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 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 ? = 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.) z = P-value = 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.) We must assume that the 15 cuttings in our sample are an SRS. We must assume that the sample has an underlying distribution that is uniform. 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

$\sigma =9,\overline{X}=145$

a) $CI=\overline{X}\pm Z_{\frac{\alpha }{2}}(\frac{\sigma }{\sqrt{n}})$

for 90% confidence $Z_{\frac{\alpha }{2}}=1.645$

$CI=145\pm 1.645(\frac{9}{\sqrt{15}})$

$CI=145\pm 3.82$

$CI=(141.18,148.82)$

b)