/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 The following data show the numb... [FREE SOLUTION] | 91Ó°ÊÓ

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The following data show the number of occupants in passenger cars observed during one hour at a busy intersection in Los Angeles (75). Suppose it can be assumed that these data follow a geometric distribution, \(p_{x}(k ; p)=(1-p)^{k-1} p, k=1,2, \ldots\) Estimate \(p\) and compare the observed and expected frequencies for each value of \(X\). $$ \begin{array}{cc} \hline \text { Number of Occupants } & \text { Frequency } \\ \hline 1 & 678 \\ 2 & 227 \\ 3 & 56 \\ 4 & 28 \\ 5 & 8 \\ 6+ & \frac{14}{1011} \\ \hline \end{array} $$

Short Answer

Expert verified
First, estimate the parameter 'p' of the geometric distribution by dividing the frequency of cars with one occupant by the total frequency. Then, calculate the expected frequencies given this geometric distribution for each 'k', and compare these with the actually observed frequencies.

Step by step solution

01

Calculate total frequency

First, you need to calculate the total frequency, which is the sum of all frequencies. This can be done by adding up all frequencies given in the table: 678 + 227 + 56 + 28 + 8 + (14/1011).
02

Find the expected frequency

Next, calculate the expected frequencies for each 'k'. Under the geometric distribution assumption, the expected frequency is given by \(N * p_{x}(k ; p)\), where N is the total frequency obtained in Step 1, and \(p_{x}(k ; p)\) is the likelihood of observing a car with 'k' occupants derived from the geometric distribution formula. However, you would first need to obtain 'p' to compute this.
03

Estimate the parameter 'p'

In a geometric distribution, 'p' is the likelihood of the first success. Considering the given data, a 'success' is defined as finding a car with just one occupant. Therefore, the estimate of 'p' can be obtained by dividing the frequency of cars with one occupant, 678, by the total frequency obtained in Step 1.
04

Compute Expected Frequencies

Now that you have an estimate for 'p', you can compute the expected frequencies for each 'k'. Again, the expected frequency for a given 'k' is \(N * p_{x}(k ; p)\), where \(p_{x}(k ; p) = (1-p)^{k-1} p\), the geometric distribution formula.
05

Compare observed and expected frequencies

Finally, compare the observed frequencies provided in the table with the expected frequencies calculated in Step 4 for each 'k'. This provides a performance measure of the geometric distribution model in fitting the observed data.

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Key Concepts

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

Statistical Estimation
Statistical estimation plays a crucial role in analyzing data and making inferences about a population based on a sample. Estimation involves using sample data to calculate an estimate of an unknown population parameter. In the context of the geometric distribution, the parameter of interest is the probability 'p' of observing the first success on a given trial.

For our exercise, we need to estimate 'p', which represents the likelihood of observing a car with only one occupant at the intersection. The method employed in the exercise to estimate 'p' is the Maximum Likelihood Estimation (MLE), which determines the value of 'p' that makes the observed data most probable. The estimate is calculated by dividing the observed frequency of the first success (in this case, cars with one occupant) by the total number of observations. This estimated value is then used to compute the expected frequencies for each other number of car occupants.

Understanding estimation is vital because it represents a bridge between sample data and the whole population, enabling researchers and statisticians to understand the broader implications of their observations. A clear grasp of statistical estimation principles is beneficial for students to accurately interpret results and make educated decisions based on data.

Expected Frequency
The concept of expected frequency refers to the predicted number of times an event is anticipated to occur in a specific number of trials. It's a central element in probability and statistics and is calculated based on the probability of the event and the total number of trials.

In the given geometric distribution exercise, expected frequency is computed using the estimated probability 'p' and the total frequency of observed events. For instance, with our estimated value of 'p', the expected frequency for each category of car occupancy (from one occupant to six or more occupants) is determined by the formula: \(N * p_{x}(k; p)\), where 'N' is the total frequency, and \(p_{x}(k; p)\) is the geometric probability function. The expected frequencies enable us to predict how often we would expect to see cars with various occupancies if the geometric distribution model is a good fit for our data.

Students often find it easier to understand expected frequencies when connected to concrete examples such as this one. By relating this concept to observed events, like the number of car occupants, it helps to solidify their understanding of how often events are expected to occur under a given probability model.

Observed Frequency
In contrast to expected frequency, observed frequency is the actual count of occurrences that have been documented within a sample. It reflects real-world measurements rather than theoretical predictions. In a perfect model, observed frequencies would match expected frequencies, but in practice, this rarely happens due to natural variability and sampling error.

The exercise involves recording the number of cars with various occupant counts passing through an intersection. This real data is then tabulated, with each category (from one occupant to six or more) having a corresponding observed frequency. Comparing the observed frequencies to the expected frequencies calculated from the geometric distribution allows students to assess the model's accuracy in describing the phenomena in question.

Appreciating the difference between observed and expected frequencies equips students with critical analytical skills. Being able to scrutinize how well a statistical model fits observed data is essential for validating study results and for understanding the limitations and potential inaccuracies of statistical models.

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Most popular questions from this chapter

Suppose \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample of size \(n\) drawn from a Poisson pdf where \(\lambda\) is an unknown parameter. Show that \(\hat{\lambda}=\bar{X}\) is unbiased for \(\lambda\). For what type of parameter, in general, will the sample mean necessarily be an unbiased estimator? (Hint: The answer is implicit in the derivation showing that \(\bar{X}\) is unbiased for the Poisson \(\lambda\).)

. Calculate the method of moments estimate for the parameter \(\theta\) in the probability function $$ p_{X}(k ; \theta)=\theta^{k}(1-\theta)^{1-k}, \quad k=0,1 $$ if a sample of size 5 is the set of numbers \(0,0,1,0,1\).

. The exponential pdf is a measure of lifetimes of devices that do not age (see Question 3.11.11). However, the exponential pdf is a special case of the Weibull distribution, which can measure time to failure of devices where the probability of failure increases as time does. A Weibull random variable \(Y\) has pdf \(f_{Y}(y ; \alpha, \beta)=\alpha \beta y^{8-1} e^{-\alpha y^{2}}\), \(0 \leq y, 0<\alpha, 0<\beta\). Find the maximum likelihood estimator for \(\alpha\) assuming that \(\beta\) is known.

A random sample of size \(n\) is taken from the pdf $$ f_{Y}(y ; \theta)=\frac{2 y}{\theta^{2}}, \quad 0 \leq y \leq \theta $$ Find an expression for \(\hat{\theta}\), the maximum likelihood estimator for \(\theta\).

Find the maximum likelihood estimate for \(\theta\) in the pdr $$ f y(y ; \theta)=\frac{2 y}{1-\theta^{2}}, \quad \theta \leq y \leq 1 $$ if a random sample of size 6 yielded the measurements \(0.70,0.63,0.92,0.86,0.43\), and \(0.21\).

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