Question

Chi-Square Goodness of Fit Test (how your work) (NO HAND WRITING ITS HARD TO READ)

A researcher has four prototypes for a small Unmanned Aerial System controller and wants to know which one that users prefer. The researcher asks 120 participants to use each controller and annotate which one they liked best. At the conclusion of the study, the researcher hopes to have a clearly identified prototype that is best preferred among the four. The table below represents the number of times each prototype was chosen by the participants.

Prototype A	Prototype B	Prototype C	Prototype D
14	34	61	11

Answer 1

H0: Null Hypothesis: There is no preferred prototype

HA: Alternative Hypothesis: There is preferred prototype

The Test Statistic is:

Decision Rule:

Take = 0.05

df = 4 - 1 = 3

By Technology, critical value of = 7.815

Reject null hypothesis if > 7.815

Test Statistic is got as follows:

Prototype	Observed (O)	Expected (E)	(O - E)2/E
Prototype A	14	30	8.5333
Prototype B	34	30	0.5333
Prototype C	61	30	32.033
Prototype D	11	30	12.033
Total =			53.133

So,

Test Statistic = 53.133

Since calculated value of = 53.133 is greater than critical value of = 7.815, the difference is significant. Reject null hypothesis.

Conclusion:
The data support the clam that there is preferred prototype.