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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.

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

\(\nu = 57\)

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.