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An article in the Journal of Agricultural Science ["The Use of Residual Maximum Likelihood to Model Grain Quality Characteristics of Wheat with Variety, Climatic and Nitrogen Fertilizer Effects" (1997, Vol. 128, pp. \(135-142\) ) ] investigated means of wheat grain crude protein content (CP) and Hagberg falling number (HFN) surveyed in the United Kingdom. The analysis used a variety of nitrogen fertilizer applications \((\mathrm{kg} \mathrm{N} / \mathrm{ha}),\) temperature \(\left({ }^{\circ} \mathrm{C}\right),\) and total monthly rainfall \((\mathrm{mm})\). The following data below describe temperatures for wheat grown at Harper Adams Agricultural College between 1982 and \(1993 .\) The temperatures measured in June were obtained as follows: $$ \begin{array}{llllll} 15.2 & 14.2 & 14.0 & 12.2 & 14.4 & 12.5 \\ 14.3 & 14.2 & 13.5 & 11.8 & 15.2 & \end{array} $$ Assume that the standard deviation is known to be \(\sigma=0.5\) (a) Construct a \(99 \%\) two-sided confidence interval on the mean temperature. (b) Construct a \(95 \%\) lower-confidence bound on the mean temperature. (c) Suppose that you wanted to be \(95 \%\) confident that the error in estimating the mean temperature is less than 2 degrees Celsius. What sample size should be used? (d) Suppose that you 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 should be used?

Short Answer

Expert verified
(a) 99% CI: (14.29, 15.07). (b) 95% LCB: 14.43. (c) 1 sample. (d) 11 samples.

Step by step solution

01

Calculate the sample mean (\( \bar{x} \))

To find the sample mean, add up all the temperatures and divide by the number of temperatures. There are 11 data points: 15.2, 14.2, 14.0, 12.2, 14.4, 12.5, 14.3, 14.2, 13.5, 11.8, 15.2. The sum of these temperatures is 161.5. So, the sample mean is \( \bar{x} = \frac{161.5}{11} \approx 14.68 \) degrees Celsius.
02

Calculate the Confidence Interval (Part a)

For a 99% confidence interval with known standard deviation, we use the formula \( \bar{x} \pm z \frac{\sigma}{\sqrt{n}} \) where \( z \) is the z-score for 99% confidence. For 99%, \( z \approx 2.576 \). With \( \sigma = 0.5 \) and \( n = 11 \), the confidence interval is \( 14.68 \pm 2.576 \cdot \frac{0.5}{\sqrt{11}} \approx 14.68 \pm 0.39 \). Thus, the interval is \( (14.29, 15.07) \).
03

Calculate Lower Confidence Bound (Part b)

For a 95% lower confidence bound, use \( \bar{x} - z \frac{\sigma}{\sqrt{n}} \), with \( z \approx 1.645 \) for 95%. So, it is \( 14.68 - 1.645 \cdot \frac{0.5}{\sqrt{11}} = 14.68 - 0.25 = 14.43 \). Thus, the lower confidence bound is 14.43.
04

Determine Sample Size for Margin of Error (Part c)

For a 95% confidence, to make the margin of error \( E = 2 \), use the formula \( n = \left( \frac{z \sigma}{E} \right)^2 \). For 95% \( z \approx 1.96 \), and \( \sigma = 0.5 \), \( n = \left( \frac{1.96 \cdot 0.5}{2} \right)^2 \approx 0.24 \). Thus, you need at least 1 sample measurement.
05

Determine Sample Size for Confidence Interval Width (Part d)

