Question

Which way of dispensing champagne, the traditional vertical method or a tilted beer-like pour,preserves more of the tiny gas bubbles that improve flavor and aroma? The following data was reported in the article “On the Losses of Dissolved \(C{O_2}\) during Champagne Serving” (J. Agr. Food Chem., 2010: 8768–8775)

\(\begin{array}{*{20}{c}}{ Temp \left( {^^\circ C} \right)}&{ Type of Pour }&n&{ Mean (g/L)}&{ SD }\\{18}&{ Traditional }&4&{4.0}&{.5}\\{18}&{ Slanted }&4&{3.7}&{.3}\\{12}&{ Traditional }&4&{3.3}&{.2}\\{12}&{ Slanted }&4&{2.0}&{.3}\\{}&{}&{}&{}&{}\end{array}\)

Assume that the sampled distributions are normal.

a. Carry out a test at significance level \(.01\) to decide whether true average\(C{O_2}\)loss at \(1{8^o}C\) for the traditional pour differs from that for the slanted pour.

b. Repeat the test of hypotheses suggested in (a) for the \(1{2^o}\) temperature. Is the conclusion different from that for the \(1{8^o}\) temperature? Note: The \(1{2^o}\) result was reported in the popular media

Short answer

the solution is

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 \(1{2^^\circ }C\).

Step-by-step solution

given

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 degree of freedom

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 \(1{8^^\circ }C\).

b)

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 degree of freedom

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 \(1{2^^\circ }C\).

conclusion

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 \(1{2^^\circ }C\).