13. Chi-Square Tests & Goodness of Fit

Goodness of FIt Test Using TI-84

13. Chi-Square Tests & Goodness of Fit

Goodness of FIt Test Using TI-84

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Multiple Choice

A student performs Goodness of Fit Test using technology to see if the proportion of each candy flavor in a bag matches the expected distribution. They get the following results: $chi squared equals 18.99$ & $p equals 0.0019$ . What can they conclude about the claimed distribution of candy flavors?

Fail to reject the $H_0$ since there is not enough evidence to suggest $H_{a}$ is true. The claimed distribution is a good fit.

Fail to reject the $H_0$ since there is not enough evidence to suggest $H_{a}$ is true. The claimed distribution is a bad fit.

Reject the $H_0$ since there is enough evidence to suggest $H_{a}$ is true. The claimed distribution is a good fit.

Reject the $H sub 0$ since there is enough evidence to suggest $H_{a}$ is true. The claimed distribution is a bad fit.

Verified step by step guidance

Identify the null hypothesis $ H_0 $ and the alternative hypothesis $ H_a $. Here, $ H_0 $ states that the observed candy flavor proportions match the claimed distribution, and $ H_a $ states that they do not match.

Note the test statistic value $ \chi^2 = 18.99 $ and the p-value $ p = 0.0019 $ obtained from the goodness of fit test using technology.

Recall the decision rule for hypothesis testing: if the p-value is less than the chosen significance level (commonly $ \alpha = 0.05 $), reject the null hypothesis $ H_0 $; otherwise, fail to reject $ H_0 $.

Compare the p-value $ 0.0019 $ to the significance level $ 0.05 $. Since $ 0.0019 < 0.05 $, this indicates strong evidence against $ H_0 $.

Conclude that there is enough evidence to reject $ H_0 $ and support $ H_a $, meaning the claimed distribution of candy flavors is not a good fit for the observed data.

4:49m

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Struggling with Statistics?

A student performs Goodness of Fit Test using technology to see if the proportion of each candy flavor in a bag matches the expected distribution. They get the following results: χ2=18.99\chi^2=18.99 & p=0.0019p=0.0019. What can they conclude about the claimed distribution of candy flavors?

Watch next

Goodness of FIt Test Using TI-84 practice set

A student performs Goodness of Fit Test using technology to see if the proportion of each candy flavor in a bag matches the expected distribution. They get the following results: $chi squared equals 18.99$ & $p equals 0.0019$ . What can they conclude about the claimed distribution of candy flavors?