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Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from \(N\left(\mu, \sigma^{2}\right)\), where both parameters \(\mu\) and \(\sigma^{2}\) are unknown. A confidence interval for \(\sigma^{2}\) can be found as follows. We know that \((n-1) S^{2} / \sigma^{2}\) is a random variable with a \(\chi^{2}(n-1)\) distribution. Thus we can find constants \(a\) and \(b\) so that \(P\left((n-1) S^{2} / \sigma^{2}

Short Answer

Expert verified
a) Proven using properties of distributions and inequalities. b) Confidence interval is calculated by first determining \(a\) and \(b\) from chi-square distribution with \(n-1\) degrees of freedom, then substituting these, \(n=9\) and \(s^2 = 7.93\) into the formula. c) If known, \(\mu\) would cause the degrees of freedom to be \(n\) instead of \(n-1\), thus changing the calculations of a, b and altering the formula of the confidence interval.

Step by step solution

01

Proof of the Probability Statement

The given inequality is: \((n-1) S^{2} / b<\sigma^{2}<(n-1) S^{2} / a\). Now to prove this, remember the properties of probability and inequalities. Properties of distributions can be used here. Since \(a\) and \(b\) are values on the \(\chi^{2}\) distribution, which is strictly positive and \(a 1/b\). Thus, the statement is proven.
02

Calculation of Confidence Interval

Given, \(n=9\) and \(s^{2}=7.93\). Use these values to calculate \(a\) and \(b\) first. These are determined by calculating the \(\chi^{2}\) distribution quantile function in R; \(a=qchisq(0.025, n-1)\) and \(b=qchisq(0.975, n-1)\). Then substitute these along with the given \(n\) and \(s\) into the confidence interval formula, \((n-1) S^{2} / b<\sigma^{2}<(n-1) S^{2} / a\). This will provide the 95% confidence interval.
03

Modification with Known Mean

If the population mean, \(\mu\), is known, then we wouldn't estimate it from the data and degrees of freedom would be \(n\) instead of \(n-1\). Accordingly, the denominator in the \(\chi^{2}\) distribution would be \(n\) and the estimation of \(s^{2}\) would be different. . Thus the constants would be calculated as : \(a=qchisq(0.025, n)\) and \(b=qchisq(0.975, n)\). The confidence interval formula will then be altered.

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Most popular questions from this chapter

This exercise obtains a useful identity for the cdf of a Poisson cdf. (a) Use Exercise \(3.3 .5\) to show that this identity is true: $$ \frac{\lambda^{n}}{\Gamma(n)} \int_{1}^{\infty} x^{n-1} e^{-x \lambda} d x=\sum_{j=0}^{n-1} e^{-\lambda} \frac{\lambda^{j}}{j !} $$ for \(\lambda>0\) and \(n\) a positive integer. Hint: Just consider a Poisson process on the unit interval with mean \(\lambda\). Let \(W_{n}\) be the waiting time until the \(n\) th event. Then the left side is \(P\left(W_{n}>1\right)\). Why? (b) Obtain the identity used in Example \(4.3 .3\), by making the transformation \(z=\lambda x\) in the above integral.

In Exercise 4.2.11, the sampling was from the \(N(0,1)\) distribution. Show, however, that setting \(\mu=0\) and \(\sigma=1\) is without loss of generality. Hint: First, \(X_{1}, \ldots, X_{n}\) is a random sample from the \(N\left(\mu, \sigma^{2}\right)\) if and only if \(Z_{1}, \ldots, Z_{n}\) is a random sample from the \(N(0,1)\), where \(Z_{i}=\left(X_{i}-\mu\right) / \sigma\). Then show the confidence interval based on the \(Z_{i}\) 's contains 0 if and only if the confidence interval based on the \(X_{i}\) 's contains \(\mu\).

Consider the sample of data (data are in the file ex4.4.3data.rda): \(\begin{array}{rrrrrrrrrrr}13 & 5 & 202 & 15 & 99 & 4 & 67 & 83 & 36 & 11 & 301 \\ 23 & 213 & 40 & 66 & 106 & 78 & 69 & 166 & 84 & 64 & \end{array}\) (a) Obtain the five-number summary of these data. (b) Determine if there are any outliers. (c) Boxplot the data. Comment on the plot.

Let \(y_{1}

A die was cast \(n=120\) independent times and the following data resulted: \begin{tabular}{c|cccccc} Spots Up & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline Frequency & \(b\) & 20 & 20 & 20 & 20 & \(40-b\) \end{tabular} If we use a chi-square test, for what values of \(b\) would the hypothesis that the die is unbiased be rejected at the \(0.025\) significance level?

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