Question

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 1

Answer:

a) The null hypothesis is H₀ : µ₁ ≥ µ₂

The alternative hypothesis is Hₐ : µ₁ ≤ µ₂ (left tailed)

b) using t-test, t= -2.69

c) the p-value = 0.0041

Step-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 1

Mean (µ₁)= 73.8

Standard deviation (σ₁) = 5.2

n₁ = 60

For machine 2

Mean (µ₂)= 76.1

Standard deviation (σ₂) = 541

n₂ = 60

a) The null hypothesis is H₀ : µ₁ ≥ µ₂

The alternative hypothesis is Hₐ : µ₁ ≤ µ₂ (left tailed)

b) Using t-test

t = (µ₁ - µ₂) / √ (σ₁^2/ n₁) + (σ₂^2/ n₂)

t= (73.8 – 76.1) / √ (5.2^2/ 60 )+ (4.1^2/ 60)

t= (-2.31) / √0.7309

t= -2.31/ 0.8549

t= -2.69

c) The excel function for the p-value = TDIST (2.69, 118, 1)

P-value = 0.0041

Since 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