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Chapter 31: Problem 1

A group of students uses a pendulum experiment to measure \(g\), the acceleration of free fall, and obtains the following values (in \(\left.\mathrm{m} \mathrm{s}^{-2}\right): 9.80,9.84,9.72,9.74\) \(9.87,9.77,9.28,9.86,9.81,9.79,9.82 .\) What would you give as the best value and standard error for \(g\) as measured by the group?

Short Answer

Expert verified
The best value for \( g \) is 9.754 \ \pm \ 0.047 \ \mathrm{m/s^2}.

Step by step solution

01

- Calculate the Mean Value

First, sum all the given measurements and then divide by the number of measurements to find the mean value.Sum of measurements: \[9.80 + 9.84 + 9.72 + 9.74 + 9.87 + 9.77 + 9.28 + 9.86 + 9.81 + 9.79 + 9.82 = 107.30\]Number of measurements: 11Mean value: \[ \bar{g} = \frac{107.30}{11} = 9.754 \ \mathrm{m/s^2}\]
02

- Calculate the Deviations

For each measurement, subtract the mean value to find the deviation.Deviations: \[(9.80 - 9.754), (9.84 - 9.754), (9.72 - 9.754), (9.74 - 9.754), (9.87 - 9.754), (9.77 - 9.754), (9.28 - 9.754), (9.86 - 9.754), (9.81 - 9.754), (9.79 - 9.754), (9.82 - 9.754)\]
03

- Square the Deviations

Square each of the deviations calculated in Step 2.Squared deviations: \[(0.046^2), (0.086^2), (0.034^2), (0.014^2), (0.116^2), (0.016^2), (0.474^2), (0.106^2), (0.056^2), (0.036^2), (0.066^2)\]
04

- Calculate the Variance

Sum all the squared deviations and then divide by the number of measurements to determine the variance.Sum of squared deviations: \[0.002116 + 0.007396 + 0.001156 + 0.000196 + 0.013456 + 0.000256 + 0.224676 + 0.011236 + 0.003136 + 0.001296 + 0.004356 = 0.268276\]Variance: \[ \sigma^2 = \frac{0.268276}{11} = 0.024387636 \]
05

- Calculate the Standard Deviation

Take the square root of the variance to find the standard deviation.Standard deviation: \[ \sigma = \sqrt{0.024387636} \approx 0.1562 \ \mathrm{m/s^2} \]
06

- Calculate the Standard Error

Divide the standard deviation by the square root of the number of measurements to find the standard error.Standard error: \[ SE = \frac{0.1562}{\sqrt{11}} \approx 0.047 \ \mathrm{m/s^2} \]
07

- Final Result

The best value for \( g \) is the mean value with its standard error.Best value for \( g \) as measured by the group: \[ 9.754 \pm 0.047 \ \mathrm{m/s^2} \]

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Key Concepts

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

Standard Error
The standard error (SE) is a measure of the precision of the mean value calculated from a sample. In this pendulum experiment, the SE tells us how accurately we can estimate the mean acceleration due to gravity, given the variability in our measurements. We calculate the SE by dividing the standard deviation by the square root of the number of measurements.

For instance, in this experiment:
  • Standard deviation: 0.1562 \(\mathrm{m/s^2}\)
  • Number of measurements: 11
The formula for SE is:
\( SE = \frac{0.1562}{\sqrt{11}} \approx 0.047 \ \(\mathrm{m/s^2}\) \),showing how closely our mean value represents the real gravitational acceleration.
Mean Value Calculation
The mean value is a central measure in statistics, representing the average of a set of numbers. To find the mean acceleration due to gravity in this experiment, we add all the measured values and divide by the number of observations.

Here's how we did it:
  • Total sum of measurements: 107.30 \(\mathrm{m/s^2}\)
  • Number of measurements: 11
The mean value calculation is:
\( \bar{g} = \frac{107.30}{11} = 9.754 \ \(\mathrm{m/s^2}\) \)
This represents our best estimate of the gravitational acceleration based on the collected data.
Standard Deviation
Standard deviation (SD) is a measure that quantifies the amount of variation or dispersion in a set of values. In our experiment with the pendulum, it helps us understand how the individual measurements of gravity differ from the mean value.

To calculate the SD, follow these steps:
  • Calculate the deviation of each measurement from the mean.
  • Square each deviation.
  • Find the average of these squared deviations (this is the variance).
  • Take the square root of the variance.
For instance, the SD in our experiment is:
\( \sigma = \sqrt{0.024387636} \approx 0.1562 \ \(\mathrm{m/s^2}\) \). This value indicates the variability of our measurements around the mean.
Variance Calculation
Variance is a statistical measurement of the spread between numbers in a data set. It shows how far each number in the set is from the mean and thus from every other number in the set.

For the pendulum experiment, we calculated the variance by following these steps:
  • Find the deviation of each measurement from the mean.
  • Square each deviation.
  • Sum all the squared deviations.
  • Divide by the number of measurements.
The variance in our experiment is:
\( \sigma^2 = \frac{0.268276}{11} = 0.024387636 \). This value helps us understand the overall spread of our experimental measurements.
Deviation Calculation
Deviation is the difference between a specific measurement and the mean value of the entire data set. In the pendulum experiment, each deviation tells us how far off a particular measurement is from our calculated mean gravitational acceleration.

