two-tailed test:

The rejection region at 5% level of significance can be obtained using the EXCEL formula. “=T.INV.2T(0.05,18)”

Thus, the critical value of Student’s t is ttab= 2.10092.

Decision rule:

If |t| > ttab, then reject the null hypothesi...

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### Hypothesis Testing 