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Chapter 6: Problem 19

Constructing Normal Quantile Plots.Use the given data values to identify the corresponding z scores that are used for a normal quantile plot, then identify the coordinates of each point in the normal quantile plot. Construct the normal quantile plot, then determine whether the data appear to be from a population with a normal distribution. Brain Volumes A sample of human brain volumes (cm \(^{3}\) ) is obtained from those listed in Data Set 8 "IQ and Brain Size" in Appendix B: 1027,1029,1034,1070,1079,1079,963,1439

Short Answer

Expert verified
Pair each sorted value with z-scores, plot the points, and examine their linearity to determine normality.

Step by step solution

01

- List the data in ascending order

First, sort the given brain volume data values in ascending order. The sorted data is: 963, 1027, 1029, 1034, 1070, 1079, 1079, 1439
02

- Assign ranks to each value

Assign ranks to each of the sorted data values. For the given data (which has 8 values), the ranks are as follows: 1, 2, 3, 4, 5, 6, 7, 8.
03

- Calculate the cumulative probability for each rank

For each rank, calculate the cumulative probability using the formula \(\frac{i - 0.5}{n}\) where i is the rank and n is the total number of data points (8 in this case).
04

- Determine corresponding z-scores

Find the z-scores corresponding to each of the cumulative probabilities using the standard normal distribution table. The calculated cumulative probabilities and their z-scores are approximately: - Rank 1: Cumulative Probability = \( \frac{0.5}{8} = 0.0625\) , z = -1.54 - Rank 2: Cumulative Probability = \( \frac{1.5}{8} = 0.1875\) , z = -0.88 - Rank 3: Cumulative Probability = \( \frac{2.5}{8} = 0.3125\) , z = -0.50 - Rank 4: Cumulative Probability = \( \frac{3.5}{8} = 0.4375\) , z = -0.16 - Rank 5: Cumulative Probability = \( \frac{4.5}{8} = 0.5625\) , z = 0.16 - Rank 6: Cumulative Probability = \( \frac{5.5}{8} = 0.6875\) , z = 0.50 - Rank 7: Cumulative Probability = \( \frac{6.5}{8} = 0.8125\) , z = 0.88 - Rank 8: Cumulative Probability = \( \frac{7.5}{8} = 0.9375\) , z = 1.54
05

- Identify coordinates for normal quantile plot

Pair each sorted data value with its corresponding z-score to create coordinate pairs: - (963, -1.54) - (1027, -0.88) - (1029, -0.50) - (1034, -0.16) - (1070, 0.16) - (1079, 0.50) - (1079, 0.88) - (1439, 1.54)
06

- Construct the normal quantile plot

Plot the points on a graph where the x-axis represents the z-scores and the y-axis represents the sorted data values. Draw a straight line to see if the points roughly follow a linear pattern.
07

- Interpret the plot

Examine the plot to determine if the points approximately form a straight line. If they do, it suggests the data is normally distributed. If the points significantly deviate from a straight line, the data may not be normally distributed.

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Key Concepts

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

standard normal distribution
The standard normal distribution is a crucial concept in statistics and data analysis. It is a special case of the normal distribution with a mean of 0 and a standard deviation of 1. This type of distribution allows us to calculate probabilities and z-scores, which are essential for comparing different data sets. The z-score measures how many standard deviations a data point is from the mean. In this exercise, understanding the standard normal distribution helps us find z-scores for the given brain volume data. This is possible because standard normal distribution tables or functions tell us the z-scores corresponding to any cumulative probabilities between 0 and 1.
cumulative probability
Cumulative probability is the probability that a random variable is less than or equal to a specific value. In this exercise, we deal with cumulative probabilities to identify the z-scores for our normal quantile plot. For each rank (which indicates the position of the data point when the data is sorted in ascending order), we calculate the cumulative probability using the formula \(\frac{i - 0.5}{n}\), where i is the rank and n is the total number of data points. For example, for rank 1, the cumulative probability is calculated as \(\frac{0.5}{8} = 0.0625\). These cumulative probabilities tell us how much of the data falls below a particular point. By using a standard normal distribution table, we can then convert these probabilities into z-scores.
data visualization
Data visualization is the graphical representation of data to understand patterns, trends, and insights more efficiently. In this exercise, we use a normal quantile plot (also called a Q-Q plot) to visualize whether the brain volume data follows a normal distribution. This type of plot helps us see how the data deviates from the expected normal distribution. By plotting the z-scores on the x-axis and the sorted data values on the y-axis, we can easily determine if the points form a straight line. If the plotted points follow a linear pattern, it indicates that the data is normally distributed. Hence, data visualization through this plot provides an intuitive way to assess the normality of our dataset.
normal distribution assessment
Normal distribution assessment is the process of evaluating whether or not a dataset follows a normal distribution. This assessment is vital because many statistical tests assume normality in the data. In this exercise, we assess normality by constructing a normal quantile plot. By plotting each sorted data value against its corresponding z-score, we can visually inspect if the data points align along a straight line. An approximate straight line suggests that the data are normally distributed. If the points significantly deviate from this line, the data may not be normally distributed. This step is essential in making accurate inferences about the population from which the sample is drawn.

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Most popular questions from this chapter

Use the data in the table below for sitting adult males and females (based on anthropometric survey data from Gordon, Churchill, et al.). These data are used often in the design of different seats, including aircraft seats, train seats, theater seats, and classroom seats. (Hint: Draw a graph in each case.) $$\begin{array}{|l|l|l|l|} \hline & \text { Mean } & \text { St. Dev. } & \text { Distribution } \\ \hline \text { Males } & 23.5 \mathrm{in} . & 1.1 \mathrm{in} . & \text { Normal } \\ \hline \text { Females } & 22.7 \mathrm{in} . & 1.0 \mathrm{in} . & \text { Normal } \\ \hline \end{array}$$ Find the probability that a male has a back-to-knee length less than 21 in.

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of \(1 .\) In each case, draw a graph, then find the probability of the given bone density test scores. If using technology instead of Table \(A-2,\) round answers to four decimal places. Between -1.00 and 5.00

Unbiased Estimators Data Set 4 "Births" in Appendix B includes birth weights of 400 babies. If we compute the values of sample statistics from that sample, which of the following statistics are unbiased estimators of the corresponding population parameters: sample mean; sample median; sample range; sample variance; sample standard deviation; sample proportion?

When women were finally allowed to become pilots of fighter jets, engineers needed to redesign the ejection seats because they had been originally designed for men only. The ACES-II ejection seats were designed for men weighing between 140 lb and 211 lb. Weights of women are now normally distributed with a mean of 171 lb and a standard deviation of 46 lb (based on Data Set 1 "Body Data" in Appendix B). a. If I woman is randomly selected, find the probability that her weight is between 140 lb and 211 lb. b. If 25 different women are randomly selected, find the probability that their mean weight is between 140 lb and 211 lb. c. When redesigning the fighter jet ejection seats to better accommodate women, which probability is more relevant: the result from part (a) or the result from part (b)? Why?

The probability of a baby being born a boy is \(0.512 .\) Consider the problem of finding the probability of exactly 7 boys in 11 births. Solve that problem using (1) normal approximation to the binomial using Table \(A-2 ;\) (2) normal approximation to the binomial using technology instead of Table \(A-2 ;\) (3) using technology with the binomial distribution instead of using a normal approximation. Compare the results. Given that the requirements for using the normal approximation are just barely met, are the approximations off by very much?
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