Question

Tensile strength tests were carried out on two different grades of wire rod, resulting in the accompanying data. Grade Sample Size Sample Mean (kg/mm2) Sample SD AISI 1064 m = 126 x = 105.4 s1 = 1.1 AISI 1078 n = 126 y = 124.7 s2 = 2.1 (a) Does the data provide compelling evidence for concluding that true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2? Test the appropriate hypotheses using a significance level of 0.01. State the relevant hypotheses. HO: u1064 - 1078 = -10 Ha: u 1064 - u1078 > -10 HO: 1064 - u1078 = -10 Ha: u1064 - u1978 < -10 HO: 1064 - 1678 = -10 Ha: 1064 - u 1078 2-10 HO: u 1064 - 1078 = -10 Ha: 1064 - 1078 # -10 HO: 1064 - 1078 = -10 Ha: 1064 - 1078 S-10 Calculate the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) Z= P-value =

Answer 1

a)

HO: μ1064 - μ1078 = -10 Ha: μ1064 - 10 μη78

		pop 1	pop 2
sample mean x =		105.40	124.70
std deviation σ=		1.100	2.100
sample size n=		126	126
std error σx1-x2=√(σ21/n1+σ22/n2)    =			0.211
test stat z =(x1-x2-Δo)/σx1-x2     =			-44.04
p value : =			0.0000	(from excel:1*normsdist(-44.04)

10 kg/mm2. Reject HO. The data suggests that the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2. Rain

b)

Point estimate of difference '=x1-x2 =			-19.300
for 95 % CI value of z=			1.960	from excel:normsinv((1+0.95)/2)
margin of error E=z*std error =			0.414
lower bound=(x1-x2)-E =			-19.71
Upper bound=(x1-x2)+E =			-18.89
from above 95% confidence interval for population mean =(-19.71 , -18.89)