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Suppose IQ scores were obtained for 20 randomly selected sets of twins. The 20 pairs of measurements yield x=98.13​, y=100.6​, r=0.904​, ​P-value=0.000, ŷ=14.68+0.88x​, where x represents the IQ score of the twin born first. Find the best predicted value of ModifyingAbove y with caret given that the twin born first has an IQ of 104​? Use a significance level of 0.05.

Answer :

The predicted value of the twin born first has an IQ of 104 is 106.2..

We have given that,

sample size n = 10

x = 98.13

y = 100.6

r = 0.904

p-value = 0.000

y'= 14.68 + 0.88x

significance level = 0.05

and we have to find predicted value of the twin born first has an IQ of 104.

We know that x represents the IQ score of the twin born first.

Therefore value of x is 104.

Hence the predicted value P will be,

P = 14.68 + 0.88(104)

  = 14.68 + 91.52

P = 106.2

Therefore the best predicted value of the twin born first has an IQ of 104 is 106.2.

The positive predicted value is the probability that a test gives result for a true statistic. The negative predictive value is the probability that a test gives a false result for a false statistic.

To know more about predicted value here

https://brainly.com/question/19756249

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