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 = 127	x = 106.9	s1 = 1.2
AISI 1078	n = 127	y = 126.8	s2 = 2.2

(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.

H₀: μ₁₀₆₄ − μ₁₀₇₈ = −10
H_a: μ₁₀₆₄ − μ₁₀₇₈ > −10H₀: μ₁₀₆₄ − μ₁₀₇₈ = −10
H_a: μ₁₀₆₄ − μ₁₀₇₈ ≠ −10    H₀: μ₁₀₆₄ − μ₁₀₇₈ = −10
H_a: μ₁₀₆₄ − μ₁₀₇₈ ≤ −10H₀: μ₁₀₆₄ − μ₁₀₇₈ = −10
H_a: μ₁₀₆₄ − μ₁₀₇₈ ≥ −10H₀: μ₁₀₆₄ − μ₁₀₇₈ = −10
H_a: μ₁₀₆₄ − μ₁₀₇₈ < −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 =

State the conclusion in the problem context.

Reject H₀. The data does not suggest that the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm².Fail to reject H₀. The data suggests that the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm².    Reject H₀. The data suggests that the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm².Fail to reject H₀. The data does not suggest that the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm².

(b) Estimate the difference between true average strengths for the two grades in a way that provides information about precision and reliability. (Use a 95% confidence interval. Round your answers to two decimal places.)

   ,  

kg/mm²

Answer 1

H₀: μ₁₀₆₄ − μ₁₀₇₈ = −10
H_a: μ₁₀₆₄ − μ₁₀₇₈ < −10

		pop 1	pop 2
sample mean x =		106.90	126.80
std deviation σ=		1.200	2.200
sample size n=		127	127
std error σx1-x2=√(σ21/n1+σ22/n2)    =			0.222
test stat z =(x1-x2-Δo)/σx1-x2     =			-44.52
p value : =			0.0000	(from excel:1*normsdist(-44.52)

  Reject H₀. The data suggests that the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm²

b)

Point estimate of difference '=x1-x2 =			-19.900
for 95 % CI value of z=			1.960	from excel:normsinv((1+0.95)/2)
margin of error E=z*std error =			0.436
lower bound=(x1-x2)-E =			-20.34
Upper bound=(x1-x2)+E =			-19.46
from above 95% confidence interval for population mean =(-20.34 , -19.46)