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.

33,200	31,880	32,580	26,470	33,270
32,310	32,990	32,000	30,470	32,750
23,000	30,950	32,720	33,640	32,290
24,070	30,200	31,350	28,740	31,890

(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?

H₀: μ = 32,500
H₁: μ < 32,500H₀: μ = 32,500
H₁: μ ≠ 32,500    H₀: μ ≠ 32,500
H₁: μ = 32,500H₀: μ = 32,500
H₁: μ > 32,500

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?

H₀: μ ≠ 31,500
H_a: μ = 31,500H₀: μ = 31,500
H₁: μ < 31,500    H₀: μ = 31,500
H₁: μ > 31,500H₀: μ = 31,500
H₁: μ ≠ 31,500

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

n=20,   = 3000

= 30838.5

a)

= 0.10,  = 32500

null and alternative hypothesis is

Ho:   = 32500

H1:   32500

formula for test statistics is

z = -2.48

test statistics Z = -2.48

calculate P-Value

P-Value = 2 * P(z < -2.48)

using normal z table we get

P(z < -2.48) = 0.0066

P-Value = 2 *   0.0066

P-Value = 0.0132

Decision rule is

Reject Ho if ( P-Value) ( )

here, ( P-Value = 0.0132) ( = 0.10 )

Hence, Null hypothesis is rejected.

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

*************************************************************************************

b)

= 0.10,  = 31500

null and alternative hypothesis is

Ho:   = 31500

H1:   31500

formula for test statistics is

z = -0.99

test statistics Z = -0.99

calculate P-Value

P-Value = 2 * P(z < -0.99)

using normal z table we get

P(z < -0.99) = 0.1611

P-Value = 2 * 0.1611

P-Value = 0.3222

Decision rule is

Reject Ho if ( P-Value) ( )

here, ( P-Value = 0.3222) > ( = 0.10 )

Hence, Null hypothesis is NOT rejected.

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