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Here are the names of 12 orchestra conductors and their performance times in minutes for Beethoven's Ninth Symphony: \(\begin{array}{llll}\text { Bermstein } & 71.03 & \text { Furtwängler } & 74.38 \\ \text { Leinsdorf } & 65.78 & \text { Ormandy } & 64.72 \\ \text { Solti } & 74.70 & \text { Szell } & 66.22 \\ \text { Bohm } & 72.68 & \text { Karajan } & 66.90 \\ \text { Masur } & 69.45 & \text { Rattle } & 69.93 \\\ \text { Steinberg } & 68.62 & \text { Tennstedt } & 68.40\end{array}\) a. Check to see that normality is a reasonable assumption for the performance time distribution. b. Compute a \(95 \%\) CI for the population standard deviation, and interpret the interval. c. Supposedly, classical music is \(100 \%\) determined by the composer's notation, including all timings. Based on your results, is this true or false?

Short Answer

Expert verified
The distribution can reasonably be assumed normal. The 95% CI suggests variability exists, less so than "100%" by notation.

Step by step solution

01

Organize the Data

List the performance times for the 12 conductors: \[ [71.03, 74.38, 65.78, 64.72, 74.70, 66.22, 72.68, 66.90, 69.45, 69.93, 68.62, 68.40] \]
02

Check Normality

Plot the data using a histogram or a Q-Q plot. Check for symmetry and a bell-shaped distribution in the histogram or for the data points to follow the line in the Q-Q plot. These are indicators of normal distribution.
03

Calculate Sample Standard Deviation

Find the mean of the sample, \( \bar{x} \), then compute the standard deviation, \( s \), using the formula: \[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n}(x_i - \bar{x})^2} \] where \( n = 12 \).
04

Compute 95% Confidence Interval for Standard Deviation

Use the Chi-Squared distribution. The confidence interval for the standard deviation \( \sigma \) is given by: \[ \left( \sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}} \right) \]where \( \chi^2_{\alpha/2} \) and \( \chi^2_{1-\alpha/2} \) are the critical values from the Chi-Squared distribution with \( n-1 \) degrees of freedom.
05

Interpretation of the CI

Interpret the interval in the context of the problem: the calculated interval suggests a range within which the true population standard deviation lies with 95% confidence.
06

Analyze the Role of the Composer's Notation

Determine whether the composer’s notation "100% determines" the timing. Consider if the variability in conductors' performance time is explained by factors other than notation. Large variance might suggest each conductor's interpretation plays a role.

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Key Concepts

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

Normal Distribution
The normal distribution, often depicted as a bell curve, is useful when analyzing data because many real-world phenomena follow this pattern. It is characterized by its symmetry around the mean. Each side of the mean mirrors the other, forming the well-known bell shape. In this exercise, the performance times of 12 orchestra conductors must be assessed for normality.
To determine if the data is normally distributed, we can use visual tools such as a histogram or a Q-Q plot. A histogram should show a single peak roughly in the middle, while a Q-Q plot, which compares the quantiles of our dataset against the quantiles of a theoretical normal distribution, should display points lying close to a straight line.
Identifying normal distribution in our datasets allows us to use various statistical tools effectively, such as computing a confidence interval, which we'll explore later.
Sample Standard Deviation
The sample standard deviation is a measure of how spread out the numbers in a data set are. It provides an estimate of the average distance of each data point from the mean of the dataset. In statistics, this concept is pivotal as it aids in understanding the variability of a sample.
To calculate the sample standard deviation, we start by finding the mean of the sample data set. It's essentially the average performance time for our conductors. Then, each performance time's deviation from this mean is squared, summed up, and divided by the total number of conductors minus one (n-1). Finally, we take the square root of this result. This mathematical process gives us the sample standard deviation. Understanding the sample standard deviation is crucial for further calculations, such as building the confidence interval, which provides greater insight into the data's precision and reliability.
Chi-Squared Distribution
The Chi-Squared distribution is an important concept in statistics used primarily with categorical data but is also pivotal in creating confidence intervals for variance and standard deviation, especially when the underlying population is normally distributed.
In this exercise, we employ the Chi-Squared distribution to compute the 95% confidence interval for the standard deviation of performance times. It involves using specific critical values from the Chi-Squared distribution. Calculating these critical values requires knowing the degrees of freedom, which in this case is 11 (calculated as the number of conductors minus one). With these values, we plug them into the formula to estimate the range within which we believe the true population standard deviation lies with a 95% level of confidence. Understanding this concept helps highlight the potential variability of our data.
Interpretation of Statistical Results
Interpreting statistical results is where we make meaningful conclusions from our analysis. In our exercise, after calculating the confidence interval for the standard deviation of performance times, interpretation helps us understand the real-world implications.
The interval provides a range in which the true population standard deviation is expected to lie, with 95% confidence. If this interval is quite wide, it indicates substantial variability among the conductors' performance times. Conversely, a narrow interval suggests consistency amongst them. This analysis allows us to question whether external factors such as each conductor's interpretation could be influencing performance times more than the composer's strict notation seemed to imply. If variability is high, it could indicate that external factors significantly impact performance, challenging the notion that the composer's notation solely dictates performance timing.

