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Here are the IQ scores for the 15 first-grade girls from the study mentioned in Example 14.8. \(\begin{array}{lllllll}102 & 96 & 106 & 118 & 108 & 122 & 115 \\ 109 & 113 & 82 & 110 & 121 & 110 & 99\end{array}\) Assume the same prior distribution used in Example 14.8, and assume that the data is a random sample from a normal distribution with mean \(\mu\) and \(\sigma=15\). a. Find the posterior distribution of \(\mu\). b. Find a \(95 \%\) credibility interval for \(\mu\). c. Add four observations with average 110 to the data and find a \(95 \%\) confidence interval for \(\mu\) using the 19 observations. Compare with the result of (b). d. Change the prior so the prior precision is very small but positive, and then recompute (a) and (b). e. Find a \(95 \%\) confidence interval for \(\mu\) using the 15 observations and compare with the credibility interval of (d).

Short Answer

Expert verified
Posterior mean and intervals reflect both added data and changing prior assumptions, influencing credibility and confidence intervals.

Step by step solution

01

Calculate the Sample Mean and Sample Size

Calculate the sample mean \( \bar{x} \) of the given IQ scores. The data has 15 observations, so the sample size \( n = 15 \). This will be crucial for determining the posterior distribution of \( \mu \).
02

Determine the Posterior Distribution of \( \mu \)

Given a normal prior with known variance, employ the Bayesian update formula for normal data. The posterior distribution of \( \mu \) is also normal, with the mean given by \( \mu_{post} = \frac{n\bar{x} + \kappa_0\mu_0}{n + \kappa_0} \) and variance \( \sigma^2_{post} = \frac{\sigma^2}{n + \kappa_0} \), where \( \sigma^2 = 15^2 \) and \( \kappa_0, \mu_0 \) are the prior parameters from Example 14.8.
03

Find the 95% Credibility Interval for \( \mu \)

To compute a 95% credibility interval, extract the 2.5th and 97.5th percentiles from the posterior distribution of \( \mu \), which is normally distributed. Use \( \mu_{post} \) and \( \sigma_{post} \) to set these bounds.
04

Extend the Data Set with Four Additional Observations

Include four new IQ observations averaging 110. Now, the sample size increases to 19. Re-compute the sample mean and apply the updated sample size to determine the new post-intervention posterior distribution.
05

Determine a 95% Confidence Interval with New Data

Compute the traditional 95% confidence interval for \( \mu \) using these 19 observations. Compare this interval with the credibility interval from Step 3.
06

Adjust the Prior Distribution

Modify the prior to have a very small, positive precision. This essentially flattens the prior influence, emulating an uninformative prior.
07

Recalculate Posterior Distribution for Adjusted Prior

With the new prior, recompute \( \mu_{post} \) and \( \sigma^2_{post} \) as described in Step 2. The effect of a less informative prior is that the posterior more heavily reflects the sample data.
08

95% Credibility Interval with Adjusted Prior

Re-determine the 95% credibility interval using the less informative prior and compare this interval to that obtained in Step 3.
09

Traditional Confidence Interval with Original Data

Using only the original 15 observations, compute a standard 95% confidence interval for \( \mu \). Compare this interval to the credibility interval found in Step 8.

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Key Concepts

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

Posterior Distribution
In Bayesian statistics, the posterior distribution is how we update our beliefs about a parameter after observing evidence or data. In this exercise, the parameter of interest is the mean IQ score, \( \mu \). When you have a prior distribution and data from a normal distribution, the posterior distribution also takes a normal form.
For this exercise, you start with a prior distribution on \( \mu \), which incorporates existing beliefs or knowledge about \( \mu \). As you observe the IQ scores, you update this prior to form the posterior distribution. The formula for the updated mean of the posterior, \( \mu_{post} \), is given by the weighted average of the prior mean and the sample mean, weighted by their respective precisions. The variance of the posterior helps us understand the spread, and it decreases as we gather more data, reflecting more certainty.
  • Parameters of the posterior:
    - \( \mu_{post} = \frac{n\bar{x} + \kappa_0\mu_0}{n + \kappa_0} \)
  • - \( \sigma^2_{post} = \frac{\sigma^2}{n + \kappa_0} \)
The posterior combines what we believed before and what the new data suggests, providing a comprehensive view of what \( \mu \) could be.
Credibility Interval
A credibility interval in Bayesian analysis offers a range in which the parameter, like the mean IQ \( \mu \), is believed to lie with a certain probability, say 95%. It tells us more directly what we deem "believable" based on the data and prior information.
To find the 95% credibility interval, you look at the areas that cover 95% of the probability under the posterior distribution curve. Specifically, the interval is created by cutting off the 2.5% tails on each side of the posterior's distribution, as it's assumed to be normally distributed.
  • Key points:
    - The interval reflects both prior and data-informed belief.
    - Provides a probability-based range for \( \mu \).
This interval is helpful for making informed decisions, as it embodies a probabilistic understanding of how likely certain values of \( \mu \) are given the observed data.
Confidence Interval
Unlike the Bayesian credibility interval, a confidence interval is a frequentist concept and tells us how confident we can be that a particular range captures the true parameter value a certain percentage of the time, like 95%.
In this exercise, the confidence interval is obtained using the classical approach under the assumption that the sample is drawn from a normal distribution. This method focuses on data only and doesn't incorporate prior beliefs.
  • Key steps to calculate:
    - Compute the sample mean.
    - Estimate the standard error (with known \( \sigma \)).
    - Use the critical value from a normal distribution corresponding to the desired confidence level.
This confidence interval offers a more objective statistical inference, but it doesn't tell us about the "credibility" of the interval's specific ranges based on prior data.
Prior Distribution
The prior distribution represents your beliefs about a parameter, like the mean IQ \( \mu \), before observing any new data. In Bayesian statistics, it's a crucial part of forming the posterior distribution, as it represents pre-existing knowledge or assumptions.
When this exercise incorporated a prior probability, it was combined with the likelihood from the observed IQ data to improve the understanding of \( \mu \). If the prior is adjusted to have very low precision, it becomes nearly "uninformative," giving less weight to our initial assumptions, thus making the data considerably more influential in determining the posterior.
  • Importance of the prior:
    - Influences the starting point for data interpretation.
    - Affects the strength of the posterior, especially with limited data.
    - Can be customized based on knowledge extent or confidence.
The choice of prior can be subjective, but it is what makes Bayesian inference flexible and adaptable to various contexts by allowing one to incorporate different levels of prior beliefs.

