Question

Please help with BOTH

1)

Tensile strength tests were carried out on two different grades of wire rod, resulting in the accompanying data. Grade AISI 1064 AISI 1078 Sample Size m = 129 n = 129 Sample Mean (kg/mm) x= 109.3 y = 129.9 Sample SD 51 = 1.1 52 = 2.0 (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/mm? Test the appropriate hypotheses using a significance level of 0.01. State the relevant hypotheses. O Ho: 41064 -41078 = -10 H: 1064 – 41078 < -10 Ho: 41064 – 41078 = -10 HA: 41064 – 41078 Ho: 41064 - 41078 = -10 H: 41064 – 41078 > -10 $-10 O Ho: 41064 - 41078 = -10 H: 1064 – 41078 2-10 O Ho: 41064 – 41078 = -10 H: U1064 – 41078 + -10 Calculate the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) P-value = State the conclusion in the problem context. Reject Ho. 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. 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. Fail to 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. Fail to reject Ho. The data does not suggest that the true average strength for the 1078 grade exceeds that for the 1054 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.) (C ) kg/mm²

2)

Answer 1

Solution-1: AISI 1064: 129 пу 109.3 S1 1.1 AISI 1078: 129 п2 129.9 X2 = 2.0 S2 Part-A: Null Hypothesis: Alternative Hypothesis: Ha: 1064H1078-10 Ho: 1064 H1078 =-10 Test Statistic: [ 2)-(1064 H1078) $1 п1 П2 [109.3 129.9 (-10) t 1.12 129 t-52.75 2.02 129 Degrees of Freedom:

S2 + n2 п df 2 ,2 1 1 + n2 S1 1 пу - 1(п 2 2.02 1.12 + 129 129 df (2.0 129 1 129 198 (Rounding down) 1.12 + 129 1 129 1 1 df - 198.9 P-value: From t-distribution table, the area in the lower tail to the left of (t Hence, p-value 0.0000 -52.75) and for (df 198), is given by 0.0000 Level of significance a 0.01 Since p-value < a, we reject the null hypothesis. Conclusion: 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 Part-B: For 95% confidence interval, a/2 0.05/2 =0.025 From t-distribution table, the value of t having an area of (a/2 = 0.025) in its upper tail and for (df 198) is given by to.025 1.972 Margin of Error: MoE to.025 X + П2 1.12 2.02 MoE 1.972 x + 129 129

MoE 0.396 95% confidence interval for the difference between true average strengths for the two grades (u1064 (109.3 129.9) t 0.396 20.6t 0.396 1078) Lower limit of 95% confidence interval -20.6 - 0.396 -21.00 Upper limit of 95% confidence interval -20.6 0.396 -20.20 Hence, 95% confidence interval = (-21.00, -20.20) Solution-2: Part-A: 4.0, s2 6.0 т 3 10, п — 10, s, %3D Degrees of Freedom: 2. т п df 2 2 1 1 m-1 т п 2 6.02 4.02 10 10 df 2 (4.02 10 (6.02 10 1 10 df 15.7 15 (Rounding down) 1 1 10 1 Part-B: 18, s1 = 4.0, s2 = 6.0 10, n m

Degrees of Freedom: 2 2 S2 т п df 2 2 1 1 1 1m т п п 4.02 10 6.02 18 df 6.02 18 1 18 24.9 24 (Rounding down) 4.02 10 1 10 df Part-C: m 10, n 18, s1 3.0, s2 = 6,0 Degrees of Freedom: 2 т п df 2 2 (s2 1 (s2 1 1 + 1 т т п (3.02 10 6.02 18 df 2 (3.02 1 10 =25.8 25 (Rounding down) 1 1 6.02 18 1 10 18 df Part-D: 21, s1 4.0, S2 12, n 6.0 m=

Degrees of Freedom: 2 2 т п df 2 .2 Si 2 ) 1 1 1 1 т т п п 6.02 12 21 4.02 df 2 2 6.02 4.02 1 1 21 1218 12 1 12 1 30 (Rounding down) df 30.1