Question

Business sensor technology provides a way for companies to learn about their customers, employees, and operations: data captured from sensors can be used to improve engagement, Sales, productivity, safety, and much more. A PwC survey of Global business and IT executives found that 25% of automotive executives: 27% of energy, utilities, and mining executives, 30% of hospitality and leisure executives; 33% of industrial products executives; and 52% of retail and consumer executives say their companies are currently investing in business sensor technology.

Suppose these results were based on 500 business and IT executives in each of the five industries: Automotive, Energy, Utilities, and Mining; Hospitality and Leisure; industrial products; and Retail and consumer.

A) at the 0.05 level of significance is there evidence of a difference among the industries with respect to the proportion of executives that say their companies are currently investing in business sensor technology?

B) Compute the p-value and interpret its meaning.

C) If appropriate, use the Marascuilo procedure and a= 0.05 to determine which companies differ in their currently investing in business sensor technology.

* Solve manually

Answer 1

A)

Hypotheses are:

H0:  There is no evidence of a difference among the industries with respect to the proportion of executives that say their companies are currently investing in business sensor technology.

Ha:  There is a evidence of a difference among the industries with respect to the proportion of executives that say their companies are currently investing in business sensor technology.

Here we need to use chi square test. Following is the 2 by 5 contingency table from the the given information :

		Automotive	Energy, utility and minning	Hospitality and leisure	Industrial products	Retail and consumer	Total
Used Sensor technology	Yes	500*.25=125	135	150	165	260	835
	No	500-125=375	365	350	335	240	1665
	Total	500	500	500	500	500	2500

Expected frequencies will be calculated as follows:

Following table shows the expected frequencies:

		Automotive	Energy, utility and minning	Hospitality and leisure	Industrial products	Retail and consumer	Total
Used Sensor technology	Yes	167	167	167	167	167	835
	No	333	333	333	333	333	1665
	Total	500	500	500	500	500	2500

Following table shows the calculations for chi square test statistics:

O	E	(O-E)^2/E
125	167	10.56287425
135	167	6.131736527
150	167	1.730538922
165	167	0.023952096
260	167	51.79041916
375	333	5.297297297
365	333	3.075075075
350	333	0.867867868
335	333	0.012012012
240	333	25.97297297
Total		105.4647462

Following is the test statistics:

Degree of freedom: df =( number of rows -1)*(number of columns-1) = (2-1)*(5-1)=4

The critical value is 9.488

Since test statistics is greater than critical value so we reject the null hypothesis.

B)

The p-value is: 0.0000

Excel function used for p-value: "=CHIDIST(11.119,2)"

C)

The formula for calculating critical range Marascuilo procedure is:

Here we have 5 population so degree of freedom is df=5-1=4

The critical value is 9.488

Following table shows the absolute difference and CR for each of the (5* (5-1)) /2 = 10 pairs:

pi	ni	pj	nj	D=Absolute(pi-pj)	CR	D> CR	Significant difference
0.25	500	0.27	500	0.02	0.085429325	No	No
0.25	500	0.3	500	0.05	0.086850216	No	No
0.25	500	0.33	500	0.08	0.088054492	No	No
0.25	500	0.52	500	0.27	0.091073649	Yes	Yes
0.27	500	0.3	500	0.03	0.087892716	No	No
0.27	500	0.33	500	0.06	0.089082901	No	No
0.27	500	0.52	500	0.25	0.09206834	Yes	Yes
0.3	500	0.33	500	0.03	0.090446413	No	No
0.3	500	0.52	500	0.22	0.093388273	Yes	Yes
0.33	500	0.52	500	0.19	0.094509276	Yes	Yes