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Suppose I.Q scores were obtained from randomly selected couples for 20 such pairs of people the linear correlation coefficient is 0.786 and the equation of the regression line is y^=16.02+0.81x where x represents the I.Q score of the wife also the 20 x values have a mean 98.93 and the 20y values have a mean of 86.5 what is the best predicted I.Q of the husband given that the wife has an I.Q of 107 use a significant level 0.05 The best predicted I.Q of the husband is _ (round to two decimal places as needed)

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probably greater the wife's :D
I get that with 95% confindence the corresponding husbands IQ is (98.049, 98.216)

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can i ask how you got that please there is a medal for you
I'll try, given that the correlation coefficient is the co-variance divided by the product of the independent variances, and that the independent variances are crucial for the 2 sample T interval that I'm aiming to find. Using the information from the regression line, I find the independent variances, then I have variances and means and should easily be able to find the interval. Given the strong correlation coefficient and the apparent difference in means, I can see that the interval should be in the upper nineties, this is only important as a check of results.
how do you enter this into the Ti84 calc??
for a classic two sample t interval try, "stat" then left twice for "tests" the interval tests are below the hypothesis tests.
can you type that using the numbers....Im lost...sorry.
there are no numbers, first the "stat" key which is located to the immediate right of the variable key. then press right twice (or left once on mine) for the test menu. The two sample T interval "2-SampTint" is the tenth result. By the way, I use a 83+ should be the same though.
yea they are the same pretty i got to that point, now what do i put for say x1, Sx1, n1, x2, Sx2, n2
and then it asks if its pooled? What the heck does that mean lol
its refering to the pooling of variances, 2sampTint works by combining x-y and (dev(x)+dev(y))/n+m-2 when its pooled, its a much more difficult process when not pooled; for the sake of simplicity pool, besides when working with a small sample like 40 its best we don't have much accuracy of information which means pooling is not going to effect the result.
this is what i got y=16.02+81x y=16.02(107) y=16.02+86.67 y=102.69 =102.69 rounded would be 102.70
@coureges I am not sure what numbers to plug in though.....I have the same question since @andijo76 and I are in the same class...
That's correct algebraically, um... it appears that interval testing is unnecessary. I am using a calculus based method. To obtain a confidence interval using algebra simply multiply that 102.7 by .05 and add/ subtract it from 102.7. That yields (97.565, 107.835); of course if you don't need the confidence interval than you should probably go with 102.7.
so i am correct?
Yes, Sorry for dragging you into calculus.
no that is okay I am just happy i got it right lol
thats the way you do it
now to fig out which numbers go where in the calc

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