Here's the process we followed:
  • Calculate the mean value: 9.754 \(\mathrm{m/s^2}\).
  • For each measurement, subtract the mean value to get individual deviations.
For instance, we found the deviations as follows:
\( (9.80 - 9.754), (9.84 - 9.754), (9.72 - 9.754), \ldots \). This provides a first look at how each measurement varies from the average.

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Most popular questions from this chapter

Each of a series of experiments 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 unknown. In the \(i\) th experiment, \(i=1,2, \ldots, N\), 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 $$ \hat{n} \hat{p}=\frac{1}{N} \sum_{i=1}^{N} x_{i} \quad \text { and } \quad(1-\hat{p})^{N}=\prod_{i=1}^{N}\left(1-\frac{x_{i}}{\hat{n}}\right). $$

On a certain (testing) steeplechase course there are 12 fences to be jumped, and any horse that falls is not allowed to continue in the race. In a season of racing a total of 500 horses started the course and the following numbers fell at each fence: \(\begin{array}{lrrrrrrrrrrrr}\text { Fence: } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \text { Falls: } & 62 & 75 & 49 & 29 & 33 & 25 & 30 & 17 & 19 & 11 & 15 & 12\end{array}\) Use this data to determine the overall probability of a horse's falling at a fence, and test the hypothesis that it is the same for all horses and fences as follows. (a) Draw up a table of the expected number of falls at each fence on the basis of the hypothesis. (b) Consider for each fence \(i\) the standardised variable $$ z_{i}=\frac{\text { estimated falls }-\text { actual falls }}{\text { standard deviation of estimated falls }} $$ and use it in an appropriate \(\chi^{2}\) test. (c) Show that the data indicates that the odds against all fences being equally testing are about 40 to 1 . Identify the fences that are significantly easier or harder than the average.

This exercise is intended to illustrate the dangers of applying formalised estimator techniques to distributions that are not well behaved in a statistical sense. The following are five sets of 10 values, all drawn from the same Cauchy distribution with parameter \(a\). \(\begin{array}{lrrrrr}\text { (i) } & 4.81 & -1.24 & 1.30 & -0.23 & 2.98 \\ & -1.13 & -8.32 & 2.62 & -0.79 & -2.85\end{array}\) \(\begin{array}{lrrrrr}\text { (ii) } & 0.07 & 1.54 & 0.38 & -2.76 & -8.82 \\\ & 1.86 & -4.75 & 4.81 & 1.14 & -0.66\end{array}\) (iii) \(\begin{array}{rrrrr}0.72 & 4.57 & 0.86 & -3.86 & 0.30 \\ -2.00 & 2.65 & -17.44 & -2.26 & -8.83\end{array}\) \(\begin{array}{lrrrrr}\text { (iv) } & -0.15 & 202.76 & -0.21 & -0.58 & -0.14 \\\ & 0.36 & 0.44 & 3.36 & -2.96 & 5.51\end{array}\) \(\begin{array}{lllrlr}\text { (v) } & 0.24 & -3.33 & -1.30 & 3.05 & 3.99 \\ & 1.59 & -7.76 & 0.91 & 2.80 & -6.46\end{array}\) Ignoring the fact that the Cauchy distribution does not have a finite variance (or even a formal mean), show that \(\hat{a}\), the ML estimator of \(a\), has to satisfy $$ s(\hat{a})=\sum_{i=1}^{10} \frac{1}{1+x_{i}^{2} / \hat{a}^{2}}=5 $$ Using a programmable calculator, spreadsheet or computer, find the value of \(\hat{a}\) that satisfies (*) for each of the data sets and compare it with the value \(a=1.6\) used to generate the data. Form an opinion regarding the variance of the estimator, Show further that if it is assumed that \((E[\hat{a}])^{2}=E\left[\hat{a}^{2}\right]\), then \(E[\hat{a}]=v_{2}^{1 / 2}\), where \(v_{2}\) is the second (central) moment of the distribution, which for the Cauchy distribution is infinite!

The following are the values and standard errors of a physical quantity \(f(\theta)\) measured at various values of \(\theta\) (in which there is negligible error): \(\begin{array}{lcccc}\theta & 0 & \pi / 6 & \pi / 4 & \pi / 3 \\ f(\theta) & 3.72 \pm 0.2 & 1.98 \pm 0.1 & -0.06 \pm 0.1 & -2.05 \pm 0.1 \\ \theta & \pi / 2 & 2 \pi / 3 & 3 \pi / 4 & \pi \\ f(\theta) & -2.83 \pm 0.2 & 1.15 \pm 0.1 & 3.99 \pm 0.2 & 9.71 \pm 0.4\end{array}\) Theory suggests that \(f\) should be of the form \(a_{1}+a_{2} \cos \theta+a_{3} \cos 2 \theta .\) Show that the normal equations for the coefficients \(a_{i}\) are $$ \begin{aligned} 481.3 a_{1}+158.4 a_{2}-43.8 a_{3} &=284.7 \\ 158.4 a_{1}+218.8 a_{2}+62.1 a_{3} &=-31.1 \\ -43.8 a_{1}+62.1 a_{2}+131.3 a_{3} &=368.4 \end{aligned} $$ (a) If you have matrix inversion routines available on a computer, determine the best values and variances for the coefficients \(a_{i}\) and the correlation between the coefficients \(a_{1}\) and \(a_{2}\) (b) If you have only a calculator available, solve for the values using a GaussSeidel iteration and start from the approximate solution \(a_{1}=2, a_{2}=-2\) \(a_{3}=4\).

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 \(\mathrm{R}\) 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 \(y(x)\).
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