Question

Are you smarter than your older brother? In a study of birth order and intelligence, IQ tests were given to

18

- and

19

-year-old  men to estimate the size of the difference, if any, between the mean IQs of firstborn sons and secondborn sons. The following data for

10

firstborn sons and

10

secondborn sons are consistent with the means and standard deviations reported in the article. It is reasonable to assume that the samples come from populations that are approximately normal. Can you conclude that the mean IQ of firstborn sons is greater than the mean IQ of secondborn sons? Let

μ1

denote the mean IQ of firstborn sons and

μ2

denote the mean IQ of secondborn sons. Use the  

=α0.10

level and the

P

-value method with the table.

Firstborn
106	116	116	112	130
123	130	103	128	109

Secondborn
107	120	123	111	102
104	125	95	100	98

Send data to Excel

Part 1 of 6

Your Answer is correct

State the appropriate null and alternate hypotheses.

H0:=μ1μ2
H1:>μ1μ2
This is a	▼right-tailed	test.

Part 2 of 6

Your Answer is correct

Compute the test statistic. Round the answer to three decimal places.

=t

  

1.889

Part: 2 / 6

2 of 6 Parts Complete

Part 3 of 6

How many degrees of freedom are there, using the simple method?

The degrees of freedom using the simple method is .

Answer 1

Answer:

This is a right-tailed test

Test statistic, t = 1.889

The degrees of freedom using the simple method is 18

Explanation:

Hypothesis

The null hypothesis states that there is no difference in the mean IQ of firstborn and the secondborn sons and the alternative hypothesis tests the claim that the mean IQ of firstborn sons is greater than the secondborn sons.

T-Test

Two sample t-test is used to compare the mean IQs assuming equal variance.

The test is performed in excel by following these steps,

Step 1: Write the data values in excel. The screenshot is shown below,

Step 2: DATA > Data Analysis > t-Test: Two-Sample Assuming Equal Variances. The screenshot is shown below,

Step 3: Select the Variable 1 Range: Firstborn column, Variable 2 Range: Seconfborn column, Hypothesized Mean Difference = 0 then OK. The screenshot is shown below,

The result is obtained. The screenshot is shown below,

From the table,

Test statistic

t = 1.889

p-value

p-value (one-tail) = 0.0751

conclusion

Since the p-value is less than 0.10, the null hypothesis is rejected at a 10% significance level.