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 4 0.5 18 Traditional Slanted Traditional 4 12 4 12 Slanted 4 Assume that the sampled distributions are normal. USE SALT P-value = 4.0 3.7 3.4 2.0 0.4 0.2 0.3 (a) Carry out a test at significance level 0.01 to decide whether true average CO₂ loss at 18°C for the traditional pour differs from that for the slanted pour. (Use ₁ for the traditional pour and ₂ for the slanted pour.) State the relevant hypotheses. о но H1 - H2=0 H₂: H₁ - H₂> 0 о но H1-12=0 Hai H1 - H₂ #0 ọ Hoi Hy Hy 20 Ha: H1 - H2 - 0 Ọ Hoi Hy Hy = 0 На Н2 - H2 ко Hoi H₁ H₂ <0 Hải Hy Hy = 0 Calculate the test statistic and P-value. (Round your test statistic to one decimal place and your P-value to three decimal places.) State the conclusion in the problem context. O Reject Ho. 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 Reject Ho. 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 Ho. 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 Ho. 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.
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 4 0.5 18 Traditional Slanted Traditional 4 12 4 12 Slanted 4 Assume that the sampled distributions are normal. USE SALT P-value = 4.0 3.7 3.4 2.0 0.4 0.2 0.3 (a) Carry out a test at significance level 0.01 to decide whether true average CO₂ loss at 18°C for the traditional pour differs from that for the slanted pour. (Use ₁ for the traditional pour and ₂ for the slanted pour.) State the relevant hypotheses. о но H1 - H2=0 H₂: H₁ - H₂> 0 о но H1-12=0 Hai H1 - H₂ #0 ọ Hoi Hy Hy 20 Ha: H1 - H2 - 0 Ọ Hoi Hy Hy = 0 На Н2 - H2 ко Hoi H₁ H₂ <0 Hải Hy Hy = 0 Calculate the test statistic and P-value. (Round your test statistic to one decimal place and your P-value to three decimal places.) State the conclusion in the problem context. O Reject Ho. 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 Reject Ho. 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 Ho. 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 Ho. 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.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
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![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.
Type of Pour n Mean (g/L) SD
Traditional 4
0.5
4
0.4
Slanted
Traditional
Slanted
4
0.2
4
0.3
Assume that the sampled distributions are normal.
Temp (°C)
18
18
12
12
USE SALT
(a) Carry out a test at significance level 0.01 to decide whether true average CO₂ loss at 18°C for the traditional pour differs from that for the slanted pour. (Use μ₁ for the traditional pour and
State the relevant hypotheses.
оно ну-H2=0
Hai H₁ - H₂ > 0
ⒸH₁: H₁-H₂ = 0
Ha Hy Hy #0
O Ho: H₁ - H₂>O
На: Н1 - H2 = 0
оно: ну-H2=0
На: H1 -H2 ко
4.0
3.7
3.4
2.0
о но ну-нако
Hai H₁ - 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.)
t =
P-value =
State the conclusion in the problem context.
O Reject Ho. 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 Reject Ho. 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 Ho. 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 Ho. 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.
H₂ for the slanted pour.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdc9c7494-1a96-4ff8-a6c5-87955a5c8cb7%2F5fe1db34-a44c-43d4-aebb-e785d4625bac%2Fs90c175_processed.png&w=3840&q=75)
Transcribed Image Text: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.
Type of Pour n Mean (g/L) SD
Traditional 4
0.5
4
0.4
Slanted
Traditional
Slanted
4
0.2
4
0.3
Assume that the sampled distributions are normal.
Temp (°C)
18
18
12
12
USE SALT
(a) Carry out a test at significance level 0.01 to decide whether true average CO₂ loss at 18°C for the traditional pour differs from that for the slanted pour. (Use μ₁ for the traditional pour and
State the relevant hypotheses.
оно ну-H2=0
Hai H₁ - H₂ > 0
ⒸH₁: H₁-H₂ = 0
Ha Hy Hy #0
O Ho: H₁ - H₂>O
На: Н1 - H2 = 0
оно: ну-H2=0
На: H1 -H2 ко
4.0
3.7
3.4
2.0
о но ну-нако
Hai H₁ - 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.)
t =
P-value =
State the conclusion in the problem context.
O Reject Ho. 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 Reject Ho. 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 Ho. 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 Ho. 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.
H₂ for the slanted pour.)
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