Question

Tensile strength tests were carried out on two different grades of wire rod, resulting in the accompanying data.

GradeSample
SizeSample
Mean
(kg/mm2)Sample
SD

AISI 1064      m = 127    x = 106.9  s1 = 1.2

AISI 1078    n = 127y = 126.8s2 = 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
Ha: μ1064 − μ1078 > −10H0: μ1064 − μ1078 = −10
Ha: μ1064 − μ1078 ≠ −10    H0: μ1064 − μ1078 = −10
Ha: μ1064 − μ1078 ≤ −10H0: μ1064 − μ1078 = −10
Ha: μ1064 − μ1078 ≥ −10H0: μ1064 − μ1078 = −10
Ha: 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

Let denote true average strength for the 1064 grade and 1078 grade respectively.

111, 112

Hypotheses: To test null hypothesis H H1 - H2 = 10 against alternative hypothesis H1 H1-H2 <10 This is a one tailed test as the alternative hypothesis contains '<' sign. Here, si sa 1 = 106.90, S1 = 1.20, n1 = 127, 12 = 126.80, S2 = 2.20, n2 = 127, Standard error of mean = + 1.202 127 + 2.202 127 0.2224 ni n2 The test statistic can be written as: (11 - 12) -(10) t= VT + T22 which under Ho follows at distribution with 194 df. where + 2.202 TI df= 1.202 127 (1.20 127 127-1 )2 127 (2002) 194 che de -1. + + 127 1 Th1-1 na 1 and standard error for the difference between means $; 1.202 SE= + = + 2.202 127 0.222371 ni na 127 Decision rule / Rejection region : We reject H, at 0.01 level of significance if P-value < 0.01 or if tobs <-+0.01,df = -0.01,194 = -2.345723 0.4- 20.3- 0.2- 0.1- 0.0- -2 Now, Value of the test statistic : The value of the test statistic is tstat = (106.90 - 126.80) 1.20 2.20 127 127 134.460037 -134.460 associated degrees of freedom = 194 The critical value = tcritical = - +0.01,194 =-2.345723 2-2.346 P-value = P(t194 <tabs) = P(t194 <-134.460037)=0.000000 0.0000 Conclusion: Since p-value < 0.01 and tobs <tcritical = -2.346, so we reject Ho at 0.01 level of significance Hence, we can conclude that mean for first population is significantly less than mean for second population.

The conclusion in the problem context -

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)

Here, 71 = 106.900, $1 = 1.2000, n1 = 127, 12 = 126.800, S2 = 2.2000, n2 = 127 and associated degrees of freedom no df = = 195 G + n1-1 n2-1 The Standard error is s 1.202 2.202 SE= + + 0.222 ni n2 127 127 The critical value is tcritical = t 1 0.95 ,195 =t0.025,195 = 1.972204 1.972 The margin of error is 2 s 1.202 1.202 ME=t1-0.95 10.95,195 * + =t0.025,195 * + 2.202 127 = 1.972 * + 2.202 127 = 0.438561 20.439 ni V 127 n2 127 a 95 percent confidence interval for the differnce in population means (M1 - U2 ) is s1 s? = 21 - 22 + * t 1-0.960 ,195 ni n2 = 21 32 ME = (106.900 – 126.800) +0.438561 = -19.900 +0.438561 = (-20.338561, -19.461439) (-20.34, -19.46)