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Chapter 27: Problem 11

According to a particular theory, two dimensionless quantities \(X\) and \(Y\) have equal values. Nine measurements of \(X\) gave values of \(22,11,19,19,14,27,8\), 24 and 18 , whilst seven measured values of \(Y\) were \(11,14,17,14,19,16\) and 14. Assuming that the measurements of both quantities are Gaussian distributed with a common variance, are they consistent with the theory? An alternative theory predicts that \(Y^{2}=\pi^{2} X\); is the data consistent with this proposal?

Short Answer

Expert verified
The data is more consistent with the second theory if \(\bar{Y^2} \approx \pi^2 \bar{X}\).

Step by step solution

01

- Calculate the Mean of X

First, compute the mean of the measurements of quantity X. The given values are: 22, 11, 19, 19, 14, 27, 8, 24, 18. The mean (\( \bar{X} \)) is calculated as: \[ \bar{X} = \frac{22 + 11 + 19 + 19 + 14 + 27 + 8 + 24 + 18}{9} \]
02

- Calculate the Mean of Y

Next, compute the mean of the measurements of quantity Y. The given values are: 11, 14, 17, 14, 19, 16, 14. The mean (\( \bar{Y} \)) is calculated as: \[ \bar{Y} = \frac{11 + 14 + 17 + 14 + 19 + 16 + 14}{7} \]
03

- Compare the Means of X and Y

Check if the means of X and Y are equal, which would support the first theory. Compare \( \bar{X} \) and \( \bar{Y} \).
04

- Calculate Y^2

Compute the square of each measured value of Y: \[ 11^2, 14^2, 17^2, 14^2, 19^2, 16^2, 14^2 \]and their mean.
05

- Calculate \( \pi^2 X \)

Multiply each measured value of X by \( \pi^2 \) and compute their mean:
06

- Compare the Means of Y^2 and \( \pi^2 X \)

Compare the mean of \( Y^2 \) with the mean of \( \pi^2 \bar{X} \). If they are approximately equal, it supports the second theory.

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Key Concepts

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

Gaussian distribution
A Gaussian distribution, also known as the normal distribution, is a type of continuous probability distribution for a real-valued random variable. It is the most commonly encountered distribution in statistics. The probability density function of a Gaussian distribution is symmetric and has a bell-shaped curve.

The bell curve is centered around the mean and the width of the curve is determined by the variance. In this exercise, we assume that the measurements of both quantities X and Y are Gaussian distributed with a common variance. This assumption simplifies the analysis, enabling us to use properties of the Gaussian distribution in hypothesis testing.

Key properties of the Gaussian distribution:
  • Symmetry: The mean, median, and mode of the distribution are all equal and located at the center of the distribution.
  • 68-95-99.7 Rule: Approximately 68% of the data lies within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations.
  • Described by two parameters: the mean (µ) and the variance (σ²).
mean calculation
The mean of a set of measurements is the arithmetic average, which represents the central value of the data set. Calculating the mean involves summing all the values and then dividing by the number of values.

For quantity X: The given values are 22, 11, 19, 19, 14, 27, 8, 24, and 18. The mean (\( \bar{X} \)) is calculated as: \[ \bar{X} = \frac{22 + 11 + 19 + 19 + 14 + 27 + 8 + 24 + 18}{9} \] For quantity Y: The given values are 11, 14, 17, 14, 19, 16, and 14. The mean (\( \bar{Y} \)) is calculated as: \[ \bar{Y} = \frac{11 + 14 + 17 + 14 + 19 + 16 + 14}{7} \] Comparing these means will help us determine if the theory that X and Y are equal is supported by the data.
theory comparison
In the context of this problem, two hypotheses are proposed for the relationship between X and Y. The first theory suggests that the values of X and Y are equal. To test this, we compare the mean of X (\( \bar{X} \)) and the mean of Y (\( \bar{Y} \)) from the measurements. If these means are close, it supports the first theory.

The second theory predicts that the relationship between Y and X is given by \[ Y^2 = \frac{\text{Observed } \bar{X}}{\text{Predicted Values }} = \frac{\text{ } X_{\text{Values } * \text{Pi}^{2}} }{} \] To test this theory, we need to calculate the mean of the squared values of Y (\( \bar{Y^2} \)) and compare it with \[ \text{Predicted Values of X }^2 = X_{\text {Values} * \text {Pi } ^2}{} {} \]. If \[ \bar{Y^2}{} && equals \text{Observed Calculated Values} \] , this supports the second theory.
variance
Variance is a measure of how much the values in a data set differ from the mean. It represents the degree of spread in the data.

In the Gaussian distribution mentioned earlier, the variance (σ²) helps describe the width of the bell curve. A small variance indicates that the data points are close to the mean, while a large variance indicates that the data points are spread out over a wide range of values.

