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given

\(\begin{array}{l}{{\bar x}_1} = 4.0\\{{\bar x}_2} = 3.7\\{s_1} = 0.5\\{s_2} = 0.3\end{array}\)

\(\begin{array}{l}{n_1} = {n_2} = 4\\\alpha = 0.01\end{array}\)

Claim differs

Either the null hypothesis or the alternative hypothesis is the assertion. The null hypothesis and alternative hypothesis are diametrically opposed. The value given in the claim must be included in the null hypothesis.

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

determine the test static

\(t = \frac{{{{\bar x}_1} - {{\bar x}_2}}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }} = \frac{{4.0 - 3.7}}{{\sqrt {\frac{{0.{5^2}}}{4} + \frac{{0.{3^2}}}{4}} }} \approx 1.029\)

Determine degree of freedom

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{0.{5^2}}}{4} + \frac{{0.{3^2}}}{4}} \right)}^2}}}{{\frac{{{{\left( {0.{5^2}/4} \right)}^2}}}{{4 - 1}} + \frac{{{{\left( {0.{3^2}/4} \right)}^2}}}{{4 - 1}}}} \approx 4\)

The P-value is the chance of getting the test statistic's result, or a number that is more severe. The P-value in the appendix containing the t-value in the row \(df = 4\) is the number (or interval) in the column title of Student's T distribution:

\(P > 2 \times 0.10 = 0.20\)

The null hypothesis is rejected if the P-value is less than or equal to the significance level:

\(P > 0.01 \Rightarrow Fail to reject {H_0}\)

There is insufficient evidence to support the allegation that the true average \(C{O_2}\) loss for the traditional pour differs from the true average \(C{O_2}\) loss for the traditional pour at \(1{8^^\circ }C\).

given

\(\begin{array}{l}{{\bar x}_1} = 3.3\\{{\bar x}_2} = 2.0\\{s_1} = 0.2\\{s_2} = 0.3\end{array}\)

\(\begin{array}{l}{n_1} = {n_2} = 4\\\alpha = 0.01\end{array}\)

Claim differs

Either the null hypothesis or the alternative hypothesis is the assertion. The null hypothesis and alternative hypothesis are diametrically opposed. The value given in the claim must be included in the null hypothesis.

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

determine the test static

\(t = \frac{{{{\bar x}_1} - {{\bar x}_2}}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }} = \frac{{3.3 - 2.0}}{{\sqrt {\frac{{0.{2^2}}}{4} + \frac{{0.{3^2}}}{4}} }} \approx 7.211\)

Determine degree of freedom

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{0.{2^2}}}{4} + \frac{{0.{3^2}}}{4}} \right)}^2}}}{{\frac{{{{\left( {0.{2^2}/4} \right)}^2}}}{{4 - 1}} + \frac{{{{\left( {0.{3^2}/4} \right)}^2}}}{{4 - 1}}}} \approx 5\)

The P-value is the chance of getting the test statistic's result, or a number that is more severe. The P-value in the appendix containing the t-value in the row \({\rm{d f = 5}}\) is the number (or interval) in the column title of Student's T distribution

\(P < 2 \times 0.0005 = 0.001\)

The null hypothesis is rejected if the P-value is less than or equal to the significance level:

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

There is sufficient evidence to support the allegation that the true average \(C{O_2}\) loss for the traditional pour differs from the true average \(C{O_2}\) loss for the traditional pour at \(1{2^^\circ }C\).

a)There is insufficient evidence to support the allegation that the true average \(C{O_2}\) loss for the traditional pour differs from the true average \(C{O_2}\) loss for the traditional pour at \(1{8^^\circ }C\).

b) There is sufficient evidence to support the allegation that the true average \(C{O_2}\) loss for the traditional pour differs from the true average \(C{O_2}\) loss for the traditional pour at \(1{2^^\circ }C\).