Question

Head ability is the ability of a cylindrical piece of material to be shaped into the head of a bolt, screw, or other cold-formed part without cracking. The article "New Methods for Assessing Cold Heading Quality" (Wire J. Intl., Oct. 1996: 66-72) described the result of a head ability impact test applied to 30 specimens of aluminum killed steel and 30 specimens of silicon killed steel. The sample mean head ability rating number for the steel specimens was 6.43, and the sample mean for aluminum specimens was 7.09. Suppose that the sample standard deviations were 1.08 and 1.19, respectively. Do you agree with the article's authors that the difference in head ability ratings is significant at the 5% level (assuming that the two head ability distributions are normal)?

Short answer

Reject null hypothesis. Agree with the author of the article.

Step-by-step solution

To find the article's authors that the difference in head ability ratings is significant at the 5% level 

Assumption that two head ability distributions are normal makes two-sample t test best for this case.

Given two normal distributions, the random variable (standardized)

\(T = \frac{{\bar X - \bar Y - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{S_1^2}}{m} + \frac{{S_2^2}}{n}} }}\)

has approximately students t distribution with degrees of freedom\(\nu ,\)where\(\nu \)is

\(\nu = \frac{{{{\left( {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}}\)

has to be rounded down to the nearest integer.

The two-sample t test for testing\({H_0}:{\mu _1} - {\mu _2} = {\Delta _0}\) uses the following value of test statistic

\(t = \frac{{\bar x - \bar y - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }}\)

For the adequate alternative hypothesis the adequate area under the \({t_\nu }\) curve is the P value.

To find the article's authors that the difference in head ability ratings is significant at the 5% level 

The hypotheses of interest are\({H_0}:{\mu _1} - {\mu _2} = 0\) versus\({H_a}:{\mu _1} - {\mu _2} \ne 0\), where\({\mu _1}\), and\({\mu _2}\)are the true mean head ability rating for aluminum killed steel specimens and the true mean for silicon killed steel, respectively. For given values

\(\begin{array}{l}\bar x = 6.43;{s_1} = 1.08;m = 30 \\\bar y = 7.09;{s_2} = 1.19;n = 30\end{array}\)

The t statistic value is

\(t = \frac{{\bar x - \bar y - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }} = \frac{{6.43 - 7.09 - 0}}{{\sqrt {\frac{{1.0{8^2}}}{{30}} + \frac{{1.1{9^2}}}{{30}}} }} = \frac{{ - 0.66}}{{\sqrt {0.03888 + 0.0472} }} = - 2.25\)

To compute the P value, you need to compute the degrees of freedom first

\(\nu = \frac{{{{\left( {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}} = \frac{{{{(0.03888 + 0.0472)}^2}}}{{\frac{{{{(0.03888)}^2}}}{{29}} + \frac{{{{(0.0472)}^2}}}{{29}}}} = 57.5\)

and the nearest round down integer is 57 , thus

\(\nu = 57\)

To find the article's authors that the difference in head ability ratings is significant at the 5% level 

The alternative hypothesis is\({H_a}:{\mu _1} - {\mu _2} \ne 0\), this the P value is two times the area under the\({t_{57}}\)curve to the right of |t|. Thus

\(P = 2 \times P(T > 2.25) = 2 \times 0.014 = 0.028\)

where T is the students statistic with 57 degrees of freedom, and the value was computed by a software (you could have taken the approximate value from the table - estimate it).

The P value is

\(P = 0.028 < 0.05 = \alpha \)

thus

reject null hypothesis

Yes, you should agree with the author of the article.