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Author Question: In a chi-square goodness-of-fit test, the value of the test statistic is found to be 11.89 and there ... (Read 43 times)

petersan

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on: Feb 26, 2023
In a chi-square goodness-of-fit test, the value of the test statistic is found to be 11.89 and there are 5 degrees of freedom. Usingα= 0.05, the null hypothesis would be:

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not rejected because thep-value for this test is less than 0.05.


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rejected because thep-value for this test is less than 0.05.


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not rejected because thep-value for this test is greater than 0.05.


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rejected because thep-value for this test is greater than 0.05.

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Marked as best answer by petersan on Feb 26, 2023

dianagarcia

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Reply #1 on: Feb 26, 2023
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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petersan

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Reply 2 on: Feb 26, 2023
Gracias!

olderstudent

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Reply 3 on: Yesterday
Thanks for the timely response, appreciate it

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