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.

33,210 / 31,910 / 32,580 / 26,470 / 33,270 / 32,280 / 32,980 / 32,070 / 30,480 / 32,700 / 23,000 / 30,960 / 32,670 / 33,690 / 32,290 / 24,000 / 30,220 / 31,280 / 28,690 /

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? -H0: μ ≠ 32,500 H1: μ = 32,500

H0: μ = 32,500 H1: μ > 32,500

-H0: μ = 32,500 H1: μ ≠ 32,500

-H0: 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 H1: μ ≠ 31,500

H0: μ = 31,500 H1: μ > 31,500

H0: μ = 31,500 H1: μ < 31,500

H0: μ ≠ 31,500 Ha: 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

A) Null and alternative hypothesis is

Ho: mu=32500 Ha: mu 32500

Test statistic

Z = (xbar -32500)/ (sigma/sqrt(n))

Where xbar is sample mean and sigma is sample standard deviation and n is sample size.

Z= (30834.5-32500) / (3033.11/sqrt(20))

= -1665.5/ 678.55

= -2.45

P value for 2 tailed z test with z value -2.45 = 0.0143

Since p value is less than alpha = 0.10, we reject Ho

Thus, there is enough evidence to conclude that the mean strength differs from 32500

B) Null and alternative hypothesis is

Ho: mu=31500 Ha: mu 31500

Test statistic

Z = (30834.5-31500)/(3033.11/sqrt(20))

= -665.5/678.55

= -0.98

P value of 2 tailed a test with z= -0.98 is 0.3271

Since p value is greater than alpha = 0.1, we fail to reject Ho

There is not enough evidence to conclude that wood's mean strength differs from 31500