The test described shows whether or not there is a ? Relationship Preference а. b. Pages Dependence None of these С. 605-607 d.

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The test described shows whether or not there is a ?
а.
Relationship
b.
Preference
Pages
Dependence
None of these
с.
605-607
d.
Transcribed Image Text:The test described shows whether or not there is a ? а. Relationship b. Preference Pages Dependence None of these с. 605-607 d.
As with the chi-square goodness-of-fit test, if there is little difference between the observed
problem.) The expected values are computed from the observed values, and they are
pos b The observed values are obtained from the sample data. (That is, they are given in the
n mua ulo ues, then the test statistic will be large and the null hypothesis will be rejected. In this
605
Section 11-2 Tests Using Contingency Tables
based on the assumption that the two variables are independent.
The formula for computing the expected values for each cell is
mus wo
Expected value =
(row sum)(column sum)
grand total
values and the expected values, then the value of the test statistic will be small and the ner
hypothesis will not be rejected. Hence, the variables are independent of each other.
However, if there are large differences in the observed values and the expected var-
t brso
case, there is enough evidence to say that the variables are dependent on or related to caen
other. This test is always right-tailed.
The degrees of freedom for the chi-square test of independence are based on the size
of the contingency table. The formula for the degrees of freedom is (R – 1)(C - 1),
that is, (rows – 1)(columns – 1). For a 2 x 3 table, the degrees of freedom would be
o (2 – 1)(3 – 1) = 1(2) = 2.
As an example, suppose a new postoperative procedure is administered to a number
of patients in a large hospital. The researcher can ask the question, Do the doctors feel dif-
ferently about this procedure from the nurses, or do they feel basically the same way?
Note that the question is not whether they prefer the procedure but whether there is a
difference of opinion between the two groups.
To answer this question, a researcher selects a sample of nurses and doctors and
a tabulates the data in table form, as shown.
(50020
00
Prefer new
procedure
Prefer old
procedure
No
Group
preference
Nurses
loctors
100
80
bsole zlos snibnng
20
30
50
120
As t survey indicates, 100 nurses prefer the new procedure, 80 prefer the old pro-
cedure, and 20 have no preference; 50 doctors prefer the new procedure, 120 like the old
procedure, and 30 have no preference. Since the main question is whether there is a dif-
ference in opinion, the null hypothesis is stated as follows:
00
Ho: The opinion about the procedure is independent of the profession.
The alternative hypothesis is stated as follows:
ogobooosol asion H: The opinion about the procedure is dependent on the profession.
o won od
If the null hypothesis is not rejected, the test means that both professions feel basi-
cally the same way about the procedure and that the differences are due to chance. If the
null hypothesis is rejected, the test means that one group feels differently about the pro-
cedure from the other. Remember that rejection does not mean that one group favors the
procedure and the other does not. Perhaps both groups favor it or both dislike it, but in dif-
ferent proportions.
Recall that the degrees of freedom for any contingency table are (rows – 1) times
(columns - 1); that is, d.f. = (R – 1)(C – 1). In this case, (2 – 1)(3 – 1) = (1)(2) = 2
The reason for this formula for d.f. is that all the expected values except one are free to varu
in each row and in each column. The critical value for a = 0.05 from Tahble G is 5 00
To test the null hypothesis by using the chi-square independence test, you must com-
pute the expected frequencies, assuming that the null hypothesis is true. These freOuen
cies are computed by using the observed frequencies given in the table
08)
ar
Transcribed Image Text:As with the chi-square goodness-of-fit test, if there is little difference between the observed problem.) The expected values are computed from the observed values, and they are pos b The observed values are obtained from the sample data. (That is, they are given in the n mua ulo ues, then the test statistic will be large and the null hypothesis will be rejected. In this 605 Section 11-2 Tests Using Contingency Tables based on the assumption that the two variables are independent. The formula for computing the expected values for each cell is mus wo Expected value = (row sum)(column sum) grand total values and the expected values, then the value of the test statistic will be small and the ner hypothesis will not be rejected. Hence, the variables are independent of each other. However, if there are large differences in the observed values and the expected var- t brso case, there is enough evidence to say that the variables are dependent on or related to caen other. This test is always right-tailed. The degrees of freedom for the chi-square test of independence are based on the size of the contingency table. The formula for the degrees of freedom is (R – 1)(C - 1), that is, (rows – 1)(columns – 1). For a 2 x 3 table, the degrees of freedom would be o (2 – 1)(3 – 1) = 1(2) = 2. As an example, suppose a new postoperative procedure is administered to a number of patients in a large hospital. The researcher can ask the question, Do the doctors feel dif- ferently about this procedure from the nurses, or do they feel basically the same way? Note that the question is not whether they prefer the procedure but whether there is a difference of opinion between the two groups. To answer this question, a researcher selects a sample of nurses and doctors and a tabulates the data in table form, as shown. (50020 00 Prefer new procedure Prefer old procedure No Group preference Nurses loctors 100 80 bsole zlos snibnng 20 30 50 120 As t survey indicates, 100 nurses prefer the new procedure, 80 prefer the old pro- cedure, and 20 have no preference; 50 doctors prefer the new procedure, 120 like the old procedure, and 30 have no preference. Since the main question is whether there is a dif- ference in opinion, the null hypothesis is stated as follows: 00 Ho: The opinion about the procedure is independent of the profession. The alternative hypothesis is stated as follows: ogobooosol asion H: The opinion about the procedure is dependent on the profession. o won od If the null hypothesis is not rejected, the test means that both professions feel basi- cally the same way about the procedure and that the differences are due to chance. If the null hypothesis is rejected, the test means that one group feels differently about the pro- cedure from the other. Remember that rejection does not mean that one group favors the procedure and the other does not. Perhaps both groups favor it or both dislike it, but in dif- ferent proportions. Recall that the degrees of freedom for any contingency table are (rows – 1) times (columns - 1); that is, d.f. = (R – 1)(C – 1). In this case, (2 – 1)(3 – 1) = (1)(2) = 2 The reason for this formula for d.f. is that all the expected values except one are free to varu in each row and in each column. The critical value for a = 0.05 from Tahble G is 5 00 To test the null hypothesis by using the chi-square independence test, you must com- pute the expected frequencies, assuming that the null hypothesis is true. These freOuen cies are computed by using the observed frequencies given in the table 08) ar
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