I need help with all parts of this queestion 6, the select options for C is greater thna or less than, and is or is not (a) Do the data in the table present sufficient evidence to indicate a difference in the average prices of light tuna in water versus oil? Test using a = 0.05.State the null and alternative hypotheses.O Ho: (M₁M₂) = 0 versus H₂: (M₁ - H₂) 0State the test statistic.t =State the rejection region. (If the test is one-tailed, enter NONE for the unused region. Round your answers to three places.)tState the conclusion.O Ho is rejected. There is insufficient evidence to conclude that there is a difference in the average prices of light tuna in water versus oil.O Ho is not rejected. There is sufficient evidence to conclude that there is a difference in the average prices of light tuna in water versus oil.O Ho is not rejected. There is insufficient evidence to conclude that there is a difference in the average prices of light tuna in water versus oil.O Ho is rejected. There is sufficient evidence to conclude that there is a difference in the average prices of light tuna in water versus oil.(b) What is the p-value for the test?p-value =(c) The MINITAB analysis uses the pooled estimate of o². Is the assumption of equal variances reasonable? Why or why not? (Round your answer to three decimal places.)The ratio of the larger of the two sample variances to the smaller iswhich is ---Select--than 3, so the assumption of equal variances ---Select--- reasonable. Data on the estimated average price for a variety of different brands of tuna, based on prices paid nationally in supermarkets, are shown in the following table. Use the MINITAB printout to answer the questions in parts (a)-(c). (Use the exact values found in theMINITAB output.)Light Tunain Water0.97 0.511.94 1.39 1.94 0.681.21 1.14 1.32 0.600.83 0.61 1.77 0.630.67 0.620.67 0.69 1.21 0.620.58 0.68Light Tunain Oil2.54 0.64SampleLight WaterLight OilTwo-Sample T-Test and CI: Light Water, Light OilDescriptive StatisticsN14Difference-0.256Estimation for DifferenceT-Value-1.180.6911TestNull hypothesisPooled StDev0.536Alternative hypothesisMean0.8941.149DF23StDev0.4020.672H₂ M₁H₂: M₂95% CI fordifference(-0.703, 0.192)M₂ = 0M₂ #0SE Mean0.110.20P-Value0.249

Question

I need help with all parts of this queestion 6, the select options for C is greater thna or less than, and is or is not (a) Do the data in the table present sufficient evidence to indicate a difference in the average prices of light tuna in water versus oil? Test using a = 0.05.State the null and alternative hypotheses.O Ho: (M₁M₂) = 0 versus H₂: (M₁ - H₂) <0O Ho: (M₁M₂) < 0 versus H₂: (M₁-M₂) >0O Ho: (M₁M₂) = 0 versus H₂: (μ₁ −μ₂) # 0O Ho: (M₁M₂) 0 versus H₂: (μ₁ −μ₂) = 0O Ho: (M₁M₂) = 0 versus H₂: (M₁M₂) >0State the test statistic.t =State the rejection region. (If the test is one-tailed, enter NONE for the unused region. Round your answers to three places.)t>State the conclusion.O Ho is rejected. There is insufficient evidence to conclude that there is a difference in the average prices of light tuna in water versus oil.O Ho is not rejected. There is sufficient evidence to conclude that there is a difference in the average prices of light tuna in water versus oil.O Ho is not rejected. There is insufficient evidence to conclude that there is a difference in the average prices of light tuna in water versus oil.O Ho is rejected. There is sufficient evidence to conclude that there is a difference in the average prices of light tuna in water versus oil.(b) What is the p-value for the test?p-value =(c) The MINITAB analysis uses the pooled estimate of o². Is the assumption of equal variances reasonable? Why or why not? (Round your answer to three decimal places.)The ratio of the larger of the two sample variances to the smaller iswhich is ---Select--than 3, so the assumption of equal variances ---Select--- reasonable. Data on the estimated average price for a variety of different brands of tuna, based on prices paid nationally in supermarkets, are shown in the following table. Use the MINITAB printout to answer the questions in parts (a)-(c). (Use the exact values found in theMINITAB output.)Light Tunain Water0.97 0.511.94 1.39 1.94 0.681.21 1.14 1.32 0.600.83 0.61 1.77 0.630.67 0.620.67 0.69 1.21 0.620.58 0.68Light Tunain Oil2.54 0.64SampleLight WaterLight OilTwo-Sample T-Test and CI: Light Water, Light OilDescriptive StatisticsN14Difference-0.256Estimation for DifferenceT-Value-1.180.6911TestNull hypothesisPooled StDev0.536Alternative hypothesisMean0.8941.149DF23StDev0.4020.672H₂ M₁H₂: M₂95% CI fordifference(-0.703, 0.192)M₂ = 0M₂ #0SE Mean0.110.20P-Value0.249

Accepted Answer

Hello! As you have posted more than 3 sub parts, we are answering the first 3 sub-parts.  In case…