Question

An article in the Journal of Agricultural Science [“The Use of Residual Maximum Likelihood to Model Grain Quality Characteristics of Wheat with Variety, Climatic andNitrogen Fertilizer Effects” (1997, Vol. 128, pp. 135–142)] investigated means of wheat grain crude protein content (CP) and Hagberg falling number (HFN) surveyed inthe UK. The analysis used a variety of nitrogen fertilizer applications (kg N/ha), temperature (ºC), and total monthly rainfall (mm). The data shown below describetemperatures for wheat grown at Harper Adams Agricultural College between 1982 and 1993. The temperatures measured in June were obtained as follows:

15.9 15.2 14.4 16.0 15.4 12.6
15.4 13.6 13.3 14.1 13.5


Assume that the standard deviation is known to be 1.8.

a. Construct a 99% two-sided confidence interval on the mean temperature. Round your answers to two decimal places (e.g. 98.76).

b. Construct a 95% lower-confidence bound on the mean temperature. Round your answer to two decimal places (e.g. 98.76).

c. Suppose that we wanted to be 99% confident that the error in estimating the mean temperature is less than 0.8 degrees Celsius. What sample size should be used?Round your answer up to the nearest integer.

d. Suppose that we wanted the total width of the two-sided confidence interval on mean temperature to be 1.5 degrees Celsius at 95% confidence. What sample size shouldbe used? Round your answer up to the nearest integer.

Answer 1

Given n=11, xbar= 14.49, s=1.8
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(a) Given a=0.01, t(0.005, df=n-1=10)=3.17 (check student t table)

So 99% CI is

xbar ± t*s/√n

--> 14.49 ± 3.17*1.8/sqrt(11)

--> (12.77, 16.21)
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(b) Given a=0.05, t(0.05, df=10)=1.81(check student t table)

So 95% lower-confidence bound is

xbar - t*s/√n

= 14.49 - 1.81*1.8/sqrt(11)

= 13.51

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(c) Given a=0.01, t(0.005, df=n-1=10)=3.17 (check student t table)

E=0.8

So n=(t*s/E)^2 = (3.17*1.8/0.8)^2 =50.87256

Take n=51

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(d) Given a=0.05, t(0.05, df=10)=1.81(check student t table)

(xbar + t*s/√n) -( xbar - t*s/√n)

--> (14.49 + 1.81*1.8/sqrt(n)) - (14.49 - 1.81*1.8/sqrt(n))

--> 2*1.81*1.8/sqrt(n) =1.5

--> n = (2*1.81*1.8/1.5)^2 = 18.87034

Take n=19