wo machines used to fill soft drink containers are being compared. The number of containers filledeach minute is counted for 60 minutes for each machine. During the 60 minutes, machine 1 filledan average of 73.8 cans per minute with a standard deviation of 5.2 cans per minute, and machine2 filled an average of 76.1 cans per minute with a standard deviation of 4.1 cans per minute.(a) We want to test whether machine 2 is faster than machine 1. State the null and alternatehypotheses.(b) Compute the relevant statistic.(c) Compute the P-value.

Answer:a) The null hypothesis is H₀ : µ₁ ≥ µ₂The alternative hypothesis is Hₐ : µ₁ ≤ µ₂ (left tailed)b) using t-test, t= -2.69c) the p-value = 0.0041Step-by-step explanation:Testing if the machine 2 is faster than the machine 1 is tested under the null and alternative hypothesis at a 5% level of significance.For machine 1Mean (µ₁)= 73.8Standard deviation (σ₁) = 5.2n₁ = 60For machine 2Mean (µ₂)= 76.1Standard deviation (σ₂) = 541n₂ = 60a) The null hypothesis is H₀ : µ₁ ≥ µ₂The alternative hypothesis is Hₐ : µ₁ ≤ µ₂ (left tailed)b) Using t-testt = (µ₁ - µ₂) / √ (σ₁^2/ n₁) + (σ₂^2/ n₂)t= (73.8 – 76.1) / √ (5.2^2/ 60 )+ (4.1^2/ 60)t= (-2.31) / √0.7309t= -2.31/ 0.8549t= -2.69c) The excel function for the p-value = TDIST (2.69, 118, 1)P-value = 0.0041Since the p-value is less than the 5% significance level, we reject the null hypothesis.The then conclude that machine 2 is faster than machine 1