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Use the Kolmogorov-Smirnov Statistic to find a \(95 \%\) confidence interval for \(\mathrm{F}(\mathrm{x}) . \mathrm{F}(\mathrm{x})\) is the cumulative distribution function of a population from which the following ordered samples was taken: \(8.2,10.4,10.6,11.5,12.6,12.9\), \(13.3,13.3,13.4,13.4,13.6,13.8,14.0,14.0,14.1,14.2\) \(14.6,14.7,14.9,15.0,15.4,15.6,15.9,16.0,16.2,16.3\) 17.2,17.4,17.7,18.1 .$

Short Answer

Expert verified
Unfortunately, the problem statement is incomplete and does not provide enough information to solve the problem. It requires information about the specific distribution of F(x) to compare with the given ordered sample data. If this information were provided, we could proceed with computing the Kolmogorov-Smirnov statistic, D, finding its critical value, and determining the 95% confidence interval for F(x).

Step by step solution

01

Determine the number of data points

Count the number of data points in the given ordered sample: There are 34 data points in the given sample.
02

Calculate the ECDF

Compute the sample cumulative distribution function for each data point: For each data point x in the sample, calculate the proportion of data points less than or equal to x. This proportion represents the ECDF(x).
03

Compute the Kolmogorov-Smirnov statistic D

The Kolmogorov-Smirnov statistic D is the maximum absolute difference between the ECDF(x) and the specified F(x): \[D = max|ECDF(x) - F(x)|\] Since we don't have a specified distribution F(x) in the exercise, we can't proceed with this step.
04

Find the critical value of D

We can't proceed with calculating the critical value of D since we don't have the Kolmogorov-Smirnov statistic from the previous step.
05

Determine the 95% confidence interval

We cannot determine the 95% confidence interval for F(x) as the exercise provides an incomplete problem statement. It requires information about a specific distribution to compare with the ordered sample, which is not provided. If given such information, we could proceed with Steps 3 and 4 then determine the confidence interval for F(x).

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Key Concepts

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

Cumulative Distribution Function
The cumulative distribution function (CDF) is a fundamental concept in statistics, representing the probability that a random variable takes on a value less than or equal to a specific value.

Imagine rolling a six-sided die and wanting to know the probability of rolling a 4 or less. To find this, one can look at the CDF of the die's outcomes. The CDF in this case would plot a rising curve, where at each point on the x-axis, representing a die number, the corresponding y-axis value shows the probability of rolling that number or less.

Mathematically, the CDF at any given point x is expressed as:\[ F(x) = P(X \leq x) \]where X is the random variable, and P stands for probability. It's a non-decreasing function that ranges from 0 to 1, providing a complete description of the probability distribution of a real-valued random variable.

The CDF is valuable for understanding the distribution and obtaining probabilities for intervals. For continuous variables, the area under the CDF curve between two values gives the probability of the random variable falling within that range.
Confidence Interval
When we talk about confidence intervals (CIs), we're dealing with an estimated range of values that's likely to include an unknown population parameter with a given probability. It expresses the degree of uncertainty associated with the estimate.

Suppose a study reports that the average height of a population is 170 cm with a 95% CI of 165 cm to 175 cm. This means that we can be 95% confident that the true average height is between 165 cm and 175 cm.

In constructing a CI, the following steps are generally followed:
  • Identify the sample statistic (mean, proportion, etc.)
  • Decide the confidence level (commonly 95% or 99%)
  • Calculate the standard error (variation of the sample statistic)
  • Calculate the margin of error using a critical value from a statistical distribution
  • Add and subtract the margin of error from the sample statistic to define the CI range

The '95%' in a 95% confidence interval refers to the idea that, if we were to take 100 different samples and compute a 95% confidence interval for each sample, then approximately 95 of those confidence intervals will contain the true population parameter.
Empirical Cumulative Distribution Function (ECDF)
The empirical cumulative distribution function (ECDF) provides a step-wise probability function derived from empirical data. Unlike the theoretical CDF, which is smooth and based on a known distribution, the ECDF represents the data you have collected from an experiment or survey, not an underlying probability model.

The ECDF is particularly helpful when one doesn't have the underlying theoretical model of distribution that generated the data. Here's how it works:
  • First, sort your sample data from smallest to largest values.
  • Next, plot these observed values on the x-axis of a graph.
  • For each x value, calculate the proportion of sample observations less than or equal to x; plot this as the y value.
  • Connect these points to form a step function that increases at the value of each observed sample point.

This ECDF can then be used to approximate probabilities for the underlying population distribution or to compare with a theoretical CDF for a goodness-of-fit test, such as the Kolmogorov-Smirnov test.
Statistical Hypothesis Testing
Statistical hypothesis testing is a formal method used to make a decision about a population based on sample data. It involves making an initial assumption, called the null hypothesis, and working out whether the observed data provides sufficient evidence to reject that hypothesis in favor of an alternative hypothesis.

For example, if we want to test whether a coin is fair, we would start with the null hypothesis that the coin is fair and has a 50% chance of landing heads up. We then flip the coin a number of times and observe the results. If the results deviate significantly from what we would expect with a fair coin, we might reject the null hypothesis in favor of an alternative: that the coin is not fair.

Test statistics, like the t-statistic or z-score, are calculated from sample data and are used to determine the p-value – the probability of observing the given result if the null hypothesis were true. A small p-value suggests that the observed data is unlikely under the null hypothesis, leading to its rejection.

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Most popular questions from this chapter

Given the probability distribution of the random variable \(\mathrm{X}\) in the table below, compute \(\mathrm{E}(\mathrm{X})\) and \(\operatorname{Var}(\mathrm{X})\). $$ \begin{array}{|c|c|} \hline \mathrm{x}_{\mathrm{i}} & \operatorname{Pr}\left(\mathrm{X}=\mathrm{x}_{i}\right) \\ \hline 0 & 8 / 27 \\ \hline 1 & 12 / 27 \\ \hline 2 & 6 / 27 \\ \hline 3 & 1 / 27 \\ \hline \end{array} $$

In a normal distribution, what is the \(Z\) -score equivalent of the median? What is the \(Z\) score above which only 16 percent of the distribution lies? What percentage of the scores lie below a \(Z\) score of \(+2.0 ?\)

A lot consisting of 100 fuses, is inspected by the following Procedure. Five of these fuses are chosen at random and tested; if all 5 "blow" at the correct amperage, the lot is accepted. Find the probability distribution of the number of defectives in a sample of 5 assuming there are 20 in the lot.

Suppose we have a binomial distribution for which \(\mathrm{H}_{0}\) is \(\mathrm{p}=1 / 2\) where \(\mathrm{p}\) is the probability of success on a single trial. Suppose the type I error, \(\alpha=.05\) and \(\mathrm{n}=100 .\) Calculate the power of this test for each of the following alternate hypotheses, \(\mathrm{H}_{1}: \mathrm{p}=.55, \mathrm{p}=.60, \mathrm{p}=.65, \mathrm{p}=.70\), and \(\mathrm{p}=.75 .\) Do the same when \(\alpha=.01\).

A sports magazine reports that the people who watch Monday night football games on television are evenly divided between men and women. Out of a random sample of 400 people who regularly watch the Monday night game, 220 are men. Using a \(.10\) level of significance, can be conclude that the report is false?

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