Question

Tensile-strength tests were carried out on two different grades of wire rod (“Fluidized Bed Patenting of Wire Rods,” Wire J., June \(1977: 56 - 61)\), resulting in the accompanying data

Sample

Sample Mean Sample

Grade Size (kg/mm2 ) SD

\(\overline {\underline {\begin{array}{*{20}{l}}{ AISI 1064}&{m = 129}&{\bar x = 107.6}&{{s_1} = 1.3}\\{ AISI 1078}&{n = 129}&{\bar y = 123.6}&{{s_2} = 2.0}\\{}&{}&{}&{}\end{array}} } \)

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 \(10kg/m{m^2}\) ? Test the appropriate hypotheses using a significance level of \(.01\).

b. Estimate the difference between true average strengths for the two grades in a way that provides information about precision and reliability

Short answer

the solution is

a)There is adequate evidence to support the allegation that the genuine average strength of the \(1078\)grade is more than \(10kg/m{m^2}\) more than that of the \(1064\) grade.

b)\(( - 16.4116, - 15.5884)\)

Step-by-step solution

test the appropriate hypotheses

1. given:

\(\begin{array}{l}{{\bar x}_1} = 107.6\\{{\bar x}_2} = 123.6\\{s_1} = 1.3\\{s_2} = 2.0\end{array}\)

\(\begin{array}{l}{n_1} = 129\\{n_2} = 129\\\alpha = 0.01\end{array}\)

We may use the \(z\)-test because the samples are large \(\left( {n > 30} \right)\). (instead of a t-test).

Assumption: exceed by more than \(10\)

Either the null hypothesis or the alternative hypothesis is asserted. The null hypothesis and the alternative hypothesis are diametrically opposed. An equality must be included in the null hypothesis.

\(\begin{array}{l}{H_0}:{\mu _1} - {\mu _2} = - 10\\{H_a}:{\mu _1} - {\mu _2} > - 10\end{array}\)

Determine the value of the test statistic:

\(z = \frac{{{{\bar x}_1} - {{\bar x}_2} - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }} = \frac{{107.6 - 123.6 - ( - 10)}}{{\sqrt {\frac{{1.{3^2}}}{{129}} + \frac{{2.{0^2}}}{{129}}} }} \approx - 28.57\)

If the null hypothesis is true, the \(P\)-value is the probability of getting a result more extreme or equal to the standardized test statistic \(z\). Using the normal probability table, determine the probability.

\(P = P(Z < - 28.57 or Z > 28.57) = 2P(Z < - 28.57) \approx 2(0) = 0\)

The null hypothesis is rejected if the P-value is less than the alpha significance level.

\(P < 0.01 \Rightarrow Reject {H_0}\)

There is adequate evidence to support the allegation that the genuine average strength of the \(1078\)grade is more than \(10kg/m{m^2}\) more than that of the \(1064\) grade.

given

b)

\(\begin{array}{l}{{\bar x}_1} = 107.6\\{{\bar x}_2} = 123.6\\{s_1} = 1.3\\{s_2} = 2.0\end{array}\)

\(\begin{array}{l}{n_1} = 129\\{n_2} = 129\end{array}\)

We will compute a \(95\% \) confidence interval, confidence intervals with other confidence levels can be calculated similarly.

\(c = 95\% = 0.95\)

confidence interval

for confidence interval\(1 - \alpha = 0.95,\) determine \({z_{\alpha /2}} = {z_{0.025}}\) using the normal probability table ( look up \(0.025\) in the table, the \(z\)-score is then the found \(z\)-score with opposite sign):

\({z_{\alpha /2}} = 1.96\)

The margin of error then becomes:

\({z_{\alpha /2}} = 1.96E = {z_{\alpha /2}} \times \sqrt {\frac{{\sigma _1^2}}{{{n_1}}} + \frac{{\sigma _2^2}}{{{n_2}}}} = 1.96 \times \sqrt {\frac{{1.{3^2}}}{{129}} + \frac{{2.{0^2}}}{{129}}} \)\( \approx 0.4116\)

The endpoints of the confidence interval of \({\mu _1} - {\mu _2}\) are:

\(\begin{array}{l}\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E = (107.6 - 123.6) - 0.4116 = - 16 - 0.4116 = - 16.4116\\\left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E = (107.6 - 123.6) + 0.4116 = - 16 + 0.4116 = - 15.5884\end{array}\)

conclusion

a)There is adequate evidence to support the allegation that the genuine average strength of the \(1078\)grade is more than \(10kg/m{m^2}\) more than that of the \(1064\) grade.

b)\(( - 16.4116, - 15.5884)\)