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 0.01 significance 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	79
2	86	89
3	90	84
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 left-tailed test.

This is a two-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 H₀

fail 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.

Answer 1

Solution:

We are given data on older siblings IQ(x) and yonger sibling IQ(y) and We will test the claim that, on average, older siblings have a higher IQ than their younger sibling.

we also assume that data come from a normal distribution.

we use R software to perform this test.

# Storing the data in variable X and Y,

> X = c(83,86,90,91,98,103,104,109,113,120)

 > Y = c(79,89,84,91,94,101,104,108,107,112)  # Performing the test. > t.test(x,y,alternative = "greater",var.equal = TRUE)      Two Sample t-test data: x and y t = 1.3046, df = 27, p-value = 0.1015 alternative hypothesis: true difference in means is greater than 0 95 percent confidence interval: -2.441751 Inf sample estimates: mean of x mean of y 39.10000 31.11111 

a) The claim is that the mean difference (x - y) is positive (μ_d > 0) , This is a right-tailed test.

Hence, option c is correct.

b) The value of test statistic is

t = 1.3046 with df = 27

c) the P-value of the test statistic is

p-value = 0.1015

d) Since p-value is 0.1015 which is greater than 0.01, so the conclusion regarding the null hypothesis is we fail to reject null hypothesis.

Hence optionn b is correct.

e) The appropriate concluding statement is

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