A critic of the SAT test wants to show that the correct answers are not equally

Question

A critic of the SAT test wants to show that the correct answers are not equally distributed among the five alternatives. He collects the following table of frequency information about the location of the correct answers for 100 of the items.

Complete the following set of questions using the table:

Position

        A

        B

        C  

         D         

         E

Frequency

        26

        34

       35

         31

        24

a) State in words what is null hypothesis would be:

b) Assuming each option would appear with the same frequency, what would be the expected value for each of the five cells:

A                     B                     C                     D                     E

___                  ___                  ___                  ___                  ___

           

c) Using = .01, determine the critical value:            _________

d) Compute the appropriate observed test statistic: __________

e) Sketch the sampling distribution and label the following: critical value, rejection region, observed test statistic, and the mean of the sampling distribution.

Position

        A

        B

        C  

         D         

         E

Frequency

        26

        34

       35

         31

        24

Explanation / Answer

observed frequncies are Oi

26 34 35 31 24   

------------------------------------------------------------------

expected frequencies are Ei

= ( 26 + 34 + 35 + 31 + 24 )/5 = 30

------------------------------------------------------------------

and claiming hypothesis is

null, Ho: each option would appear with the same frequency

alternative, H1: each option would appear with the diffrent frequency

level of significance, = 0.01

from standard normal table, chi square value at right tailed, ^2 /2 =9.2103

since our test is right tailed,reject Ho when ^2 o > 9.2103

we use test statistic ^2 o = (Oi-Ei)^2/Ei

from the table take sum of (Oi-Ei)^2/Ei, we get ^2 o = 3.1332

critical value

the value of |^2 | at los 0.01 with d.f, n - 1 = 3 - 1 = 2 is 9.2103

we got | ^2| =3.1332 & | ^2 | =9.2103

make decision

hence value of | ^2 o | < | ^2 | and here we do not reject Ho

^2 p_value =0.2088

ANSWERS

---------------

null, Ho: no association exists b/w them OR observations are independent

alternative, H1: exists association b/w them OR observations are dependnet

test statistic: 3.1332

critical value: 9.2103

p-value:0.2088

decision: do not reject Ho

each option would appear with the same frequency

Observed (Oi ) Expected ( Ei) Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 26 30 -4 16 0.5333 34 30 4 16 0.5333 35 30 5 25 0.8333 31 30 1 1 0.0333 24 30 -6 36 1.2

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