The image displays text related to a statistical analysis of CO2 loss in champagne based on different pouring methods and temperatures. The study compares traditional vertical pouring with a slanted, beer-like pouring method.**Table:**- **Columns:** - Temperature (18°C and 12°C) - Type of Pour (Traditional and Slanted) - Mean CO2 loss (g/L) - Standard Deviation (SD)**18°C:**- Traditional pour: Mean 4.0, SD 0.6- Slanted pour: Mean 3.7, SD 0.4**12°C:**- Traditional pour: Mean 3.4, SD 0.2- Slanted pour: Mean 2.0, SD 0.1Assumption: Normal distribution of the sample means.**Part (a):**Task: Perform a t-test at the 0.01 significance level to determine if there is a significant difference in average CO2 loss at 18°C between the two pouring methods. Hypotheses are stated for the comparison, with specific notations for traditional and slanted pours. Instructions are provided to compute the t-statistic and P-value, and choose between rejection or acceptance of the null hypothesis based on the results.**Part (b):**Task: Repeat the hypothesis test for the 12°C condition. Similar steps for formulating hypotheses, calculating the test statistic, and interpreting the conclusion are outlined as in Part (a).Educational Note: The document serves as a practical example for students learning hypothesis testing, showcasing how statistical tests can inform about differences in experimental conditions.

Given the data:Temp (°C)Type of PournMean…

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 an article. Temp (°C) Type of Pour n Mean (g/L) SD 18 0.6 4.0 3.7 18 0.4 Traditional 4 Slanted 4 Traditional 4 Slanted 4 12 0.2 12 0.1 Assume that the sampled distributions are normal. LAUSE SALT (a) Carry out a test at significance level 0.01 to decide whether true average CO, loss at 18°C for the traditional pour differs from that for the slanted pour. (Use μ, for the traditional pour and μ, for the slanted pour) State the relevant hypotheses. OH₂H₁ H₂ <0 H₂H₁-H₂ = 0 OH H₁ H₂>0 H₂H₁-H₂=0 OH₂H₁-H₂=0 OH₁¹ H₁ H₂ = 0 H1 H₂-H₂>0 OH₂H₁-H₂=0 H₂H₁ H₂ <0 Calculate the test statistic and P-value. (Round your test statistic to one decimal place and your P-value to three decimal places.) 3.4 2.0 P-value= State the conclusion in the problem context. O Reject H. The data suggest that the true average CO₂ loss at 18°C for the traditional pour differs from that for the slanted pour. O Fail to reject H. The data suggest that the true average CO, loss at 18°C for the traditional pour differs from that for the slanted pour. O Reject H. The data do not suggest that the true average CO₂ loss at 18°C for the traditional pour differs from that for the slanted pour. O Fail to reject H. The data do not suggest that the true average CO, loss at 18°C for the traditional pour differs from that for the slanted pour. (b) Repeat the test of hypotheses suggested in (a) for the 12° temperature. Is the conclusion different from that for the 18° temperature? Note: The 12° result was reported in the popular media. (Use μ, for the traditional pour and for the slanted pour.) H₂ State the relevant hypotheses. OHH₁ H₂ <0 OH₂H₁-H₂=0 OHH₁ H₂O H₂H₁-H₂=0 O H₁¹ H₁-H₂ = 0 H₂H₂-H₂ <0 OH₁¹ H₁ H₂ = 0 H₂H₁ H₂ #0 Calculate the test statistic and P-value. (Round your test statistic to one decimal place and your P-value to three decimal places.) P-value= State the conclusion in the problem context. O Reject H. The data do not suggest that the true average CO, loss at 12°C for the traditional pour differs from that for the slanted pour. O Reject H. The data suggest that the true average CO₂ loss at 12°C for the traditional pour differs from that for the slanted pour. O Fail to reject H. The data suggest that the true average CO₂ loss at 12°C for the traditional pour differs from that for the slanted pour. O Fail to reject H.. The data do not suggest that the true average CO, loss at 12°C for the traditional pour differs from that for the slanted pour.