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.

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

y=mx+

f(2x−x2)=−3x2+6x−1

step-by-step explanation:

set up the composite function and evaluate.

f(2x−x2)=−3x2+6x−1