Use the contingency table to the right to complete parts? (a) through? (c) below. |
A |
B |
Total |
||
1 |
1313 |
3737 |
5050 |
||
2 |
3737 |
3838 |
7575 |
||
Total |
5050 |
7575 |
125125 |
a. Find the expected frequency for each cell.
A |
B |
Total |
|
1 |
2020 |
3030 |
5050 |
2 |
3030 |
4545 |
7575 |
Total |
5050 |
7575 |
125125 |
?(Type integers or? decimals.)
b. Compare the observed and expected frequencies for each cell.
Choose the correct answer below.
A.
The totals for the observed values are always greater than the totals for the expected values.
B.
The totals for the observed and expected frequencies are the same.
C.
The expected values are always greater than the observed values.
D.
The observed values are always greater than the expected values.
c. Compute
chi Subscript STAT Superscript 2?2STAT.
Is it significant at
alpha?equals=0.050.05??
First let
pi?1
be the proportion of all events of interest in? A, and let
pi?2
be the proportion of all events of interest in B. Determine the hypotheses. Choose the correct answer below.
A.
Upper H 0 : pi 1 equals pi 2H0: ?1=?2
Upper H 1 : pi 1 not equals pi 2H1: ?1??2
B.
Upper H 0 : pi 1 greater than or equals pi 2H0: ?1??2
Upper H 1 : pi 1 less than pi 2H1: ?1<?2
C.
Upper H 0 : pi 1 not equals pi 2H0: ?1??2
Upper H 1 : pi 1 equals pi 2H1: ?1=?2
D.
Upper H 0 : pi 1 less than or equals pi 2H0: ?1??2
Upper H 1 : pi 1 greater than pi 2H1: ?1>?2
Calculate the test statistic.
chi Subscript STAT Superscript 2?2STATequals=nothing
?(Round to two decimal places as? needed.)
What is the? p-value?
?p-valueequals=nothing
?(Round to three decimal places as? needed.)Is this value significant at
alpha?equals=0.050.05??
A.
YesYes?,
because the? p-value is
less than or equal toless than or equal to
the alpha level.? Therefore, the null hypothesis should not be rejected.not be rejected.
B.
YesYes?,
because the? p-value is
less than or equal toless than or equal to
the alpha level.? Therefore, the null hypothesis should be rejected.be rejected.
C.
NoNo?,
because the? p-value is
greater thangreater than
the alpha level.? Therefore, the null hypothesis should not be rejected.not be rejected.
D.
NoNo?,
because the? p-value is
greater thangreater than
the alpha level.? Therefore, the null hypothesis should be rejected.be rejected.
a) as expected frequency =row total*column total/grand total
Expected | Ei=?row*?column/?total | A | B | Total |
1 | 20.000 | 30.000 | 50 | |
2 | 30.000 | 45.000 | 75 | |
Total | 50 | 75 | 125 |
b)
B.
The totals for the observed and expected frequencies are the same.
c)
A.
Upper H 0 : pi 1 equals pi 2H0: ?1=?2
Upper H 1 : pi 1 not equals pi 2H1: ?1??2
applying chi square test:
chi square | =(Oi-Ei)2/Ei | A | B | Total |
1 | 2.45 | 1.63 | 4.08 | |
2 | 1.63 | 1.089 | 2.72 | |
Total | 4.08 | 2.72 | 6.806 |
test statistic =6.81
p value =0.0091
option B:Yes?, because the? p-value is less than or equal to the alpha level.? Therefore, the null hypothesis should be rejected.
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