My Notes An agronomist examines the cellulose content of a variety of alfalfa ha

Question

My Notes An agronomist examines the cellulose content of a variety of alfalfa hay. Suppose that the cellulose conte in the e population has standard deviation 9 milligrams per gram (mg/g). A sample of 14 cuttings has mean ce lose contentX 146 mg/g a) Give a 90% confidence nterval for the mean cellulose content in the population. (Round your answers to two decimal place 42.04 (b) A previous study claimed that the mean cellulose content was At 140 mg/g, but the agronomist believes that the mean is higher than that figure. State Ho and Ha. Ho: 140 mg/g; H.: A 40 mg/g Ho: 140 mg/g; Ha: A 140 mg/g e Ho: H 40 mg/g; He: A 40 mg/g o Ho: u 40 mg/g Ho: M 40 mg/g Ho: 140 mg/g; H.: A 140 mg/g Carry out a significance test to see the new data support this belief. (Use a 0.0 Round your value for 2 to two d mal places and round your P-value to four decimal place P value- 0.0064 x Do the data support this belief? State your conclusion. eject the null hypothesis, there is significant evidence of a mean cellulose content greater than 140 mg/g. Reject the null hypothesis, ere not is significant evidence of a mean cellulose conte greater than 140 mg/g. Fail to reject the null hypothesis, the ere is significant evidence of a mea 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 al that apply. Because our sample is not too large, the standard deviation of the population and sample must be less than 10. We must assume that the sample has an underlying distribution that is unifomm. Because our sample is not too larger the population should be normally distributed, or at leas not extremely nonnormal We must assume that the 14 cuttings in our sample are an SRS.

Explanation / Answer

Solution:-

For the first question, where the value of z = 2.80
The value of P = 0.0026.

In the second question,
Ho = mu = muo
z = 1.92
Left-tailed P-value (Z < 1) = 0.9726
Right-tailed P-value (Z > 1) = 0.0274
Two-tailed P-value (|Z| > 1) = 0.0548

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