Question

Sibling IQ Scores: 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 10siblings 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)	difference (d = x − y)
1	86	84	2
2	89	92	-3
3	93	89	4
4	94	94	0
5	101	97	4
6	106	104	2
7	107	107	0
8	112	111	1
9	116	110	6
10	123	115	8
Mean	102.7	100.3	2.4
s	12.2	10.5	3.2

If you are using software, you should be able copy and paste the data directly into your software program.

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

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

(b) What is the test statistic? Round your answer to 2 decimal places.
t_d =  
To account for hand calculations -vs- software, your answer must be within 0.01 of the true answer.

(c) Use software to get the P-value of the test statistic. Round to 4 decimal places.
P-value =

Answer 1

Data Summary

	n	Mean (M)	Variance (SS)	Standard Deviation (SD)
Older Sibling IQ	10	102.7	149.3445	12.2207
Younger Sibling IQ	10	100.3	110.6778	10.5204

a)   The null and alternative hypotheses are                      
   Ho : μd ≤ 0       μd is the difference μ1 - μ2              
   Ha : μd > 0                      
   This is a right-tailed test                     
                          
                          
b)   We use independent Samples T test since population standard deviation is unknown, and sample size is small                      
   Using the formulae given below, we get                      
                          
   Degrees of Freedom                      

df1 = n1 - 1   df2 = n2 - 1   df = n1 + n2 - 2

    df1 = 9   df2 = 9                  
   df = 18                      
                          
   Pooled Variance Sp²  

                   
   Sp² = 130.0111                      
                           
   Mean Squared Error S_(M1-M2)    

               
   S_(M1-M2) = 5.0992                      
                          
   t-statistic  

                  
   td = 0.47                      
                          
                          
c)   For t = 0.47, df = 18 we find the Right Tailed p-value using Excel function t.dist                      
   p-value = 1 - t.dist(0.4707, 18, TRUE)                      
   p-value = 0.3218                      
                          
   Let the significance level α = 0.05                      
   Decision                      
   0.3218 > 0.05                      
   that is p-value > α                      
   Hence we DO NOT REJECT Ho                      
                          
   Conclusion                      
   There does not exist enough statistical evidence at α = 0.05 to show that                       
   the mean difference is positive                       
   In other words,                      
   older siblings need not have higher IQ than their younger siblings