Question

Sibling IQ Scores (Raw Data, Software Required):
There have been numerous studies involving the correlation and differences in IQ's among siblings. Here we consider a small example of such a study. We will test the claim that, on average, older siblings have a higher IQ than their younger sibling. The results are depicted for a sample of 10 siblings in the table below. Test the claim at the 0.05significance level. You may assume the sample of differences comes from a normally distributed population.

Pair ID	Older Sibling IQ (x)	Younger Sibling IQ(y)
1	83	81
2	86	89
3	90	86
4	91	91
5	98	94
6	103	101
7	104	104
8	109	108
9	113	107
10	120	112

You should be able copy and paste the data directly into your software program.

(a) The claim is that the mean difference (x - y) is positive (μ_d > 0). What type of test is this?

This is a two-tailed test.This is a left-tailed test.     This is a right-tailed test.

(b) What is the test statistic? Round your answer to 2 decimal places.
t

_d

=  

(c) What is the P-value of the test statistic? Round to 4 decimal places.
P-value =  

(d) What is the conclusion regarding the null hypothesis?

reject H₀fail to reject H₀     

(e) Choose the appropriate concluding statement.

The data supports the claim that, on average, older siblings have a higher IQ than their younger sibling. There is not enough data to support the claim that, on average, older siblings have a higher IQ than their younger sibling.     We reject the claim that, on average, older siblings have a higher IQ than their younger sibling.We have proven that, on average, older siblings have a higher IQ than their younger sibling.

Additional Materials

Answer 1

Solution:-

a) This is right tailed test.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_d = 0

Alternative hypothesis: u_d > 0

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

Pair ID	Older (X)	Younger (Y)	X - Y	(d - dbar)^2
1	83	81	2	0.16
2	86	89	-3	29.16
3	90	86	4	2.56
4	91	91	0	5.76
5	98	94	4	2.56
6	103	101	2	0.16
7	104	104	0	5.76
8	109	108	1	1.96
9	113	107	6	12.96
10	120	112	8	31.36
Sum	997	973	24	92.4
Mean	99.7	97.3	2.4	9.24

s = sqrt [ (\sum (d_i - d)² / (n - 1) ]

s = 3.204164

SE = s / sqrt(n)

S.E = 1.013246

DF = n - 1 = 10 -1

D.F = 9

b)

d = X - y

d = 2.4

t = [ (x₁ - x₂) - D ] / SE

t = 2.37

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 9 degrees of freedom is greater than 2.37.

c)

Thus, the P-value = 0.0210

Interpret results. Since the P-value (0.0210) is less than the significance level (0.05), we have to reject the null hypothesis.

d) Reject the null hypothesis.

e) The data supports the claim that, on average, older siblings have a higher IQ than their younger sibling.