Question

AMS communicates with customers who subscribe to cable television services through a special secured email system that sends messages about service changes, new features, and billing information to in-home digital set-top boxes for later display. To enhance customer service, the operations department established the business objective of reducing the amount of time to fully update each subscriber's set of messages. The department selected two candidates messaging systems and conducted an experiment in which 30 randomly chosen cable subscribers were assigned one of the two systems (15 assigned to each system). Update times were measured, and the results are organized in Table AMS 10.1 on page 388 (and stored in AMS 10-1).

EX. 1. Analyze the data in Table AMS 10.1 and write a report to the computer operations department that indicates your findings. Include an appendex in which you discuss the reason you selected a particular statistical test to compare the two independent groups of callers. Email Interface 1 Email Interface 2 4.13, 3.75, 3.93, 3.74, 3.36, 3.85, 3.26, 3.73, 4.06, 3.33, 3.96, 3.57, 3.13, 3.68, 3.63 3.71, 3.89, 4.22, 4.57, 4.24, 3.9, 4.09, 4.05, 4.07, 3.8, 4.36, 4.38, 3.49, 3.57, 4.74

EX 2. Suppose that instead of the research design described in the case, there were only 15 subscribers sampled, and the update process for each subscriber email was measured for each of the two messaging systems. Suppose that the results were organized in Table AMS 10.1- making each row in the table a pair of values for an individual subscriber. Using these suppositions, reanalyze the table AMS 10.1 data and write a report for presentation to the team that indicates your findings

Need help with EX 2!

Answer 1

(example 1)

here we use t-test with

null hypothesis H0:mean1=mean2 and alternate hypothesis H1:mean1≠mean2

statistic t=|(mean1-mean2)|/((s_p*(1/n1 +1/n2)^1/2) with df is n=n1+n2-2 and s_p²=((n1-1)s1²+(n2-1)s2²)/n

here we fail to reject the null hypothesis as two tailed p-value is more than typical level of significance alpha=0.05 and conclude that both interface are equivalent

t-test
	sample	mean	s	s2	n	(n-1)s2
	interface1	3.8413	0.3981	0.1585	15	2.2192
	interface2	3.9047	3.9047	15.2464	15	213.4499
	difference=	0.0633	sum=	15.4049	30	215.6691
	sp2=	7.7025
	sp=	2.7753
	SE=	1.0134
	t=	0.0625
one tailed	p-value=	0.4753
two tailed	p-value=	0.9506
critical	t(0.05, 28)	2.0484

(b)

here we use paired t-test with

null hypothesis H0:d=0 and alternate hypothesis H1::​d not equal=0

t=d^-/(sd/sqrt(n))=-3.04

here we fail to accept ( or reject )the null hypothesis as two tailed p-value is less than typical level of significance alpha=0.05 and conclude that both interface are not equivalent

	x	y	d=x-y
	4.13	3.75	0.38
	3.93	3.74	0.19
	3.36	3.85	-0.49
	3.26	3.73	-0.47
	4.06	3.33	0.73
	3.96	3.57	0.39
	3.13	3.68	-0.55
	3.63	3.71	0.18
	3.89	4.22	-0.65
	4.57	4.24	-4.24
	3.90	4.09	-4.09
	4.05	4.07	-4.07
	3.80	4.36	-4.36
	4.38	3.49	-3.49
	3.57	4.74	-4.74
		n=	15
		mean=	-1.6853
		sd=	2.1449
		SE=sd/sqrt(n)=	0.5538
		t=	-3.0432
		critical t=	2.1448
	two tailed	p-value=	0.0088
	one tailed	p-value=	0.0044