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Most popular questions from this chapter

Aphid infestation of fruit trees can be controlled either by spraying with pesticide or by inundation with ladybugs. In a particular area, four different groves of fruit trees are selected for experimentation. The first three groves are sprayed with pesticides 1,2 , and 3 , respectively, and the fourth is treated with ladybugs, with the following results on yield: \begin{tabular}{llll} \hline Treatment & \multicolumn{1}{l}{\(n_{i}\) (number of } & \({\bar{x}_{i}}\) (bushels/ \\ & trees) & tree) & \\ \hline 1 & 100 & \(10.5\) & \(1.5\) \\ 2 & 90 & \(10.0\) & \(1.3\) \\ 3 & 100 & \(10.1\) & \(1.8\) \\ 4 & 120 & \(10.7\) & \(1.6\) \\ \hline \end{tabular} Let \(\mu_{i}=\) the true average yield (bushels/tree) after receiving the \(i\) th treatment. Then $$ \theta=\frac{1}{3}\left(\mu_{1}+\mu_{2}+\mu_{3}\right)-\mu_{4} $$ measures the difference in true average yields between treatment with pesticides and treatment with ladybugs. When \(n_{1}, n_{2}, n_{3}\), and \(n_{4}\) are all large, the estimator \(\hat{\theta}\) obtained by replacing each \(\mu_{i}\) by \(\bar{X}_{i}\) is approximately normal. Use this to derive a large-sample \(100(1-\alpha) \%\) CI for \(\theta\), and compute the \(95 \%\) interval for the given data.

Here are the lengths (in minutes) of the 63 nineinning games from the first week of the 2001 major league baseball season: \(\begin{array}{llllllllll}194 & 160 & 176 & 203 & 187 & 163 & 162 & 183 & 152 & 177 \\ 177 & 151 & 173 & 188 & 179 & 194 & 149 & 165 & 186 & 187 \\ 187 & 177 & 187 & 186 & 187 & 173 & 136 & 150 & 173 & 173 \\ 136 & 153 & 152 & 149 & 152 & 180 & 186 & 166 & 174 & 176 \\ 198 & 193 & 218 & 173 & 144 & 148 & 174 & 163 & 184 & 155 \\ 151 & 172 & 216 & 149 & 207 & 212 & 216 & 166 & 190 & 165 \\ 176 & 158 & 198 & & & & & & & \end{array}\) Assume that this is a random sample of nineinning games (the mean differs by \(12 \mathrm{~s}\) from the mean for the whole season). a. Give a \(95 \%\) confidence interval for the population mean. b. Give a \(95 \%\) prediction interval for the length of the next nine-inning game. On the first day of the next week, Boston beat Tampa Bay 3-0 in a nine- inning game of \(152 \mathrm{~min}\). Is this within the prediction interval? c. Compare the two intervals and explain why one is much wider than the other. d. Explore the issue of nomality for the data and explain how this is relevant to parts (a) and (b).

Nine Australian soldiers were subjected to extreme conditions, which involved a 100 -min walk with a 25 -lb pack when the temperature was \(40^{\circ} \mathrm{C}\left(104^{\circ} \mathrm{F}\right)\). One of them overheated (above \(39^{\circ} \mathrm{C}\) ) and was removed from the study. Here are the rectal Celsius temperatures of the other eight at the end of the walk ("Neural Network Training on Human Body Core Temperature Data," Combatant Protection and Nutrition Branch, Aeronautical and Maritime Research Laboratory of Australia, DSTO TN-0241, 1999): \(\begin{array}{llllllll}38.4 & 38.7 & 39.0 & 38.5 & 38.5 & 39.0 & 38.5 & 38.6\end{array}\) We would like to get a \(95 \%\) confidence interval for the population mean. a. Compute the \(t\)-based confidence interval of Section 8.3. b. Check for the validity of part (a). c. Generate a bootstrap sample of 999 means. d. Use the standard deviation for part (c) to get a \(95 \%\) confidence interval for the population mean. e. Investigate the distribution of the bootstrap means to see if part (d) is valid. f. Use part (c) to form the \(95 \%\) confidence interval using the percentile method. g. Compare the intervals and explain your preference. h. Based on your knowledge of normal body temperature, would you say that body temperature can be influenced by environment?

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