Question

The void volume within a textile fabric affects comfort, flammability, and insulation properties. Permeability of a fabric refers to the accessibility of void space to the flow of a gas or liquid. We have summary information on air permeability (in cm1cm2/sec) for two different types of plain weave fabric (see below). Cotton: 51.7.49.4,50.3,47 8,477,50 4,51.4,55 2.487,50 3.49.8.46.1 Acetate: 58.9,55,5,62 1,57.8,60 3,58.6.60.8,58,5,61.2 61.8,643,59.5 (1) Find a 98% confidence interval for the true difference in mean permeability Acetate - Cotton ( Number Number Enter your answers correct to 2 decimal places (2) What assumptions did you make to answer this question? Do they seem reasonable? Tick all that apply Data are approximately normal (Seems reasonable from the normal quantile plot.) We have two independent random samples of measurements. (Whether this is satisfies depends on the study design) Data are approximately normal (But data are strongly skewed from the normal quantito plot, so this is not reasonable.) Standard deviations are equal (Seems reasonable, from the data) The sample mean difference is approximately normal (Seems reasonable since np(1-P) > 5.)

Answer 1

a)

Sample Mean 1 (X1) = 49.9 Sample Standard Deviation 1 (81) = 2.33 Sample Size 1 (N1) = 12 Sample Mean 2 (X2) = 59.94 Sample Standard Deviation 2 (82) = 2.32 Sample Size 2 (N2) = 12

Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are df = ni+n2 – 2 = 12 + 12 – 2 = 22. = 2.508. The The critical value for a = 0.02 and df = 22 degrees of freedom is to = tı-a/2;n -1 corresponding confidence interval is computed as shown below: Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows: Sp = (n1 – 1)si (n2 – 1)s ni + n2 2 (12 = 1) × 2.332 + (12 – 1) 2.322 12 + 12 – 2 2.325 Since we assume that the population variances are equal, the standard error is computed as follows: 1 1 se S + ni n2 1 1 = 2.325 x + 12 12 = 0.949

CI​=​(Xˉ2​−Xˉ1−tc​×se,Xˉ2−Xˉ1​+tc​×se)

=((59.94-49.9)−2.508×0.949,(59.94-49.9)+2.508×0.949)

=(7.659908,12.420092)​

b)

Data are approximately normal

( seems reasonable from the normal quantile plot )

We have two independent random samples of measurements

(Whether this is satisfies depends on the study design)