To achieve a width of 1.5 degrees at 95% confidence, use \( n = \left( \frac{2z \sigma}{W} \right)^2 \), where \( W = 1.5 \). With \( z = 1.96 \), \( n = \left( \frac{2 \cdot 1.96 \cdot 0.5}{1.5} \right)^2 \approx 10.73 \). Thus, at least 11 samples are needed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Standard Deviation
When exploring confidence intervals, understanding the concept of standard deviation is essential. Simply put, standard deviation measures how spread out the numbers in a dataset are from the average (mean). The smaller the standard deviation, the closer the numbers are to the mean, indicating less variability. Conversely, a larger standard deviation indicates that the numbers are more spread out from the mean. In our exercise, the standard deviation is given as \(\sigma = 0.5\), meaning the data points are close to the average mean temperature, making our dataset relatively consistent. Knowing the standard deviation allows us to determine how confident we can be in estimating the mean temperature. It is a crucial component when calculating confidence intervals, as it helps you determine the spread of the data and thus the precision of your interval estimates. Always remember: a low standard deviation implies more precise data, while a high standard deviation suggests more variability.
Sample Size Determination
Sample size determination is a critical aspect of designing any statistical study involving confidence intervals. It answers the question: How many samples do we need to accurately estimate a parameter? For instance, in part (c) of the exercise, we are asked to find how many samples are needed to be 95% confident that our mean estimation's error is less than 2 degrees Celsius. We use the formula \[ n = \left( \frac{z \sigma}{E} \right)^2 \]where \( z \) is the z-score for the desired confidence level, \( \sigma \) is the standard deviation, and \( E \) is the margin of error. In this case, \( z \approx 1.96 \), \( \sigma = 0.5 \), and \( E = 2 \), leading us to need only 1 sample. However, to achieve a specific confidence interval width, part (d) requires a different calculation with a formula accounting for the total width \( W \), where the minimum sample size is slightly higher. Determining the right sample size helps balance resource use with the precision of the interval estimate.
Deciphering the Z-score
The z-score is another vital tool for creating confidence intervals. It represents the number of standard deviations a data point is from the mean of a dataset. In constructing confidence intervals, the z-score is used to set the specific level of confidence desired, like 95% or 99%. For example, in our exercise, when calculating the 99% confidence interval, the z-score used is approximately 2.576. This value indicates how confident we want our interval estimate to be. A higher z-score means a wider range, giving us more confidence that the true mean is within our interval, though it may lack precision. On the flip side, a lower z-score, like 1.645 used for the 95% lower confidence bound, offers a narrower range but might include less certainty. Here, the z-score enables us to adjust the confidence level without changing the underlying data itself. Understanding this metric is key to interpreting statistical estimates and ensuring they align with the desired confidence levels.

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Most popular questions from this chapter

An article in the Australian Journal of Agricultural Research ["Non-Starch Polysaccharides and Broiler Performance on Diets Containing Soyabean Meal as the Sole Protein Concentrate" \((1993,\) Vol. \(44(8),\) pp. \(1483-1499)]\) determined that the essential amino acid (Lysine) composition level of soybean meals is as shown here \((\mathrm{g} / \mathrm{kg})\) $$ \begin{array}{|l|l|l|l|l|} \hline 22.2 & 24.7 & 20.9 & 26.0 & 27.0 \\ \hline 24.8 & 26.5 & 23.8 & 25.6 & 23.9 \\ \hline \end{array} $$ (a) Construct a \(99 \%\) two-sided confidence interval for \(\sigma^{2}\). (b) Calculate a \(99 \%\) lower confidence bound for \(\sigma^{2}\). (c) Calculate a \(90 \%\) lower confidence bound for \(\sigma\). (d) Compare the intervals that you have computed.

A random sample has been taken from a normal distribution and the following confidence intervals constructed using the same data: (38.02,61.98) and (39.95,60.05) (a) What is the value of the sample mean? (b) One of these intervals is a \(95 \% \mathrm{CI}\) and the other is a \(90 \%\) CI. Which one is the \(95 \%\) CI and why?

The Bureau of Meteorology of the Australian Government provided the mean annual rainfall (in millimeters) in Australia 1983-2002 as follows (http://www.bom.gov.au/ climate/change/rain03.txt): $$ \begin{array}{l} 499.2,555.2,398.8,391.9,453.4,459.8,483.7,417.6,469.2 \\ Check the assumption of normality in the population. Construct a \(95 \%\) confidence interval for the mean annual rainfall. 452.4,499.3,340.6,522.8,469.9,527.2,565.5,584.1,727.3 \\ 558.6,338.6 \end{array} $$

A normal population has a known mean of 50 and unknown variance. (a) A random sample of \(n=16\) is selected from this population, and the sample results are \(\bar{x}=52\) and \(s=8 .\) How unusual are these results? That is, what is the probability of observing a sample average as large as 52 (or larger) if the known, underlying mean is actually \(50 ?\) (b) A random sample of \(n=30\) is selected from this population, and the sample results are \(\bar{x}=52\) and \(s=8\). How unusual are these results? (c) A random sample of \(n=100\) is selected from this population, and the sample results are \(\bar{x}=52\) and \(s=8\). How unusual are these results? (d) Compare your answers to parts (a)-(c) and explain why they are the same or different.

Past experience has indicated that the breaking strength of yarn used in manufacturing drapery material is normally distributed and that \(\sigma=2\) psi. A random sample of nine specimens is tested, and the average breaking strength is found to be 98 psi. Find a \(95 \%\) two-sided confidence interval on the true mean breaking strength.

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