AP Statistics 2017

independence    hypothesizes    that    two categorical variables   are independent within  a   given   population.

A   test    for homogeneity of  proportions hypothesizes    that    the proportions of  the values  of  a   single

categorical variable    are the same    for more    than    one population. (c) is  incorrect.  It  is  a   reversal    of  the

actual  difference  between the two designs.    (d) is  incorrect.  You always  use count   data    when

computing   chi-square.

4. 
   

            The expected    value   of  the cell    with    the frog    is      .

   
   


5. Based on the design of the study this is a test of homogeneity of proportions.

I. H    0   :   The proportions of  patrons who rate    the restaurant  Excellent,  Good,   Fair,   and Poor    are the

same    for the Eastern and Western sides   of  town.

H (^) A : Not all the proportions are the same.
II . We will use the chi-square test for homogeneity of proportions.
Calculation of expected values (using the TI-83/84). yields the following results:
Since all expected values are at least 5, the conditions necessary for
this test are present.
III . , df = (2 – 1)(4 – 1) = 3 ≠ P -value > 0.25
(from Table C). χ 2 cdf(2.86,1000,3)=0.414.
IV . The P- value is larger than any commonly accepted significance level. Thus, we cannot reject H 0 .
We do not have evidence that location influences customer satisfaction.

6. 
   

            If  n = 15, then    df  =   15  –   1   =   14. In  the table   we  find    the entry   in  the column  for tail    probability of

   
   

   0.02 and the row with 14 degrees of freedom. That value is 26.87. Any value of X 2 larger than 26.87
   will yield a P -value less than 0.02.

   
   


7. 
   

I.          Let p   1    be the true    proportion  of  defects produced    on  Monday.

Let p   2    be the true    proportion  of  defects produced    on  Tuesday.

Let p   3    be the true    proportion  of  defects produced    on  Wednesday.

Let p   4    be the true    proportion  of  defects produced    on  Thursday.

Let p   5    be the true    proportion  of  defects produced    on  Friday.

H   0   :   p   1    =  p   2    =  p   3    =  p   4    =  p   5    (the   proportion  of  defects is  the same    each    day of  the week.)

H (^) A : At least one proportion does not equal the others (the proportion of defects is not the same
each day of the week).
II . We will use a chi-square goodness-of-fit test. The number of expected defects is the same for each
day of the week. Because there was a total of 150 defects during the week, we would expect, if the
null is true, to have 30 defects each day. Because the number of expected defects is greater than 5