Stat 11

March 20, 2006

Homework #6 -- SOLUTIONS

 

Critical values:

 

1.  Draw a picture showing  t*a/2, n-1  (as a point on the horizontal axis) and the probabilities  C  and  a/2  (as areas under a curve).

 

 

 

 

 

 

 

 


          The curve is the “t” distribution with n-1 degrees of freedom.  (It looks like the standard normal curve, but has fatter tails.)

 

2.  If C = 0.80, what are  a,  z*a/2,  and  t*a/2, 15 ?

 

a = 0.20  (it’s always 1-C)

z*a/2 = 1.28  (or, more accurately, 1.281552).  Look for the probability 1-a/2 = 0.90 among the probabilities in the “z” table; now 1.28 is the closest value of z.  Or, in Excel write =NORMSINV(0.90) .

t*a/2,15 = 1.341, either from the text Table D (row label 15, top-of-column label a/2 = 0.10, bottom-of-column label C=0.90) or with Excel using   =TINV(0.20,15) .

 

3.  What are  t*0.005,  100  and  z*0.005  ?   (You would use one of these critical values for a 99% CI when n = 101.)

 

          t*0.005,  100 = 2.626   (table or =TINV(0.010,100))

          z*0.005  =  2.576.

 

4.  When  n  is large,  t*a/2, n-1  is close to  z*a/2.  Is the difference greater when C is large or when C is small?

 

          The difference between t* and z* is greater when C is large.  That’s because the t distribution differs from the z distribution mainly in the extreme tails, and that’s what you’re using when C is large.

 

(Nothing to turn in on the next two problems)

 

X.  Verify that  z*a/2  (in Table D) agrees with    =NORMSINV(a/2)   (in Excel) and that t*a/2, n-1  (in Table D) agrees with    =TINV (a, n-1 )    (in Excel).

 

Y.  If possible, learn how to find  z*a/2  and  t*a/2, n-1  on your calculator.  (The t-critical values you want may be labeled as “two-tailed” critical values.)

 

Confidence Intervals for Means:

 

5.   We weighed a random sample of 15 newly hatched turtles, and found an average mass of 8.0  grams.  We know from our extensive experience with turtles that the (population) standard deviation of their masses is  3.0  grams.  If  m  is the average mass of all newly hatched turtles, what is a  95%  confidence interval for  m ?

 

MOE  = z*0.025 times s times

          = 1.960    times    3.0 grams    times   

                   = 1.518 grams

So the confidence interval is  [ - MOE,  + MOE ] 

          = [ 6.482 grams,  9.518 grams  ]

 

            Also, what is a  99%  confidence interval for  m ?

 

          Same but using z*0.005 = 2.576 instead of 1.960.

          MOE = 1.995 grams

          Confidence interval

                   = [ 6.005 grams,  9.995 grams ]

 

6.  We also weighed a random sample of  20  newly hatched salamanders, and found an average mass of  4.5  grams.  Lacking experience with salamanders, we also calculated the standard deviation of our sample, finding  s = 1.4  grams.  If  m  is the average mass of all newly hatched salamanders, what is a 95% confidence interval for  m ?

 

          The difference here is that we’re using s in place of s, so we need to use t-critical values.

          MOE = t*0.025, 19  times  s  times 

                   = 2.093   times   1.4 grams  times  

                   =  0.655 grams

          Confidence interval

                   =  [  3.845 grams,  6.255 grams  ]

Also, what is an  94%  confidence interval for  m ?   (You’ll need to go beyond the table for this one.)

 

The hard part here is getting  t*0.03, 19. 

For accuracy you need a machine; Excel gives  TINV ( 0.06, 19 ) = 2.000.   (Really,  2.000017.)  So MOE = 0.626 grams and the CI is

          [  3.874 grams,  6.126 grams  ].

 

7.  Construct a 95% confidence interval for  m,  if you can, from these measurements:

 

            80.0    82.0    82.5    83.0    83.0   84.0   86.5             ()

                       

MOE = t*0.025,  6 ´ 1.979 ´ (1/sqrt(7))  = 1.830

CI = [ 81.170,  84.830 ]

 

8.  Construct a 95% confidence interval for  m,  if you can, from these measurements:

 

            4.00     4.00    4.01    4.03    4.10   4.20   4.57            ()

 

Aha, trick question!  The answer is  MOE = 0.191,  CI = [3.939, 4.321].  But you should also say “n is awfully small, and from this sample it looks as if the population is nowhere near normal, so we can’t trust this confidence interval.”  Avoid publishing it if you can, or at least avoid calling it a “95%” interval, and warn readers that it’s unreliable. 

 

Confidence intervals for proportions

 

9.  Among your sample of 400 relays, exactly 16 were defective. 

a.  What is ?

                   16 / 400  =   0.04   ( or 4 percent )

b.  Give a 90% confidence interval for  p,  the fraction of all relays that are defective.

                   SE =   =    =  0.0098.

                   MOE = z*0.05  times SE  (always use z* for proportions)

                   MOE = 1.645  times 0.0098  = 0.0161

 

                   CI  =  [ 0.0239,  0.0561 ]

 

           

10.  In another factory, only 4 of 400 relays in your sample were defective.

a.  What is ?

                   4 / 400  =   0.01   ( or 1 percent )

b.  Give a 99% confidence interval for p at this factory.

                   SE =   =    =  0.00497.

                   MOE = z*0.005  times SE  (always use z* for proportions)

                   MOE = 2.576  times 0.00497  = 0.0128

 

                   CI  =  [ – 0.0028,  0.0228 ]

 

c.  If you don’t like your answer, explain what went wrong.

          The left end of the CI is silly; proportions can’t be negative.

 

          The problem here is that when p is very close to 0 or 1, the (necessary) approximation of using  for  p  isn’t very good.  For the left endpoint, the damage is obvious.  But the right endpoint is also bad, for the same reason. 

 

          The conventional wisdom is that you shouldn’t trust confidence intervals for proportions unless your data includes at least 5 “hits” and 5 “misses.”  That’s just a rough rule.

 

          What to do about it?

                   (1)  Correct the CI to  [ 0.0000, 0.0228 ].  That’s not cheating;

                             you really do know that the true value can’t be below 0.

                   (2)  Consider using Wilson’s method for the right endpoint.  In

                             this case it means using  in place of  for

                             computing the SE (only).  That will give a slightly higher

                             right endpoint.

 

11.  The following statement is wrong:

 

            (WRONG)  For a given confidence level, halving the margin of error requires a sample twice as large.

Rounded Rectangular Callout: four times 


            Fix it.

 

 

12.  The marketing manager wants you to plan a survey to find out what fraction of adults in this county read the local sports page.  In nearby counties, the fraction varies from 15% to 40%.

 

            “I don’t want to waste money on this survey,” says the manager.  “On the other hand, I don’t want to run much risk of its being wrong by more than five percentage points.”

 

            What sample size will you recommend ?

 

Let’s take him to mean he wants the MOE to be 0.05.

 

With what confidence?  He doesn’t really say, but let’s try C = 95%.

 

So we want:

 

                   MOE = z* 

 

… and we may as well use  2  for z*, since we’re just guessing the confidence level anyway…

… and we may as well substitute  0.5  for  ,  since that’s an upper bound  (Other good choices would be to use  =0.40 or maybe halfway between  0.15 and 0.40,  but not just 0.15 because that’s too optimistic )…

…So we get

 

                   0.05 = 2 

 

          which forces  n  =  400.  Anything from, say, 250 to 500 would be reasonable.

 

 (end)