Question

The P-value for a chi-square test is always one-tailed. This is true because of all of the following reasons, EXCEPT:

Group of answer choices

Observed data that is either higher or lower than the null hypothesis predicts will result in a positive chi-square value.

Only a large positive value of chi-square provides stronger evidence to conclude H_A.

A chi-square distribution is skewed.

The chi-square statistic is always positive, unlike a z or t statistic which can be ether positive or negative.

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For a two-way classification table, we define:

n = total # of observations

r = # of rows in the table

c= # of columns in the table

For the Chi-Square Test of Independence, the degrees of freedom (df) is:

Group of answer choices

r x c - 1

n

n-1

(r-1)x(c-1)

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What are the correct hypotheses for the Chi-Square Test of Independence,?

Group of answer choices

H₀: A change in one variable does not cause a change in the other variable.

H_A: A change in one variable does cause a change in the other variable.

H₀: The variables of interest are dependent.

H_A: The variables of interest are independent.

H₀: The variables of interest are related to each other.

H_A: The variables of interest are not related.

H₀: The variables of interest are independent.

H_A: The variables of interest are dependent.

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Suppose you run a Chi-Squared Test of Independence on a 3-by-4 table and obtain a chi-squared value of 13.21. According to Table F, the P-value is:

Group of answer choices

between 0.025 and 0.05

between 0.05 and 0.10

between 0.02 and 0.025

between 0.01 and 0.02

Answer 1

The P-value for a chi-square test is always one-tailed. This is true because of all of the following reasons except A chi-square distribution is skewed. Since skewness does not affect the direction if test (like t distribution is also skwed bit we can test in both direction)

2) The degree of freedom for chi-square test is (r-1)*(c-1)., where r and c are the number of eows and coloumn respectively.

3) to Chi-square test, the null and alternative hypothesis are as

H0: The variables of interest are independent.

HA: The variables of interest are dependent. (since Chi-square test can be apply to test the dependency of two random variables)

4) For Chisquare value 13.21 with degree of freedom (3-1)*(4-1) = 6, P_value = P[Chisq>13.21] = 0.03982

therefore, P-value is between 0.025 and 0.05.