Question

Please circle all final answers and indicate which portion you are working on. Will rate for correct answers. Please show all work. Thank you!

10.2.3 GO Tutorial The diameter of steel rods manufactured on two different extrusion machines is being investigated. Two random samples of sizes nj = 15 and n2 = 17 are selected, and the sample means and sample variances are Ij = 8.70, sỉ = 0.35, 72 = 8.68, sź = 0.40, respectively. Assume that o = oz and that the data are drawn from a normal distribution. (a) Is there evidence to support the claim that the two machines produce rods with different mean diameters? Give bounds on the P-value used to make your conclusion. Use only Table V of Appendix A. P-value Conclusion: Ho LINK TO TEXT (b) Construct a 95% confidence interval for the difference in mean rod diameter. Use only Table V of Appendix A. Round your answer to 3 decimal places. LINK TO TEXT (c) Which of the following is a correct interpretation of this interval? O We are 95% confident that the true difference in mean diameters from the two machines is between -0.424 and 0.464. We are 95% confident that the true difference in mean diameters from the two machines is 0. Since O is contained in the interval, there is a 95% probability that the true difference in mean diameters from the two machines is 0. There is a 95% probability that the interval contains the true difference in mean diameters from the two machines. LINK TO TEXT

Answer 1

(a)To check whether diameter of two steel rods manufactured on different extrusion machines are same on not, we have taken two sample ,and data given are

For Sample 1

Sample size = n1 = 15

Sample mean = X1 = 8.70

Sample standard deviation = s1 = 0.592

For sample 2

Sample size = n2 = 17

Sample mean = X2 = 8.68

Sample standard deviation = s2 = 0.632

Null and alternative hypotheses need to be tested, where μ1 ​ is mean diameter of 1st rod and μ2 is mean diameter of 2nd rod,​

Ho: μ1 ​ = μ2​

Ha: μ1​ ≠ μ2​

This corresponds to a two-tailed test.

At the significance level is α = 0.05, and the degrees of freedom is

df = n1+n2-2 = 15+17-2 = 30

The degrees of freedom are computed assuming that the population variances are equal(as given in data)

The critical value for this two-tailed test is tc​=2.042, for α=0.05 and degree of freedom is 30 using excel formula =T.INV.2T(0.05,30).

tc(lower) = -2.042

tc(upper) = 2.042

The rejection region for this two-tailed test is R = t < -2.042 & t > 2.042

Test statistics

t = X1, X2 (nl-1)312+(n2-1):22 nl+n2-2 G+ m2)

$t = \frac{8.70-8.68}{\sqrt{\frac{(15-1)0.35+(17-1)0.40}{15+17-2}\left ( \frac{1}{15}+\frac{1}{17} \right )}}$

It is observed that t(test) = 0.092 is in non rejection region -2.042<t​<2.042, it is then concluded that the null hypothesis is not rejected.

p value calculated as p = 0.9273, using excel formula =T.DIST.2T(0.092,30)

p = 0.9273

since p-value is p = 0.9273 > 0.05, it is concluded that the null hypothesis is not rejected.
Conclusion

Ho - Not rejected

(b)We need to construct the 95% confidence interval for the difference between the population means μ1​−μ2

For that pooled standard deviation calculated as

Sp (nl - 1)s12 + (n2 – 1) s22 nl + n2 – 2

Sp (15 – 1)0.35+ (17 – 10.40 15 + 17 – 2

Now calculation for standard error (population variance is equal)

$s_{e} = s_{p}\sqrt{\frac{1}{n1}+\frac{1}{n2}}$

$s_{e} = 0.614\sqrt{\frac{1}{15}+\frac{1}{17}}$

Confidence interval calculated as

$CI = \left ( X1-X2 -tc*s_{e},X1-X2 +tc*s_{e}\right )$

$CI = \left ( 8.70-8.68 -2.042*0.217,8.70-8.68 +2.042*0.217\right )$$CI = \left ( -0.424,0.464\right )$

(c)So option 1st is correct

We are 95% confident that true difference in two mean diameters if two machines is between -0.424 and 0.464