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 uv1 denote the mean IQ of firstborn sons and uv2 denote the mean IQ of secondborn sons. Use the a=0.05 level and the P-value method.

Question

First Borns:
115
128
108
114
126
105
129
109
109
117
Second Borns:
96
86
106
96
84
110
129
98
118
129
Hv0 : [?]
Hv1 : [?]
This is a [right/left or two-tailed?] test

Answer 1

To analyze the problem regarding the mean IQ of firstborn sons (\( \mu_1 \)) and secondborn sons (\( \mu_2 \)), we can set up the hypotheses and determine the type of test.

Given that you want to conclude whether the mean IQ of firstborn sons is greater than that of secondborn sons, we can formulate the hypotheses as follows:

Null Hypothesis (\( H_0 \)): \[ \mu_1 \leq \mu_2 \] (This states that the mean IQ of firstborns is less than or equal to that of secondborns.)

Alternative Hypothesis (\( H_1 \)): \[ \mu_1 > \mu_2 \] (This states that the mean IQ of firstborns is greater than that of secondborns.)

This is a right-tailed test, as we are testing if the mean of one group is greater than the mean of another group.

Steps to Conduct the Test:

1. Calculate the Means and Standard Deviations

   • For the firstborns:

     • Data: 115, 128, 108, 114, 126, 105, 129, 109, 109, 117
     • Compute the mean (\( \bar{x}_1 \)) and standard deviation (\( s_1 \)).

   • For the secondborns:

     • Data: 96, 86, 106, 96, 84, 110, 129, 98, 118, 129
     • Compute the mean (\( \bar{x}_2 \)) and standard deviation (\( s_2 \)).

2. Conduct the t-test

   • Use the formula for the t-statistic for two independent samples: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]
   • Where:
     • \( n_1 \) and \( n_2 \) are the sample sizes (both 10 in this case).

3. Determine the Degrees of Freedom

   • The degrees of freedom can be calculated using: \[ df = n_1 + n_2 - 2 = 10 + 10 - 2 = 18 \]

4. Find the Critical Value and/or P-value

   • Given \( \alpha = 0.05 \), find the critical t-value for a one-tailed test with \( df = 18 \). Check a t-table or use statistical software/calculator for the exact critical value.

   • Calculate the P-value corresponding to the calculated t-statistic.

5. Decision Rule

   • If the P-value is less than or equal to \( \alpha \), reject \( H_0 \).
   • If the t-statistic is greater than the critical value, reject \( H_0 \).

By following these steps with actual calculations based on the provided data, you can reach a conclusion regarding the hypothesis about the mean IQ of firstborn sons compared to secondborn sons.

Answer 2

Compute the P-value, round the answer to 4 decimal places

Answer 3

To compute the P-value for the hypothesis test comparing the mean IQs of firstborn sons and secondborn sons, we need to calculate the means and standard deviations of the two groups first, followed by the t-statistic.

Step 1: Calculate the Means and Standard Deviations

Data for Firstborns:

• Values: 115, 128, 108, 114, 126, 105, 129, 109, 109, 117

Mean (\( \bar{x}_1 \)): \[ \bar{x}_1 = \frac{115 + 128 + 108 + 114 + 126 + 105 + 129 + 109 + 109 + 117}{10} = \frac{1250}{10} = 125.0 \]

Standard Deviation (\( s_1 \)): \[ s_1 = \sqrt{\frac{\sum (x_i - \bar{x}_1)^2}{n-1}} \] Calculating the squared differences:

• \( (115-125)^2 = 100 \)
• \( (128-125)^2 = 9 \)
• \( (108-125)^2 = 289 \)
• \( (114-125)^2 = 121 \)
• \( (126-125)^2 = 1 \)
• \( (105-125)^2 = 400 \)
• \( (129-125)^2 = 16 \)
• \( (109-125)^2 = 256 \)
• \( (109-125)^2 = 256 \)
• \( (117-125)^2 = 64 \)

Summing these squared differences: \[ 100 + 9 + 289 + 121 + 1 + 400 + 16 + 256 + 256 + 64 = 1221 \]

Calculating \( s_1 \): \[ s_1 = \sqrt{\frac{1221}{9}} \approx \sqrt{135.67} \approx 11.65 \]

Step 2: Data for Secondborns

Values: 96, 86, 106, 96, 84, 110, 129, 98, 118, 129

Mean (\( \bar{x}_2 \)): \[ \bar{x}_2 = \frac{96 + 86 + 106 + 96 + 84 + 110 + 129 + 98 + 118 + 129}{10} = \frac{1110}{10} = 111.0 \]

Standard Deviation (\( s_2 \)): Calculating the squared differences:

• \( (96 - 111)^2 = 225 \)
• \( (86 - 111)^2 = 625 \)
• \( (106 - 111)^2 = 25 \)
• \( (96 - 111)^2 = 225 \)
• \( (84 - 111)^2 = 729 \)
• \( (110 - 111)^2 = 1 \)
• \( (129 - 111)^2 = 324 \)
• \( (98 - 111)^2 = 169 \)
• \( (118 - 111)^2 = 49 \)
• \( (129 - 111)^2 = 324 \)

Summing these squared differences: \[ 225 + 625 + 25 + 225 + 729 + 1 + 324 + 169 + 49 + 324 = 2253 \]

Calculating \( s_2 \): \[ s_2 = \sqrt{\frac{2253}{9}} \approx \sqrt{250.33} \approx 15.82 \]

Step 3: Calculate the t-statistic

Now we can calculate the t-statistic using the formula: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] Substituting the computed values: \[ t = \frac{125.0 - 111.0}{\sqrt{\frac{11.65^2}{10} + \frac{15.82^2}{10}}} = \frac{14}{\sqrt{\frac{136.45}{10} + \frac{250.03}{10}}} \] Calculating inside the square root: \[ t = \frac{14}{\sqrt{13.645 + 25.003}} = \frac{14}{\sqrt{38.648}} \approx \frac{14}{6.21} \approx 2.25 \]

Step 4: Calculate the P-value

We now find the P-value corresponding to our t-statistic.

Degrees of Freedom:

The degrees of freedom for this test is: \[ df = n_1 + n_2 - 2 = 10 + 10 - 2 = 18 \]

Using a t-distribution table or a calculator:

• We have \( t \approx 2.25 \) with \( df = 18 \).

Looking up this value (or using a t-distribution calculator), we find the one-tailed P-value.

Using software or a calculator: \[ P(t \geq 2.25) \approx 0.0186 \]

Final P-value

Rounding to four decimal places, the P-value is: \[ \text{P-value} \approx 0.0186 \]

Thus, assuming the calculations are correct, you would reject the null hypothesis if \(0.0186 < 0.05\).