Question

 

1. Fusible interlinings are being used with increasing frequency to support outer fabrics and improve the shape and drape of various pieces of clothing. The article “Compatibility of Outer and Fusible Interlining Fabrics in Tailored Garments” (Textile Res. J., 1997: 137-142) gave the accompanying data on extensibility (%) at 100 gm/cm for both high-quality fabric (H) and poor-quality fabric (P) specimens:

H: 1.2, .9, .7 ,1.0, 1.7, 1.7, 1.1, .9, 1.7, 1.9 ,1.3, 2.1 ,1.6, 1.8, 1.4, 1.3 ,1.9 ,1.6, .8 2.0 ,1.7, 1.6, 2.3, 2.0

P: 1.6, 1.5, 1.1, 2.1, 1.5, 1.3, 1.0, 2.6

(a) Assuming the samples came from approximate normal populations, run a test at the ?=0.05 significance level to determine whether the true average extensibility differs for the two types of fabrics. Be sure to follow the directions given at the top of the assignment.

(b) Construct a 95% confidence interval for ??−??, the difference between the average extensibility of the two fabrics. Explain how you could draw your conclusion in part (a) by using this confidence interval.

Answer 1

(a) here we use t-test with

null hypothesis H0:µ1=µ2 and alternate hypothesis H1: µ1≠µ2

statistic t=|(mean1-mean2)|/((s_p*(1/n1 +1/n2)^1/2)=0.4166

with df is n=n1+n2-2=30 and s_p²=((n1-1)s1²+(n2-1)s2²)/n=0.2169

since the two tailed p-value is more than level of significance alpha=0.05, so we accept the null hypothesis and conclude that the true average extensibility donot differs for the two types of fabrics.

t-test
	sample	mean	s	s2	n	(n-1)s2
	H	1.5083	0.4442	0.1973	24	4.5382
	P	1.5875	0.5303	0.2812	8	1.9685
	difference=	0.0792	sum=	0.4785	32	6.5067
	sp2=	0.2169
	sp=	0.4657
	SE=	0.1901
	t=	0.4166
two tailed	p-value=	0.6800
two tialed critical	t(0.05)	2.0423

(b)

(1-alpha)*100% confidence interval for difference population mean= difference of sample mean±t(alpha/2)*SE(difference)

95% confidence interval =0.0792±t(0.05/2,30)*0.1901=0.0792±2.0423*0.1901=0.0792±0.3882=(0.3110,0.4674)

SE=(s_p*(1/n1 +1/n2)^1/2)=0.1901