An agronomist examines the cellulose content of a variety of alfalfa hay. Suppos

Question

An agronomist examines the cellulose content of a variety of alfalfa hay. Suppose that the cellulose content in the population has standard deviation = 9 milligrams per gram (mg/g). A sample of 16 cuttings has mean cellulose content x = 145 mg/g.

(a) Give a 90% confidence interval for the mean cellulose content in the population. (Round your answers to two decimal places.)

(b) A previous study claimed that the mean cellulose content was = 140 mg/g, but the agronomist believes that the mean is higher than that figure. State H0 and Ha.

H0: = 140 mg/g;   Ha: < 140 mg/g

H0: > 140 mg/g;   Ha: = 140 mg/g

H0: = 140 mg/g;   Ha: 140 mg/g

H0: < 140 mg/g;   Ha: = 140 mg/g

H0: = 140 mg/g;   Ha: > 140 mg/g

Carry out a significance test to see if the new data support this belief. (Use = 0.05. Round your value for z to two decimal places and round your P-value to four decimal places.)

Do the data support this belief? State your conclusion.

(c) The statistical procedures used in (a) and (b) are valid when several assumptions are met. What are these assumptions? (Select all that apply.)

We must assume that the sample has an underlying distribution that is uniform.

We must assume that the 16 cuttings in our sample are an SRS.

Because our sample is not too large, the standard deviation of the population and sample must be less than 10.

Because our sample is not too large, the population should be normally distributed, or at least not extremely nonnormal.

C is where I am going wrong, I said "Because our sample is not too large, the population should be normally distributed, or at least not extremely nonnormal.", but it was not correct according to webassign.

Explanation / Answer

a.

Standard error of the mean = SEM = S/N = 2.25

t(, N-1) = 1.753

Confidence interval = m +/- (t(, N-1)*SEM)

Mean = 145

Lower bound: 141.06

Upper bound: 148.94

b.

H0: = 140 mg/g;   Ha: > 140 mg/g

Z = (X - ) /
Z = (145 - 140) / 9

Z = 0.55556

p value :0.2892

c. Because our sample is not too large, the population should be normally distributed, or at least not extremely nonnormal.

We must assume that the 16 cuttings in our sample are an SRS.

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