laboratory exercises

Question

How strong a force (in pounds) is needed to pull apart pieces of wood 4 inches long and 1.5 inches square? The following are data from students performing a comparable laboratory exercise. Suppose that the strength of pieces of wood like these follow a Normal distribution with standard deviation 3000 pounds.

33,140	31,910	32,560	26,480	33,320
32,280	32,980	31,990	30,460	32,700
23,050	30,980	32,680	33,680	32,310
24,080	30,190	31,310	28,680	31,940

(a) We are interested in statistical significant evidence at the 𝛼 = 0.10 level for a claim that the mean is 32,500 pounds.

What are the null and alternative hypotheses?

What is the value of the test statistic. (Round your answer to two decimal places.)

z =

What is the P-value of the test? (Round your answer to four decimal places.)

P-value = 

What is your conclusion?

There is enough evidence to conclude that the wood's mean strength differs from 32,500 pounds.

There is not enough evidence to conclude that the wood's mean strength differs from 32,500 pounds.    

(b) We are interested in statistical significant evidence at the 𝛼 = 0.10

 level for a claim that the mean is 31,500 pounds.

What are the null and alternative hypotheses?

What is the value of the test statistic. (Round your answer to two decimal places.)

z = 

What is the P-value of the test? (Round your answer to four decimal places.)

P-value = 

What is your conclusion?

There is enough evidence to conclude that the wood's mean strength differs from 31,500 pounds.

There is not enough evidence to conclude that the wood's mean strength differs from 31,500 pounds.    

Answer 1

Part A

Null Hypothesis (H₀): μ=32500

Alternate Hypothesis (H_a): μ≠32500

Test Statistic

Z = (x̄-μ)/(σ/√n) = (30836-32500)/(3000/√20) ≈ -2.48

P-Value

P(Z<-2.48) + P(Z>2.48) = 2*P(Z<-2.48) = 2(0.0066) = 0.0132

Given our significance level of α=0.10, since 0.0132<0.10, we reject the null hypothesis. This means there is enough evidence to conclude that the wood's mean strength differs from 32,500 pounds.

Part B

Null Hypothesis (H₀): μ=31500

Alternate Hypothesis (H_a): μ≠31500

Test Statistic

Z = (x̄-μ)/(σ/√n) = (30836-31500)/(3000/√20) ≈ -0.99

P-Value

P(Z<-0.99) + P(Z>0.99) = 2*P(Z<-0.99) = 2(0.1611) = 0.3222

Given our significance level of α=0.10, since 0.3222>0.10, we fail to reject the null hypothesis. This means there is not enough evidence to conclude that the wood's mean strength differs from 31,500 pounds.

Hope the answers and explanations helped!