AP Statistics 2017

H (^) A : at least one of these proportions is incorrect.
II . We will use the χ^2 goodness-of-fit test. The problem states that the sample is a random
sample. The expected values are type A: 400; type B: 110; type AB: 40; type O: 450. Each
of these is greater than 5. The conditions needed for the test are satisfied.
III . The data are summarized in the table:
From the X 2 table (Table C, which is read very much like Table B), we see that 119.44 is
much larger than any value for df = 3. Hence, the P- value < 0.0005. (A calculator gives a P
-value of 1.02 × 10−25 —more about how to do this coming up next.)
IV . Because the P -value is so small, we reject the null hypothesis. We have very strong
evidence that the proportions of the various blood types among black Americans differ from
the proportions among white Americans.
Calculator Tip: The computations in part III of the above hypothesis test can be done on the TI-83/84
as follows: put the observed values in L1 and the expected values in L2 . Let L3=(L1-L2)^2 /L2 . Quit
the lists and compute LIST MATH sum(L3) . This will return the value of X 2 . Alternatively, you could
do STAT CALC 1-Var Stats L3 and note the value of Σx .
To find the probability associated with a value of X 2 , do the following: DISTR χ 2 cdf(lower
bound, upper bound, df) . In the above example, that might look like χ 2
cdf(119.44,1000,3)=1.01868518 × 10−25 .
The TI-83/84 and early versions of the TI-84 do not have a χ^2 goodness-of-fit test built in. Newer
versions of the TI-83/84 do have it, however. It is found in the STAT TESTS menu and is identified as
χ^2 GOF–Test .