Question

An agronomist examines the cellulose content of a variety of alfalfa hay. Suppose that the cellulose content in the population has standard deviation σ = 9 milligrams per gram (mg/g). A sample of 16 cuttings has mean cellulose content x = 144 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 μ = 140 mg/g, but the agronomist believes that the mean is higher than that figure. State H₀ and H_a.

A H₀: μ = 140 mg/g;   H_a: μ ≠ 140 mg/g

B H₀: μ = 140 mg/g;   H_a: μ > 140 mg/g

C H₀: μ > 140 mg/g;   H_a: μ = 140 mg/g

D H₀: μ = 140 mg/g;   H_a: μ < 140 mg/g

E H₀: μ < 140 mg/g;   H_a: μ = 140 mg/g

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

z	=
P-value	=

Do the data support this belief? State your conclusion.

A. Reject the null hypothesis, there is significant evidence of a mean cellulose content greater than 140 mg/g.

B. Reject the null hypothesis, there not is significant evidence of a mean cellulose content greater than 140 mg/g.    

C. Fail to reject the null hypothesis, there is significant evidence of a mean cellulose content greater than 140 mg/g.

D. Fail to reject the null hypothesis, there not is significant evidence of a mean cellulose content greater than 140 mg/g.

(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 population should be normally distributed, or at least not extremely nonnormal.

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

Answer 1

Answer)

As the population standard deviation is known, here we can use standard normal z table to estimate the interval and also conduct the test.

Sample size (n) = 16

Mean = 144

S.d = 9

A)

From z table, critical value z for 90% confidence interval is 1.645

Margin of error (MOE) = Z*(S.D/√N) = 1.645*(9/√16) = 3.70125

Confidence interval is given by

(Mean - MOE, Mean + MOE)

(140.29875, 147.70125)

(140.30, 147.70)

B)

Ho : u = 140

Ha : u > 140

(Option B)

C)

Z = (obtained mean - claimed mean)/(s.d/√n)

Z = (144-140)/(9/√16)

Z = 1.78

From z table, p(z>1.78) = 0.0375

P-value is = 0.0375

As the obtained p-value is less than the given significance level of alpha = 0.05

We reject the null hypothesis

Option A is correct.

C)

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

That is simple random sample

Because to assume the normality

Sample should be random