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 = 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 μ = 140 mg/g, but the agronomist believes that the mean is higher than that figure. State H0 and Ha.
H0: μ > 140 mg/g; Ha: μ = 140 mg/g
H0: μ = 140 mg/g; Ha: μ ≠ 140 mg/g
H0: μ < 140 mg/g; Ha: μ = 140 mg/g
H0: μ = 140 mg/g; Ha: μ < 140 mg/g
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 = _______
Do the data support this belief? State your conclusion.
Reject the null hypothesis, there is significant evidence of a mean cellulose content greater than 140 mg/g.
Reject the null hypothesis, there not is significant evidence of a mean cellulose content greater than 140 mg/g.
Fail to reject the null hypothesis, there is significant evidence of a mean cellulose content greater than 140 mg/g.
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.)
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.
We must assume that the sample has an underlying distribution that is uniform.
We must assume that the 15 cuttings in our sample are an SRS.
a)
90% confidence interval for the mean cellulose content in the population=141.18 , 148.82
b)
H0: μ = 140 mg/g; Ha: μ > 140 mg/g
z =(145-140)*sqrt(15)/9=2.15
p value =0.0158
Reject the null hypothesis, there is significant evidence of a mean cellulose content greater than 140 mg/g.
c)
Because our sample is not too large, the population should be normally distributed, or at least not extremely nonnormal.
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