Sibling IQ Scores: There have been numerous studies involving the correlation an

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 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) difference (d = x y) 1 85 83 2 2 88 93 -5 3 92 86 6 4 93 93 0 5 100 96 4 6 105 103 2 7 106 106 0 8 111 110 1 9 115 109 6 10 122 114 8 Mean 101.7 99.3 2.4 s 12.2 10.6 3.8 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 (d > 0). What type of test is this? This is a left-tailed test. This is a right-tailed test. This is a two-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 0.012 of the true answer. (c) Use software to get 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 younger siblings. There is not enough data to support the claim that, on average, older siblings have a higher IQ than younger siblings. We reject the claim that, on average, older siblings have a higher IQ than younger siblings. We have proven that, on average, older siblings have a higher IQ than younger siblings

Explanation / Answer

I am using R software to solve this problem.

At first, we need to import the data into R as below:

PairID <- 1:10
OlderSiblingIQ <- c(85,88,92,93,100,105,106,111,115,122)
YoungerSiblingIQ <- c(83,93,86,93,96,103,106,110,109,114)
difference <- c(2,-5,6,0,4,2,0,1,6,8)

InputData <- data.frame(PairID,OlderSiblingIQ,YoungerSiblingIQ,difference)

a) Alternative hypothesisi will decide if it is a left tailed test or right tailed test. This is a right tailed test as we are testing the claim that the mean difference is positive (d > 0)

b) To calculate the test statistic, we can use right tailed t.test() function in R as below:

t.test(InputData$OlderSiblingIQ,InputData$YoungerSiblingIQ,alternative = "greater")

   Welch Two Sample t-test

data: InputData$OlderSiblingIQ and InputData$YoungerSiblingIQ
t = 0.46866, df = 17.659, p-value = 0.3225
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
-6.489407 Inf
sample estimates:
mean of x mean of y
101.7 99.3

So the test statistic is 0.46866

c) As we can see above the p value for the test statistic is  0.3225

d) As p value is greater than 0.05, we fail to reject the null hypothesis.

e) Appropriate concluding statement will be:

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

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.