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Denote with\({\mu _1}\)the true average compression strength (lb) for srawberry drink and with\({\mu _2}\)for cola drink. The hypotheses of interest are\({H_0}:{\mu _1} - \)\({\mu _2} = 0\)versus\({H_a}:{\mu _1} - {\mu _2} < 0\). The answer should be based on the\(P\)value.

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 calculated is the \(P\) value.

In order to use this statistic the assumption of normally had to be made. There is enough information to calculate the \(t\) value as

\(\begin{array}{l}t = \frac{{\bar x - \bar y - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }} = \frac{{540 - 554}}{{\sqrt {\frac{{{{21}^2}}}{{15}}} + \frac{{{{15}^2}}}{{15}}}} = \frac{{ - 14}}{{\sqrt {29.4 + 15} }}\\ = - 2.1\end{array}\)

and the corresponding degrees of freedom of the statistic can be computed using formula given above

\(\nu = \frac{{{{\left( {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}} = \frac{{{{44.4}^2}}}{{\frac{{{{29.4}^2}}}{{14}} + \frac{{{{15}^2}}}{{14}}}} = \frac{{1971.36}}{{77.81}} = 25.36\)

and round down to the nearest integer to get the degrees of freedom

\(\nu = 25\)

The \(P\) value for the lower tailed test is the probability

\(P = P(T < 2.1) = 0.023\)

where the value can be found in the appendix of the book and statistic \(T\) is students statistic with \(25\) degrees of freedom. Since

\(P = 0.023 < 0.05\)

reject the null hypothesis

From the data you can conclude that extra corbonation of cola results in a higher average compression strength with significance level \(\alpha = 0.05\$ .\)