Question

In: Statistics and Probability

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

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 15 cuttings has mean cellulose content x = 146 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 Ha.

a. H0: μ = 140 mg/g;   Ha: μ < 140 mg/g

b. H0: μ = 140 mg/g;   Ha: μ ≠ 140 mg/g

c. H0: μ < 140 mg/g;   Ha: μ = 140 mg/g

d. H0: μ > 140 mg/g;   Ha: μ = 140 mg/g

e. H0: μ = 140 mg/g;   Ha: 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.)

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

b. We must assume that the 15 cuttings in our sample are an SRS.

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

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

Solutions

Expert Solution

(a) The 90% Confidence Interval

Given = 146 mg/g, = 9 mg/g, n = 15

The Zcritical at = 0.1 is

The Confidence Interval is given by ME, where

The Lower Limit = 146 - 3.82 = 142.18

The Upper Limit = 146 + 3.82 = 149.82

The 90% Confidence Interval is (142.18 , 149.82)

__________________________________________________________________________

(b) Hypothesis Test for a single mean

The Hypothesis: Option e

H0: = 146 mg/g

Ha: > 146 mg/g

The Test Statistic: The test statistic is given by the equation:

Z observed = 2.58

The p Value: The p value for Z = 2.58, p value = 0.0049

______________________________________________

Since p value < alpha (0.05), we reject H0.

Therefore Option a: Reject H0. There is significant evidence of a mean cellulose content greater than 140 mg/g.

_______________________________________________

(c) The Assumption that need to be met are (a) Because the sample is not too large, the population should be normally distributed or at least not extremely non normal

and (b) We must assume that the 15 cuttings in our sample are an SRS (Simple random sampling)


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