Question

Alyssa is interested in testing whether pet owners and non-pet owners differ in health status. To test this question, Alyssa collects data from 100 randomly selected people and assesses whether or not they own a pet and if they are healthy or unhealthy (the data are shown in the table). What decision should Alyssa reach regarding her hypothesis? How big is the effect size? Complete the six steps of hypothesis testing. Show your work for each step. Report your results and conclusion in APA style. Assume that alpha is .05.

	Healthy	Unhealthy
Pet Owner	33	22
Non-Pet Owner	14	31

Answer 1

The following cross tablulation have been provided. The row and column total have been calculated and they are shown below:

	Healthy	Unhealthy	Total
Pet Owner	33	22	55
Non-Pet Owner	14	31	45
Total	47	53	10

= The expected values are computed in terms of row and column totals. In fact, the formula is Eij RX, where R; corresponds to the total sum of elements in row i, C; corresponds to the total sum of elements in column j, and T is the grand total. The table below shows the calculations to obtain the table with expected values: Expected Values Healthy Unhealthy Total Pet Owner 47 x 55 100 25.85 53 x 55 100 29.15 55 Non-Pet Owner 47 x 45 100 = 21.15 53x45 100 = 23.85 45 Total 47 53 100 Based on the observed and expected values, the squared distances can be computed according to the following formula: (E - OP/E. The table with squared distances is shown below: Squared Distances Healthy Unhealthy Pet Owner (33–25.85) 25.85 1.978 22-29.15) 29.15 1.754 Non-Pet Owner (14-21.15) 21.15 = 2.417 (31-23.85) 23.85 = 2.144

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

H0​: The two variables are independent

Ha​: The two variables are dependent

This corresponds to a Chi-Square test of independence.

Rejection Region

Based on the information provided, the significance level is a = 0.05, the number of degrees of freedom is df = (2-1) (2-1) = 1, so then the rejection region for this test is R= {x? : x² > 3.841}. Test Statistics The Chi-Squared statistic is computed as follows: x? (Oi; - E;;)2 = 1.978 +2.417 +1.754 + 2.144 = 8.292 Eij 2=1 Decision about the null hypothesis Since it is observed that x= 8.292 > x = 3.841, it is then concluded that the null hypothesis is rejected. Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the two variables are dependent, at the 0.05 significance level. The corresponding p-value for the test is p = Pr(xí > 8.292) = 0.004.

***please comment if you have any doubts.Happy to help you.Thank you. Please Like.