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StatisticsQ&A LibraryListed 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 1204Question

Asked May 21, 2019

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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

Step 1

**Solution:**

**Null and alternative hypotheses:**

Null hypothesis: µ_{1 }− µ_{2 }= 0 (equivalently µ_{1 }= µ_{2})

Alternative hypothesis: µ_{1 }− µ_{2 }≠ 0 (equivalently µ_{1 }≠ µ_{2})

**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: *x*_{1}-bar = 1121.7, *s*_{1 }= 138.2743, *n*_{1} = 10

For 2nd born: *x*_{2}-bar = 1130.2, *s*_{2 }= 117.4477, *n*_{2} = 10

Step 3

**Degrees of freedom :**

*df = n*1 + *n*2 – 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 *t*tab= 2.10092.

**Decision rule:**

If |*t|* > *t*tab, then reject the null hypothesi...

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