Question

A random sample of 125 students is chosen from a population of 4,000 students. If the mean IQ in the sample is 100 with a standard deviation of 8, what is the 98% confidence interval for the students' mean IQ score?

100−120

98.33−101.67

92−108

108.66−111.34

Answer 1

well, the interval will have to have a mean of 100 so the first and last ones dont fit

Answer 2

ok, how do i go about solving this?

Answer 3

you need to be able to determine a zscore, from a table or calculator such that:\[CI=mean \pm z(sd/\sqrt{n})\]

Answer 4

since the top is either 108 or not 108: mean + z(sd/sqrt(n)) would suffice

Answer 5

ive got a a TI nSpire cx CAS so it will be perfect

Answer 6

TI inspire, i never was able to go that fancy. not sure if they have the same setup as the ti83

if you can find the invnorm function, we would be set

Answer 7

im still learning this calculator.. what does it look like?

Answer 8

http://www.wolframalpha.com/input/?i=100+confidence+interval&a=*MC.~-_*Formula.dflt-&a=*FSVar.Q.*1.3-.100-_**TInterval.xbar-.dflt-&f3=0.98&f=TInterval.c%5Cu005f0.98&f4=125&f=TInterval.n_125&f5=8&f=TInterval.s%5Cu005f8

Answer 9

i also have a TI Nspire cx CAS

Answer 10

it looks like: invnorm

Answer 11

the t interval is prolly not what is required for this

Answer 12

since we want 98% about the mean, that leaves 2% left over to divide between the tails. 2/2 = 1 so, the invnorm(.9900) = 2.326 according to the wolf

Answer 13

when in calculator mode press menu and than 6: (statistics) then 5 (Distributions) then 3 (inversenormal).

then it will ask for the area, the mu and the sigma.

alternatively you can just type invNorm(area,mu,sigma)

Answer 14

the top of the interval would therefore be: mean + 2.326(sd/sqrtn)

Answer 15

sigma would have to be altered to the standard error: sigma/sqrtn

Answer 16

im getting not 108 :)

Answer 17

hmm, in hind sight .... 100+8 is not going to the the same as 100 + 2.326(8/sqrt(125)) so the not 108 would be the most logical option without any real accurate mathing

Answer 18

yes indeed, for future reference though the exact input would be \( invNorm(0.99,100,\frac{ 8 }{ \sqrt{125} }\)) for the upper boundary and \( invNorm(0.01,100,\frac{ 8 }{ \sqrt{125} }\)) for the lower boundary