Question

1.Conduct an analysis and hypothesis test of your choice on the data you collected. Write a 250-500 word research summary of the findings generated in the assignments for Topics 2 through 5. The research summary should address the following.

a. Explain what type of analysis and hypothesis test was conducted on the data collected.

b. Summarize the survey results based on the results of the data you analyzed.

c. Include the Excel analysis as part of the document.

Data available below: Time stamp 1.  If given the opportunity to work from home would you? Yes/No/Maybe

2. Do you consider working from home more of a employee convenience or employer benefit?

3. Who benefits more from the "Work from Home" opportunity?

8/11/2019 11:21:21	Yes	Employee Convenience	Employee
8/11/2019 11:22:28	Yes	Employer Benefit	Employer
8/11/2019 11:24:03	Maybe	Employee Convenience	Employee
8/11/2019 11:26:37	Maybe	Employee Convenience	Employee
8/11/2019 11:29:04	Yes	Employee Convenience	Employee
8/11/2019 11:36:23	Yes	Employer Benefit	Employee
8/11/2019 11:36:55	Yes	Employee Convenience	Employee
8/11/2019 11:43:05	Yes	Employee Convenience	Employee
8/11/2019 11:59:13	Yes	Employer Benefit	Employee
8/11/2019 12:14:33	Maybe	Employee Convenience	Employee
8/11/2019 12:22:02	Yes	Employer Benefit	Employer
8/11/2019 12:39:02	Yes	Employee Convenience	Employee
8/11/2019 12:47:51	Yes	Employee Convenience	Employee
8/11/2019 13:12:20	Yes	Employee Convenience	Employer
8/11/2019 13:49:33	Yes	Employer Benefit	Employer
8/11/2019 14:07:45	Maybe	Employee Convenience	Employer
8/11/2019 15:16:18	Yes	Employer Benefit	Employer
8/11/2019 18:55:11	Yes	Employer Benefit	Employer
8/11/2019 19:07:52	Yes	Employer Benefit	Employer
8/11/2019 20:01:33	Maybe	Employee Convenience	Employee
8/11/2019 20:03:24	Yes	Employer Benefit	Employer
8/11/2019 20:06:52
8/11/2019 21:38:25	Yes	Employer Benefit	Employer
8/12/2019 5:52:13	Yes	Employer Benefit	Employer
8/12/2019 6:56:04	Yes	Employee Convenience	Employee
8/12/2019 12:16:08	Yes	Employer Benefit	Employer

Answer 1

a.

Based on the data available, the Chi square test of independence is performed to test the association between work from home (Yes/Maybe) and reason to work from home (employee convenience/employer benefit)

b.

The chi square test of independence is performed in part c. The results of the test hypothesis are,

The null and alternative hypothesis for the test are defined as,

H0​: The two variables are independent

H1: The two variables are dependent

Result: based on the result obtained in part c, the null hypothesis is rejected. Hence we can conclude that both the variables work from home (Yes/Maybe) and reason to work from home are dependent.

(Note: Since one the category of variable work from home (No) doesn't take any value, it discarded from the variable, to test the hypothesis)

c.

The chi square test of independence is performed in following steps,

Step 1: The observed values are,

Observed values
	Employee Convenience	Employer Benefit	Total
Yes	8	12	20
Maybe	5	0	5
Total	13	12	25

Step 2: The expected values are obtained using the formula,

Expected values
	Employee Convenience	Employer Benefit	Total
Yes			20
Maybe			5
Total	13	12	25

Step 3: The Chi-Square Value is obtained using the formula

Observed,	Expected,
8	10.4	-2.4	5.76	0.5538
12	9.6	2.4	5.76	0.6
5	2.6	2.4	5.76	2.2154
0	2.4	-2.4	5.76	2.4
				5.7692

Step 4: The critical value for the chi square statistic is obtained from chi square distribution table for,

Where, r is the number of rows and c is the number of column in the table.

For significance level = 0.05 and degree of freedom = 1, the critical value is,

Step 5: Since the chi square value is greater than the critical value, It can be concluded that the null hypothesis is rejected at 5% significance level.

Hence both the variables work from home (Yes/Maybe) and reason to work from home are dependent.

Chi square test in excel

The square test is performed in excel using the function =CHISQ.TEST(actual_range, expected_range). The screenshot is shown below,

P-value = 0.0163

Since the P value is less than 0.05 at 5% significance level, It can be concluded that the null hypothesis is rejected at 5% significance level.