13. Chi-Square Tests & Goodness of Fit
Independence Test Using TI-84
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A student performs an independence test using technology to see if pet ownership is independent of relationship status. They get the following results: & . What can they conclude about pet ownership and relationship status?
A
Reject since there is enough evidence to suggest that pet ownership and relationship status are dependent.
B
Fail to reject since there is not enough evidence to suggest that pet ownership and relationship status are dependent.
C
Reject since there is not enough evidence to suggest that pet ownership and relationship status are dependent.
D
Fail to reject since there is enough evidence to suggest that pet ownership and relationship status are dependent.
Verified step by step guidance1
Identify the null hypothesis \( H_0 \) and the alternative hypothesis \( H_a \). Here, \( H_0 \): pet ownership and relationship status are independent, and \( H_a \): pet ownership and relationship status are dependent.
Look at the given chi-square test statistic \( \chi^2 = 0.545 \) and the p-value \( p = 0.7614 \). The p-value represents the probability of observing the test results under the assumption that \( H_0 \) is true.
Choose a significance level \( \alpha \), commonly 0.05, to decide whether to reject \( H_0 \).
Compare the p-value to the significance level \( \alpha \). If \( p \leq \alpha \), reject \( H_0 \); if \( p > \alpha \), fail to reject \( H_0 \).
Since the p-value \( 0.7614 \) is greater than \( 0.05 \), conclude that there is not enough evidence to reject \( H_0 \). Therefore, pet ownership and relationship status are likely independent based on this test.
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