Question

ints] DETAILS DEVORESTATO 9.E.00 5 sile strength tests were carried out on two different grades of wire rod, resulting in the accompanying data Sample Sample Mean Sample Grade Size (kg/mm) SD AISI 1064 m=127 = 104.9 s. - 1.1 AISI 1078 y = 125.8 s, -2.1 (a) Does the data provide compelling evidence for concluding that true werage strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mma? Test the appropriate hypotheses using a significance level of 0.01. State the relevant hypotheses H-10 H-10 OH--10 = 127 OH--10 H-10 O Here -10 H-10 ON-10 Calculate the test statistie and value. (Round your test statistic to two decimal places and your value to four decimal places) value State the conclusion in the problem contact O Reject. The data does not suggest that the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 km O Fail to reject. The data does not suggest that the true verage strength for the 1078 grade exceeds that for the 1064 grade by more than 10 ! O Fail to reject H, The data suggests that the true werage strength for the 1078 grade exceeds that for the 1064 grade by more than to O Reject. The data suggests that the true weruge strength for the 1078 grade exceeds that for the 106 grade by more than 10 km () Estimate the difference between true werage strengths for the two grades is a way that provides information about precies de confidence interval. Round your answers to two decimal places) You may need to use the appropriate table is the Appendix of Tales to answer this questi DETALS DEVORESTATS SE022 NVA 7. 1-/12 Points A structor has given a short qui consisting of the points and on the second part super hit

Answer 1

a) The correct hypothesis is option 1 that is

$H_{0}:\mu _{1064}-\mu _{1078}=-10 vs H_{a}:\mu _{1064}-\mu _{1078}<-10$

b) We have to calculate the test statistic and p-vale for which we run the given steps in minitab-

Steps for Fisher’s t test-

1. Enter two series of observations in a two columns.

2. Click on “Stat” then “Basic statistics” then “2-sample t”

3. Check “sumarrized data” and write the given values as for sample 1: sample size=127,sample mean = 104.9 and sample standard deviation = 1.1 and. for sample 2: sample size=127,sample mean = 125.8 and sample standard deviation = 2.1

4. Select “assume equal variances”.

5. Click on “options” and set confidence level = 99.0, “hypothesized difference”=0.0 and alternative as “less than” and click OK.

6. Click OK.

Thus we get the values-

z = -51.82 and p-value = 0.000

c) We have to state the conclusion in the problem context.

We know if p-value<=0.01 then the null is rejected. Here, p-value<0.01 so we reject the null and conclude that option 4 is correct that is 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.

d) we have to estimate the difference between the true average strength for the two grades in way that provides information about decision and reliability that is we have to find the 95% confidence interval.

We run the above given steps for 95% confidence level and alternative as not equal.

Thus we get the confidence interval as (-21.314, -20.486).