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Most popular questions from this chapter

The single-factor ANOVA model considered in Chapter 11 assumed the observations in the \(i\) th sample were selected from a normal distribution with mean \(\mu_{i}\) and variance \(\sigma^{2}\), that is, \(X_{i j}=\mu_{i}+\varepsilon_{i j}\) where the \(\varepsilon\) 's are normal with mean 0 and variance \(\sigma^{2}\). The normality assumption implies that the \(F\) test is not distribution-free. We now assume that the \(\varepsilon\) 's all come from the same continuous, but not necessarily normal, distribution, and develop a distribution-free test of the null hypothesis that all \(I \mu_{i}\) 's are identical. Let \(N=\sum J_{i}\), the total number of observations in the data set (there are \(J_{i}\) observations in the \(i\) th sample). Rank these \(N\) observations from 1 (the smallest) to \(N\), and let \(\bar{R}_{i}\) be the average of the ranks for the observations in the ith sample. When \(H_{0}\) is true, we expect the rank of any particular observation and therefore also \(\bar{R}_{i}\) to be \((N+1) / 2\). The data argues against \(H_{0}\) when some of the \(\bar{R}_{i}\) 's differ considerably from \((N+1) / 2\). The Kruskal-Wallis test statistic is $$ K=\frac{12}{N(N+1)} \sum J_{i}\left(\bar{R}_{i}-\frac{N+1}{2}\right)^{2} $$ When \(H_{0}\) is true and either (1) \(I=3\), all \(J_{i} \geq 6\) or (2) \(I>3\), all \(J_{i} \geq 5\), the test statistic has approximately a chi-squared distribution with \(I-1\) df. The accompanying observations on axial stiffness index resulted from a study of metal-plate connected trusses in which five different plate lengths-4 in., 6 in., 8 in., 10 in., and 12 in. were used ("Modeling Joints Made with LightGauge Metal Connector Plates," Forest Products \(J ., 1979: 39-44)\). \(\begin{array}{lllll}i=1(4 \text { in. }): & 309.2 & 309.7 & 311.0 & 316.8 \\\ & 326.5 & 349.8 & 409.5 & \\ i=2(6 \text { in. }): & 331.0 & 347.2 & 348.9 & 361.0 \\ & 381.7 & 402.1 & 404.5 & \\ i=3(8 \text { in. }): & 351.0 & 357.1 & 366.2 & 367.3 \\ & 382.0 & 392.4 & 409.9 & \\ i=4(10 \text { in. }): & 346.7 & 362.6 & 384.2 & 410.6 \\ & 433.1 & 452.9 & 461.4 & \\ i=5(12 \text { in. }): & 407.4 & 410.7 & 419.9 & 441.2 \\ & 441.8 & 465.8 & 473.4 & \end{array}\) Use the \(K-W\) test to decide at significance level \(.01\) whether the true average axial stiffness index depends somehow on plate length.

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The article "Measuring the Exposure of Infants to Tobacco Smoke" (New Engl. J. Med., 1984: 1075-1078) reports on a study in which various measurements were taken both from a random sample of infants who had been exposed to household smoke and from a sample of unexposed infants. The accompanying data consists of observations on urinary concentration of cotinine, a major metabolite of nicotine (the values constitute a subset of the original data and were read from a plot that appeared in the article). Does the data suggest that true average cotinine level is higher in exposed infants than in unexposed infants by more than 25 ? Carry out a test at significance level \(.05\).\begin{array}{lrrrrrrrr} \text { Unexposed } & 8 & 11 & 12 & 14 & 20 & 43 & 111 & \\ \text { Exposed } & 35 & 56 & 83 & 92 & 128 & 150 & 176 & 208 \end{array}

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