Question

Suppose you know that the IQ scores of all incoming freshman are normally distributed with a standard deviation of 15. A simple random sampple of 100 freshman was collected with a mean of 120. Find a 90% confidence interval for the mean IQ score of all incoming freshman. (round all values to three decimal places)

Does this sample satisfy the conditions for constructing a confidence interval?

Select an answer Yes No Why do you keep asking these questions of me?

What critical value is appropriate for this problem?

Select an answer critical t-value critical z-value

What is the value of the critical value? (just enter the number)



What is the margin of error?

E=E=

State the 98% confidence interval. Round to one decimal place.

<μ<<μ<

Answer 1

Solution :

Given that,

n = 100

= 120   

s = 15

Note that, Population standard deviation() is unknown..So we use t distribution.

Our aim is to construct 90% confidence interval.   

c = 0.90

= 1- c = 1- 0.90 = 0.10

  /2 = 0.10 2 = 0.05

Also, d.f = n - 1 = 99  

    =    =  0.05,99 = 1.66

( use t table or t calculator to find this value..)

The margin of error is given by

E =  /2,d.f. * ( / n)

=  1.66* ( 15/ 100 )

E = 2.491

Now , confidence interval for mean() is given by:

( - E ) <   <  ( + E)

( 120 - 2.491 )   <   <  ( 120 + 2.491 )

117.509 <   < 122.491

Required 90% confidence interval is ( 117.509 , 122.491 )