Question

Tensile strength tests were carried out on two different grades of wire rod, resulting in the accompanying data. Sample Size Sample SD Grade Sample Mean (kg/mm2) = 109.4 y = 121.2 AISI 1064 m = 126 $1 = 1.2 AISI 1078 n = 126 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. O Ho: 41064 – 41078 = -10 Ha: 41064 – 41078 % -10 O Ho: M1064 - M1078 = -10 Ha: M1064 - 41078 > -10 O Ho: 41064 – 41078 = -10 Ha: M1064 - 41078 * - 10 O Ho: M1064 – 41078 = -10 Ha: M1064 – 41078 < -10 O Ho: 41064 – 41078 = -10 Ha: M1064 – 41078 2-10 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. O 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. O 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 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 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

Answer:

Given that,

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

m=126, =109.4 and s1=1.2

T

n=126, $\bar{y}$ =121.2, nad s2=2.1

(a).

To test the hypothesis:

$H_0:\mu _{1064}-\mu _{1078}=-10$

H1 : 41064 – 1078 < -10(Claim

Test statistics,

$Z=\frac{(\bar{x}-\bar{y})-(-10)}{\sqrt{\frac{s_1^2}{m}+\frac{s_2^2}{n}}}$

$=\frac{(109.4-121.2)-(-10)}{\sqrt{\frac{1.2^2}{126}+\frac{2.1^2}{126}}}$

=-1.8/0.2155

Z=-8.3527

The p-value =P(Z < -8.3527), using Z-table

The P-Value is < 0.00001.

The result is significant at p < 0.01.Reject H0.

Answer: Option (c)

(b).

The 95% confidence interval for $(\mu _1-\mu _2)$ ,

$\Rightarrow \left [ (\bar{x}-\bar{y})\pm Z_{\alpha /2} \sqrt{\frac{s_1^2}{m}+\frac{s_2^2}{n}}\right ]$

$\Rightarrow \left [ (109.4-121.2)\pm 1.965\times \sqrt{\frac{1.2^2}{126}+\frac{2.1^2}{126}}\right ]$

$\Rightarrow \left [ -11.8\pm 1.965\times 0.2155\right ]$

$\Rightarrow \left [ -11.8\pm 0.4235\right ]$

=[-11.8-0.4235, -11.8+0.4235]

=[-12.2235, -11.3765]

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