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Chapter 8: Problem 100

An article in the Journal of Human Nutrition and Dietetics ["The Validation of Energy and Protein Intakes by Doubly Labeled Water and 24-Hour Urinary Nitrogen Excretion in Post-Obese Subjects" (1995, Vol. 8, pp. \(51-64\) ) ] showed the energy intake expressed as a basal metabolic rate, BMR (MJ). $$ \begin{array}{lllll} 5.40 & 5.67 & 5.79 & 6.85 & 6.92 \end{array} $$ \(\begin{array}{lllll}5.70 & 6.08 & 5.48 & 5.44 & 5.51\end{array}\) (a) Use a normal probability plot to check the normality assumption. (b) Find a \(99 \%\) two-sided confidence interval on the mean BMR

Short Answer

Expert verified
Check normality with a Q-Q plot, and the 99% CI for the mean BMR is \([5.247, 6.321]\).

Step by step solution

01

Organize the Data

First, list the given data points for BMR (MJ) in order to analyze them. We have the following data points: \( \{5.40, 5.67, 5.79, 6.85, 6.92, 5.70, 6.08, 5.48, 5.44, 5.51\} \).
02

Check Normality Assumption

To check the normality, a normal probability plot (also known as a Q-Q plot) is used. Plot each data point against the expected quantiles of a normal distribution. If the points roughly lie on a straight line, the data can be assumed to be normally distributed.
03

Compute Sample Mean

Calculate the sample mean \( \bar{x} \) of the data. \[ \bar{x} = \frac{1}{10} \sum_{i=1}^{10} x_i = \frac{5.40 + 5.67 + 5.79 + 6.85 + 6.92 + 5.70 + 6.08 + 5.48 + 5.44 + 5.51}{10} \approx 5.784 \]
04

Compute Sample Standard Deviation

Calculate the sample standard deviation \( s \) using the formula: \( s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \). After calculation, \( s \approx 0.522 \).
05

Determine 99% Confidence Interval for Mean

Use the formula for a two-sided confidence interval: \( \bar{x} \pm t_{\alpha/2} \frac{s}{\sqrt{n}} \). Given \( n = 10 \), and using the t-distribution table for \( t_{0.005, 9} \approx 3.250 \), compute the confidence interval.\[ CI = 5.784 \pm 3.250 \times \frac{0.522}{\sqrt{10}} \] \[ CI \approx 5.784 \pm 0.537 \] \[ CI \approx [5.247, 6.321] \]
06

Conclusion

The normal probability plot should confirm normality if the plotted points lie on a straight line. The 99% confidence interval for the mean BMR is \([5.247, 6.321]\).

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Key Concepts

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

Normal Probability Plot
A Normal Probability Plot, or Q-Q plot, is a tool used to visually assess if a set of data approximates a normal distribution. When you create this plot, you are plotting your sample data's quantiles against the quantiles of a normal distribution.

If the points on this plot lie close to a straight line, it suggests your data is normally distributed.
  • Start by plotting each data point on the y-axis against the corresponding quantile on the x-axis from a normal distribution.
  • Check for a linear trend in the plot: a straight line indicates normality.
  • Significant deviations from the line suggest deviations from normality.
This technique can help determine whether you can use statistical tests that assume normality, such as calculating a confidence interval using a t-distribution.
Sample Mean Calculation
The sample mean is a measure of the center of your dataset and is calculated by summing all the data points and dividing by the number of data points. For this exercise:
  • Add all the BMR values: 5.40, 5.67, 5.79, 6.85, 6.92, 5.70, 6.08, 5.48, 5.44, and 5.51, which totals to 57.84.
  • Since there are 10 values, divide by 10.
  • The calculation yields a sample mean, \( \bar{x} = 5.784 \).
This mean is crucial for summarizing the data, providing a single representative value around which the data points cluster.
Sample Standard Deviation Calculation
The sample standard deviation quantifies the amount of variation or dispersion in a set of data points. It measures how much the data points deviate from the sample mean.
  • First, compute the deviations of each data point from the mean (each data point minus 5.784).
  • Square each of these deviations.
  • Sum all these squared deviations.
  • Then divide by \( n-1 \), where \( n \) is the number of data points (in this case, 10).
  • Taking the square root of that result gives you the standard deviation, approximately \( s \approx 0.522 \).
This statistic helps understand how much the BMR data points differ from the average BMR value across the sample.
T-Distribution
The t-distribution is used in statistics for estimating population parameters when the sample size is small and the population standard deviation is unknown.
  • The t-distribution is similar to the normal distribution but has heavier tails, meaning it accounts for variability better when sample sizes are small.
  • It is specified by the degrees of freedom, which in this example is \( n-1 \) (9 for a sample of 10).
  • For a 99% confidence interval, use the critical value from the t-distribution table, denoted as \( t_{\alpha/2, n-1} \), which is 3.250 in this context.
Utilizing the t-distribution allows you to construct the confidence interval: \( \bar{x} \pm t_{\alpha/2} \frac{s}{\sqrt{n}} \), thus giving a range that is likely to contain the true mean BMR.

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Most popular questions from this chapter

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 \(90 \%\)

During the 1999 and 2000 baseball seasons, there was much speculation that the unusually large number of home runs that were hit was due at least in part to a livelier ball. One way to test the "liveliness" of a baseball is to launch the ball at a vertical surface with a known velocity \(V_{L}\) and measure the ratio of the outgoing velocity \(V_{O}\) of the ball to \(V_{L}\). The ratio \(R=V_{O} / V_{L}\) is called the coefficient of restitution. Following are measurements of the coefficient of restitution for 40 randomly selected baseballs. The balls were thrown from a pitching machine at an oak surface. $$ \begin{array}{llllll} 0.6248 & 0.6237 & 0.6118 & 0.6159 & 0.6298 & 0.6192 \\ 0.6520 & 0.6368 & 0.6220 & 0.6151 & 0.6121 & 0.6548 \\ 0.6226 & 0.6280 & 0.6096 & 0.6300 & 0.6107 & 0.6392 \\ 0.6230 & 0.6131 & 0.6223 & 0.6297 & 0.6435 & 0.5978 \\ 0.6351 & 0.6275 & 0.6261 & 0.6262 & 0.6262 & 0.6314 \\ 0.6128 & 0.6403 & 0.6521 & 0.6049 & 0.6170 & \\ 0.6134 & 0.6310 & 0.6065 & 0.6214 & 0.6141 & \end{array} $$ (a) Is there evidence to support the assumption that the coefficient of restitution is normally distributed? (b) Find a \(99 \% \mathrm{CI}\) on the mean coefficient of restitution. (c) Find a \(99 \%\) prediction interval on the coefficient of restitution for the next baseball that will be tested. (d) Find an interval that will contain \(99 \%\) of the values of the coefficient of restitution with \(95 \%\) confidence. (e) Explain the difference in the three intervals computed in parts (b), (c), and (d).

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 UK. The analysis used a variety of nitrogen fertilizer applications (kg N/ha), temperature \(\left({ }^{\circ} \mathrm{C}\right),\) and total monthly rainfall \((\mathrm{mm})\). The data shown 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 we 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 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 should be used?

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