In this exercise, we assume a common variance for both X and Y, which simplifies the computation and comparison of the means. Calculating variance involves the following steps:
  • Find the mean (µ) of the data set.
  • Subtract the mean from each data point and square the result (this is called the squared deviation).
  • Compute the average of these squared deviations; this result is the variance.
Mathematically, it can be expressed as:
\[ \text{Var} (X) = \frac{\text{sum of squared deviations}}{n} with notation \text{S^2} = \frac{\text{ }}{\text{ }}X_{\text{Value} } {}{} \].

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Most popular questions from this chapter

The following are the values obtained by a class of 14 students when measuring a physical quantity \(x: 53.8,53.1,56.9,54.7,58.2,54.1,56.4,54.8,57.3,51.0,55.1\) \(55.0,54.2,56.6\) (a) Display these results as a histogram and state what you would give as the best value for \(x\). (b) Without calculation estimate how much reliance could be placed upon your answer to (a). (c) Databooks give the value of \(x\) as \(53.6\) with negligible error. Are the data obtained by the students in conflict with this?

It is claimed that the two following sets of values were obtained (a) by randomly drawing from a normal distribution that is \(N(0,1)\) and then (b) randomly assigning each reading to one of the two sets \(\mathrm{A}\) and \(\mathrm{B}\). \(\begin{array}{lrrrrrrr}\text { Set A } & -0.314 & 0.603 & -0.551 & -0.537 & -0.160 & -1.635 & 0.719 \\ & 0.610 & 0.482 & -1.757 & 0.058 & & & \\ \text { Set B } & -0.691 & 1.515 & -1.642 & -1.736 & 1.224 & 1.423 & 1.165\end{array}\) Make tests, including \(t\) - and \(F\)-tests, to establish whether there is any evidence that either claims is, or both claims are, false.

The function \(y(x)\) is known to be a quadratic function of \(x\). The following table gives the measured values and uncorrelated standard errors of \(y\) measured at various values of \(x\) (in which there is negligible error): $$ \begin{array}{lccccc} x & 1 & 2 & 3 & 4 & 5 \\ y(x) & 3.5 \pm 0.5 & 2.0 \pm 0.5 & 3.0 \pm 0.5 & 6.5 \pm 1.0 & 10.5 \pm 1.0 \end{array} $$ Construct the response matrix \(R\) using as basis functions \(1, x, x^{2} .\) Calculate the matrix \(\mathrm{R}^{\mathrm{T}} \mathrm{N}^{-1} \mathrm{R}\) and show that its inverse, the covariance matrix \(\mathrm{V}\), has the form $$ \mathrm{V}=\frac{1}{9184}\left(\begin{array}{ccc} 12592 & -9708 & 1580 \\ -9708 & 8413 & -1461 \\ 1580 & -1461 & 269 \end{array}\right). $$ Use this matrix to find the best values, and their uncertainties, for the coefficients of the quadratic form for \(y(x)\).

An experiment consists of a large, but unknown, number \(n(\gg 1)\) of trials in each of which the probability of success \(p\) is the same, but also unkown. In the \(i\) th trial, \(i=1,2, \ldots, N\), the total number of successes is \(x_{i}(\gg 1)\). Determine the log-likelihood function. Using Stirling's approximation to \(\ln (n-x)\), show that $$ \frac{d \ln (n-x)}{d n} \approx \frac{1}{2(n-x)}+\ln (n-x) $$ and hence evaluate \(\partial\left({ }^{n} C_{x}\right) / \partial n\) By finding the (coupled) equations determining the ML estimators \(\hat{p}\) and \(\hat{n}\), show that, to order \(N^{-1}\), they must satisfy the simultaneous 'arithmetic' and 'geometric' mean constraints $$ \hat{n} \hat{p}=\frac{1}{N} \sum_{i=1}^{N} x_{i} \quad \text { and } \quad(1-\hat{p})^{N}=\prod_{i=1}^{N}\left(1-\frac{x_{i}}{\hat{n}}\right). $$

Prove that the sample mean is the best linear unbiased estimator of the population mean \(\mu\) as follows. (a) If the real numbers \(a_{1}, a_{2}, \ldots, a_{n}\) satisfy the constraint \(\sum_{i=1}^{n} a_{i}=C\), where \(C\) is a given constant, show that \(\sum_{i=1}^{n} a_{i}^{2}\) is minimised by \(a_{i}=C / n\) for all \(i\). (b) Consider the linear estimator \(\hat{\mu}=\sum_{i=1}^{n} a_{i} x_{i}\). Impose the conditions (i) that it is unbiased, and (ii) that it is as efficient as possible.
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