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Define the sets \(A_{1}=\\{x:-\infty

Short Answer

Expert verified
Without specific numeric values, it's impossible to provide a concrete short answer. But the process has been described. After you calculate the chi-square test statistic and find the critical value from the chi-square distribution table for \(7\) degrees of freedom and \(5\%\) significance level, compare these two numbers. If the test statistic is less than the critical value, then \(H_{0}\) would be accepted. Otherwise, it would be rejected.

Step by step solution

01

Calculate Expected Probabilities

First, it's needed to compute the expected probability for each set \(A_{i}\) using the given formula \(p_{i 0}=\int_{A_{i}} \frac{1}{2 \sqrt{2 \pi}} \exp\left[-\frac{(x-3)^{2}}{2(4)}\right] d x\). This requires computing an integral for each set \(A_{i}\).
02

Calculate Expected Frequencies

After calculating expected probabilities for each set \(A_{i}\), multiply each probability by the total number of observations to get the expected frequency for each set. Sum those results to ensure it equals the total observed frequency.
03

Calculate Chi-Square Statistic

After calculating expected frequencies, the next step is to compute the Chi-square statistic. It's done by using the formula \(\chi^2= \sum_{i=1}^{8} \frac{(Observed_{i} - Expected_{i})^2}{Expected_{i}}\), where Observed_{i} and Expected_{i} stand for observed and expected frequencies for each set \(A_{i}\).
04

Find Critical Value

The critical value of the chi-square distribution needed for a decision about the null hypothesis is found using significance level \(\alpha=0.05\) and degrees of freedom which is \(k-1=8-1=7\). Use a chi-square distribution table or statistical software to find this value.
05

Make a Decision About the Hypothesis

To conclude, compare the computed Chi-square statistic with the critical value. If the statistic is below the critical value, then it's recommended to accept the null hypothesis \(H_{0}\). If it's above, then it should be rejected.

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Key Concepts

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

Multinomial Distribution
When dealing with probabilities and distributions, the multinomial distribution plays a key role, especially when multiple outcomes are possible. Imagine you are tossing a dice; instead of just two outcomes like heads or tails, you have six possible results. This is where the multinomial distribution comes in handy.

In the context of the exercise, we have multiple sets of values, as described by the intervals of \(A_i\). Each of these intervals is assigned a probability \(p_{i0}\), and collectively, these probabilities are expected to follow a multinomial distribution. This means each probability specifies the likelihood of each outcome in a finite number of categories.

Key points include:
  • Each probability corresponds to a specific category or set.
  • The sum of all probabilities in a multinomial distribution equals 1.
  • These probabilities help in calculating expected outcomes, which are essential for hypothesis testing.
Understanding how to calculate these probabilities through integration is fundamental in testing our hypothesis and reaching meaningful conclusions.
Hypothesis Testing
Hypothesis testing is a statistical method that allows us to make decisions or inferences about populations based on sample data. We set up two hypotheses: the null hypothesis \(H_0\) and the alternative hypothesis \(H_a\).

The null hypothesis \(H_0\) usually suggests no effect or no difference, and it serves as the statement we wish to test. In our exercise, \(H_0\) is about the assignment of probabilities \(p_{i0}\) to our sets \(A_i\).

Key steps in hypothesis testing include:
  • Defining the hypotheses clearly.
  • Choosing a suitable test, such as the chi-square test.
  • Calculating the test statistic using observed and expected probabilities.
  • Making a decision based on the comparison of the test statistic to critical values.
When performing the test, if the test statistic falls in a region that suggests low probability of occurrence under \(H_0\), we reject the null hypothesis. This process allows us to use data to validate or refute claims statistically, which is critical in research and decision-making.
Level of Significance
The level of significance, often denoted by \(\alpha\), is a threshold used in hypothesis testing. It quantifies the probability of rejecting the null hypothesis when it is actually true, known as the Type I error.

In our exercise, we use a \(5\%\) level of significance. This means we are willing to accept a \(5\%\) chance of incorrectly rejecting the null hypothesis. In statistical terms:
  • \(\alpha = 0.05\) is chosen as a common standard, representing a balance between rigor and practicality.
  • This level determines the critical value from the chi-square distribution table for test evaluation.
  • It provides a cushion against errors, ensuring decisions are made with a known level of certainty.
The choice of \(\alpha\) is pivotal. A lower \(\alpha\) indicates stricter testing, while a higher one may allow for more flexibility in rejecting \(H_0\). By following this critical level, we safeguard the integrity and reliability of our statistical conclusions.

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Most popular questions from this chapter

Let \(f(x)=\frac{1}{6}, x=1,2,3,4,5,6\), zero elsewhere, be the pmf of a distribution of the discrete type. Show that the pmf of the smallest observation of a random sample of size 5 from this distribution is $$ g_{1}\left(y_{1}\right)=\left(\frac{7-y_{1}}{6}\right)^{5}-\left(\frac{6-y_{1}}{6}\right)^{5}, \quad y_{1}=1,2, \ldots, 6 $$ zero elsewhere. Note that in this exercise the random sample is from a distribution of the discrete type. All formulas in the text were derived under the assumption that the random sample is from a distribution of the continuous type and are not applicable. Why?

Assume a binomial model for a certain random variable. If we desire a \(90 \%\) confidence interval for \(p\) that is at most \(0.02\) in length, find \(n\). Hint: Note that \(\sqrt{(y / n)(1-y / n)} \leq \sqrt{\left(\frac{1}{2}\right)\left(1-\frac{1}{2}\right)}\).

It is known that a random variable \(X\) has a Poisson distribution with parameter \(\mu\). A sample of 200 observations from this distribution has a mean equal to 3.4. Construct an approximate \(90 \%\) confidence interval for \(\mu\).

Let \(Y_{1}

Let \(p\) denote the probability that, for a particular tennis player, the first serve is good. Since \(p=0.40\), this player decided to take lessons in order to increase \(p\). When the lessons are completed, the hypothesis \(H_{0}: p=0.40\) is tested against \(H_{1}: p>0.40\) based on \(n=25\) trials. Let \(y\) equal the number of first serves that are good, and let the critical region be defined by \(C=\\{y: y \geq 13\\}\). (a) Determine \(\alpha=P(Y \geq 13 ; p=0.40)\). (b) Find \(\beta=P(Y<13)\) when \(p=0.60\); that is, \(\beta=P(Y \leq 12 ; p=0.60)\) so that \(1-\beta\) is the power at \(p=0.60\).

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