Question

Eleven resilient modulus observations of a ceramic mixture of type A are measured and found to have a sample average of 18.42 psi and sample standard deviation of 2.77 psi. Ten resilient modulus observations of a ceramic mixture of type B are measured and found to have a sample average of 19.28 psi and sample standard deviation of 2.41 psi.

(a) Is there sufficient evidence to support the claim that type A ceramic has larger variability than type B? Use α = 0.05.

(b) Compute the appropriate 95% confidence interval corresponding to your test in part (a).

Answer 1

(a) Given that, population standard deviation (0)=2.77 sample size (n) = 10 sample standard deviation (s)=2.41 population variance (o^2)=7.6729 sample variance (5^2) = 5.8081 null, Ho: o =2.77; alternate, H1:0 <2.77 level of significance, alpha = 0.05 from standard normal table,left tailed chisqr^2 alpha/2 3.325 since our test is left-tailed reject Ho, if chisqr^2 0 < 3.325 we use test statistic chisar^2 =(n-1)*5^2/0^2 chisor^2 cal =(10 - 1) * 5.8081 / 7.6729 = 9*5.8081/7.6729 = 6.8127 critical value the value of chisar^2 alpha at los 0.05 with d.f (n-1)=9,3.325 we got chisor^2 =6.813 & left tailed chisar^2 alpha/2 3.325 make decision hence value of chisar^2 cal > chisar^2 alpha and here we do not reject Ho chisar^2 p_value =0.3434 null, Ho: 0=2.77; alternate, H1 : 0 <2.77 test statistic: 6.8127 critical value: 3.325

p-value:0.3434 decision: do not reject Ho no suffcient evidence to support the claim that type A ceramic has larger variability than type B CONFIDENCE INTERVAL FOR STANDARD DEVIATION ci = (n-1) s^2 / 4^2 right < 0^2 < (n-1) s^2 / 4^2 left where, S = standard deviation 4^2 right = (1 - confidence level)/2 4^2 left = 1 - y^2 right n = sample size since alpha =0.05 4^2 right = (1 - confidence level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025 ^2 left = 1 - Y^2 right = 1 - 0.025 = 0.975 the two critical values y^2 left, y^2 right at 9 df are 19.0228 , 2.7 s.d( s )=2.41 sample size(n)=10 confidence interval for 0^2=[ 9 * 5.8081/19.0228 < 0^2 <9* 5.8081/2.7] = [ 52.2729/19.0228 < O^2 < 52.2729/2.7004 ] [ 2.7479 < 0^2 < 19.3575 ] and confidence interval for o = sqrt(lower) <o< sqrt(upper) = [ sqrt (2.7479) <o< sqrt(19.3575), ] = [ 1.6577 <o< 4.39971