Question

Scenario: "A counselor working with people who are developmentally disabled has read research relating accurate responses to the type of reinforcement they receive. For three weeks, the counselor provides verbal praise every time a client completes a simple task correctly and then totals the number of simple problem-solving tasks that are completed successfully. For the next three weeks, the counselor provides a small piece of candy for each successfully completed task. The data for the group is as follows:"

Verbal praise: 17 observed

Tangible rewards: 27 observed

Total observed: 44

In your paper, begin with a paragraph introducing the scenario and explaining why a chi-square test is needed for the situation. Then, address the following: Identify which chi-square test you used and which program or procedure you used for the analysis. Describe the procedure you used to calculate the chi-square. If using VassarStats (Links to an external site.), copy and paste the data entry area and output into your paper. If using Excel, submit the Excel spreadsheet separately and mention in your paper that the spreadsheet is attached. If using hand calculations, include a table showing the observed and expected frequencies, along with the row and column totals and chi-square calculations. Report the chi-square value, degrees of freedom, and the p Explain how you determined the p Is the result statistically significant? If the test is a chi-square test of independence with a statistically significant result, report the effect size using Cramér’s V. Explain the meaning of the results in terms of the scenario. Discuss any assumptions, limitations, and implications associated with the situation and analysis. Summarize the main points of the paper in a concluding paragraph.

Answer 1

This is chi-square goodness of fit test

p	Oi	Ei	(Oi-Ei)^2/Ei
0.5	17	22	1.1364
0.5	27	22	1.1364
	44	TS	2.2727
	0.05 level	crititcal	3.8415
		p-value	0.1317

Formulas in Excel

p	Oi	Ei	(Oi-Ei)^2/Ei
0.5	17	=$B$6*A2	=(B2-C2)^2/C2
0.5	27	=$B$6*A3	=(B3-C3)^2/C3
	=SUM(B2:B4)	TS	=SUM(D2:D5)
	0.05 level	crititcal	=CHISQ.INV(0.95,1)
		p-value	=CHISQ.DIST.RT(D6,1)

TS = 2.2727

df = 2-1 = 1

p-value ==CHISQ.DIST.RT(TS,1) = 0.1317

since p-value > alpha

we fail to reject the null hypothesis

we conclude that there is not sufficient evidence for difference in proportion for verbal praise and tangible rewards}