13. Chi-Square Tests & Goodness of Fit

Independence Test Using TI-84

13. Chi-Square Tests & Goodness of Fit

Independence Test Using TI-84

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Multiple Choice

A student performs an independence test using technology to see if pet ownership is independent of relationship status. They get the following results: $chi squared equals 0.545$ & $p equals 0.7614$ . What can they conclude about pet ownership and relationship status?

Reject $H_0$ since there is enough evidence to suggest that pet ownership and relationship status are dependent.

Fail to reject $H sub 0$ since there is not enough evidence to suggest that pet ownership and relationship status are dependent.

Reject $H_0$ since there is not enough evidence to suggest that pet ownership and relationship status are dependent.

Fail to reject $H_0$ since there is enough evidence to suggest that pet ownership and relationship status are dependent.

Verified step by step guidance

Identify the null hypothesis $ H_0 $ and the alternative hypothesis $ H_a $. For a test of independence, $ H_0 $ states that pet ownership and relationship status are independent, while $ H_a $ states that they are dependent.

Look at the given test statistic $ \chi^2 = 0.545 $ and the p-value $ p = 0.7614 $. The p-value represents the probability of observing a test statistic as extreme as, or more extreme than, the one calculated, assuming $ H_0 $ is true.

Compare the p-value to the significance level $ \alpha $ (commonly 0.05). If $ p \leq \alpha $, reject $ H_0 $; if $ p > \alpha $, fail to reject $ H_0 $.

Since the p-value $ 0.7614 $ is greater than the typical significance level $ 0.05 $, there is not enough evidence to reject the null hypothesis.

Conclude that we fail to reject $ H_0 $, meaning there is not enough evidence to suggest that pet ownership and relationship status are dependent; they appear to be independent based on this test.

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Struggling with Statistics?

A student performs an independence test using technology to see if pet ownership is independent of relationship status. They get the following results: χ2=0.545\chi^2=0.545 & p=0.7614p=0.7614. What can they conclude about pet ownership and relationship status?

Watch next

Independence Test Using TI-84 practice set

A student performs an independence test using technology to see if pet ownership is independent of relationship status. They get the following results: $chi squared equals 0.545$ & $p equals 0.7614$ . What can they conclude about pet ownership and relationship status?