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Exam 14: Chi Square Tests

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Question 61

Multiple Choice

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According to the work done by Arthur D. Hasler on salmon migration, salmon

A) use olfactory cues to locate their home stream
B) use detectors of the earth's magnetic field to locate their home stream
C) use a "follow-the-experienced-leader^" system to locate their home stream
D) cannot reliably locate the stream they were hatched in.

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Question 62

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The chi square distribution is

A) one curve
B) a family of curves
C) a family of symmetrical curves
D) a family of symmetrical, rectangular curves.

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Question 63

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A 6 x 8 test of independence of susceptibility to colds and susceptibility to allergies produced a chi square of 12.51. Write a conclusion about the two variables.

Goodness of fit is to independence as

Chi square test probabilities come from the F distribution.

Suppose you had on a single graph three chi square distributions with degrees of freedom of 1, 5 and 10. Choose a point on the X-axis and draw a vertical line. The distribution that would have the most area to the right of the vertical line would be the one with

Values of $\phi$ that are above are considered large.

The effect size index, $\phi$ , varies depending on the number of

The advantage of combining categories in a chi square problem is that it

Rejection of the null hypothesis for a test of independence is to rejection of the null hypothesis for a goodness-of-fit test as

In a correctly worked chi square problem, the sum of the expected frequencies is greater than the sum of the observed frequencies

A sociobiology theory predicted that the amount of help offered by three groups would depend on the degree of kinship to the helped group. The degrees of kinship were 15 percent, 10 per-cent, and 0 percent. When the frequency of help data were analyzed, the chi square value was smaller than the tabled value for the appropriate degrees of freedom. The data

Data Set 14-4:Each of the following represent data for a chi square test of independence. $\begin{array}{rlllllllll}\mathrm{W} & \mathrm{X} & \mathrm{Y} & \mathrm{Z}\\3 \quad 1 & 15 \quad 45 & 10 \quad 20 \quad 30 & 15 \quad 18 \quad 24 \\25 \quad 2 & 40 \quad 10 & 30 \quad 20 \quad 10 & 21 \quad 35 \quad 40 \\25 \quad 3 & \quad & \quad \quad & 18 \quad 20 \quad 25\end{array}$ -In Data Set 14-4, the short-cut method can be used on

Externalizers and Internalizers are two categories of personality types in a personality theory by Rotter. A large group of people at a PTA meeting were given a brief personality test and then classified as lower, middle, or upper class. Test the data below with a $\chi$ ² test. Identify the problem as one of independence or goodness of fit. Interpret the $\chi$ ² value you find. $\begin{array} { l c c c } & \text { Lower } & \text { Middle } & \text { Upper } \\\text { Internalizers } & 8 & 36 & 12 \\\text { Externalizers } & 18 & 21 & 2\end{array}$

The effect size index your text described for chi square was d.

A researcher tested his theory about the moon being made of green cheese with a goodness-of-fit test. The chi square value obtained did not allow him to reject the null hypothesis. He should conclude that

Data Set 14-4:Each of the following represent data for a chi square test of independence. $\begin{array}{rlllllllll}\mathrm{W} & \mathrm{X} & \mathrm{Y} & \mathrm{Z}\\3 \quad 1 & 15 \quad 45 & 10 \quad 20 \quad 30 & 15 \quad 18 \quad 24 \\25 \quad 2 & 40 \quad 10 & 30 \quad 20 \quad 10 & 21 \quad 35 \quad 40 \\25 \quad 3 & \quad & \quad \quad & 18 \quad 20 \quad 25\end{array}$ -Refer to Data Set 14-4. Degrees of freedom equal 4 for

In a $\chi$ ² test of independence between gender and kinds of phobias, the null hypothesis was rejected. The proper conclusion is that

For a 3 x 3 chi square test of independence, the number of degrees of freedom is 4.

After reviewing studies of chi square analyses of computer-generated small samples, your text noted that the principal problem with chi square analyses of small samples is the high likelihood of a