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Physics 270 – Experimental Physics

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Let say we are given a functional relationship between several measured variables Q(x, y, …) What is the uncertainty in Q if the uncertainties in x and y are known? Example: Power in an electric circuit is P = I 2 R LetI = 1.0 ± 0.1 A R =10.0 ± 1.0 . Determine the power and its uncertainty, assuming I and R are uncorrelated.

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We can use the variance in Q(x,y) as a function of the variances in x and y to find the fractional uncertainties associated with each variable in the function. If x and y are uncorrelated, then this is zero. The derivation of this equation is in the lab manual.

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z = xy Power in an electric circuit is P = I 2 R. Let I = 1.0 ± 0.1 A and R = 10.0 ± 0.5 . Determine the power and its variance using propagation of errors, assuming I and R are uncorrelated. P = I 2 R = 10.0 ± 2.0 W

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Z = x + y Example: I = I 1 + I 2 I 1 = 4.0 ± 0.1 A I 2 = 6.0 ± 0.2 A A A

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C = 2πR Then, ΔC = 2π ΔR For example, if R = 3.1 ± 0.2 cm, then C = 2π(3.0 cm) = 19.478 cm = 19.5 cm ΔC = 2π (0.2 cm) = 1.257 cm C = 19.5 ± 1.3 cm

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Unphysical situations can arise if we use the propagation of errors results blindly! Example: Suppose we measure the volume of a cylinder: V = 2πRL. Let R = exactly 1.0 cm, and L = 1.0 ± 0.5 cm. Using propagation of errors: σ V = 2πRσ L = π/2 cm 3 V = π ± π/2 cm 3 If the error on V (σ V ) is to be interpreted in the Gaussian sense ☞ finite probability (≈ 3%) that the volume (V) is < 0 since V is only 2σ away from than 0! ☞ Clearly this is unphysical! ☞ Care must be taken in interpreting the meaning of σ V.

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quick estimation of uncertainty for a function = 38.0° ± 0.6°

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method for determining whether experimental findings validate a theoretical prediction Observed values are those that the researcher obtains empirically through direct observation. The theoretical or expected values are developed on the basis of an established theory or a working “hypothesis.”

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For example, we might expect that if we flip a coin 200 times, that we would tally 100 heads and 100 tails. In checking this statistical expectation, we might find only 92 heads and 108 tails. Should we reject this coin as being fair? Should we just attribute the difference between expected and observed frequencies to random fluctuation?

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FaceFrequency 142 255 338 457 564 644 Consider a second example: let’s suppose that we have an unbiased, six-sided die. We roll this die 300 times and tally the number of times each side appears: Ideally, we might expect every side to appear 50 times. What should we conclude from these results? Is the die biased?

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The null hypothesis is a statement that is assumed true. The use of the chi-squared distribution is hypothesis testing follows this process: (1) a null hypothesis H 0 is stated, (2) a test statistic is calculated, the observed value of the test statistic is compared to a critical value, and (3) a decision is made whether or not to reject the null hypothesis.

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The null hypothesis rejected only when the data has a degree of statistical confidence that the null hypothesis is false, when the level of confidence exceeds a pre-determined level, usually 95 %, that causes a rejection of the null hypothesis. If experimental observations indicate that the null hypothesis should be rejected, it means either that the hypothesis is indeed false or the measured data gave an improbable result indicating that the hypothesis is false, when it is really true.

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FaceOE (O E) 2 / E Heads921000.64 Tails1081000.64 Totals200 2 =1.28 Table I: The Coin Toss Example FaceOE (O E) 2 / E 142501.28 255500.50 338502.88 457500.98 564503.92 644500.72 Totals300 2 =10.28 Table II: The 6-sided Die Example

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The chi-square distribution is tabulated and available in most texts on statistics. To use the table, one must know how many degrees of freedom df are associated with the number of categories in the sample data. This is because there is a family of chi-square distribution, each a function of the number of degrees of freedom. The number of degrees of freedom is typically equal to k 1. Coin Example: df = 1 Die Example: df = 5

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A chi-square table lists the chi-squared distribution in terms of df and in terms of the level of confidence, = 1 p. df = 0.10 = 0.05 = 0.025 df = 0.10 = 0.05 = 0.025 12.7063.8415.0241117.27519.67521.920 24.6055.9917.3781218.54921.02623.337 36.2517.8159.3481319.81222.36224.736 47.7799.48811.1431421.06423.68526.119 59.23611.07112.8331522.30724.99627.488 610.64512.59214.4491623.54226.29628.845 712.01714.06716.0131724.76927.58730.191 813.36215.50717.5351825.98928.86931.526 914.68416.91919.0231927.20430.14432.852 1015.98718.30720.4832028.41231.41034.170

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The standard practice in the world of statistics is to use a p = 95 % level of confidence in the hypothesis decision making. Thus, if the value of chi-squared that is calculated indicates a value of 1-p that is less than or equal to 0.05, then the null hypothesis should be rejected. In the coin-flip example, you can toss a coin and get 14 heads out of twenty flips and find chi-squared = 0.0577. This would indicate that such an observation can happen by chance and the coin can be considered a fair coin. Such a finding would be described by statisticians as “not statistically significant at the 5 % level.”

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If one found 15 heads out of 20 tosses, then chi- squared would be somewhat less than 0.05 and the coin would be considered biased. This would be described as “statistically significant at the 5 % level.” The significance level of the test is not determined by the p value. It is pre-determined by the experimenter. You can choose a 90 % level, a 95 % level, a 99 % level, etc.

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If I asked you to tell me the average income of all of the employees of State Farm Insurance Company, I am asking for the value of the parameter , the mean income for the entire population of State Farm employees. Since the number of employees is very large and spread around the United States, you would be forced to take a sample of that population. You would have to try and collect data from many job levels and from people from around the country. You would probably send out a survey and get substantial return of the forms to provide an acceptable sample. Then, you could report an average salary for your sample, which we’ll call. If I then ask, what is the average salary for employees of State Farm, you would tell me the value you got from your sample. You would be making a point estimate.

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A public opinion poll may report that the results have a margin of error of ± 3%, which means that readers can be 95% confident (not 68% confident) that the reported results are accurate within 3 percentage points. In physics, the same average result would be reported with an uncertainty of ± 1.5% to indicate the 68% confidence interval.

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