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Chapter 9: Problem 8

Find the critical values \(\chi_{1-\alpha / 2}^{2}\) and \(\chi_{\alpha / 2}^{2}\) for the given level of confidence and sample size. \(99 \%\) confidence, \(n=14\)

Short Answer

Expert verified
\( \chi_{0.005}^{2} = 2.167 \) and \( \chi_{0.995}^{2} = 32.357 \)

Step by step solution

01

- Understand the problem

Find the critical values \( \chi_{1-\alpha / 2}^{2} \) and \( \chi_{\alpha / 2}^{2} \) for a 99% confidence level and sample size n=14.
02

- Determine the degrees of freedom

Degrees of freedom (df) are given by \( n - 1 \). For a sample size \( n = 14 \), \( df = 14 - 1 = 13 \).
03

- Find the significance level

The confidence level is 99%, so the significance level \( \alpha = 1 - 0.99 = 0.01 \).
04

- Calculate the critical values

Find \( \chi_{\alpha / 2}^{2} \) and \( \chi_{1-\alpha / 2}^{2} \) using a chi-square table, with \( \alpha = 0.01 \) and \( df = 13 \). \( \alpha / 2 = 0.005 \). Find \( \chi_{0.005}^{2} \) and \( \chi_{0.995}^{2} \).
05

- Use the chi-square distribution table

From the chi-square table for \( df = 13 \): \( \chi_{0.005}^{2} = 2.167 \) and \( \chi_{0.995}^{2} = 32.357 \).

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Key Concepts

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

Confidence Level
The confidence level in a statistical analysis tells us how confident we can be in our results. Specifically, it represents the percentage of times a calculated interval would include the true population parameter if we repeated our study multiple times.

For example, with a 99% confidence level, we expect that 99 out of 100 calculated intervals will contain the true population parameter. In other words, we are 99% sure our interval captures the truth.

The higher the confidence level, the more sure we are about our results. Common confidence levels include 90%, 95%, and 99%.

In our problem, we used a 99% confidence level which is very high. This high confidence level implies a low risk of being wrong.
Degrees of Freedom
Degrees of freedom (df) refer to the number of values in a calculation that are free to vary. It's a crucial concept in statistics, especially when working with distributions.

When you calculate the degrees of freedom for a sample, you generally use the formula:
\[ df = n - 1 \]

Here, \( n \) is the sample size. In our example, with a sample size of 14, the degrees of freedom are:
\[ df = 14 - 1 = 13 \]

This degrees of freedom value helps us use the chi-square distribution table to find critical values. The higher the degrees of freedom, the closer the chi-square distribution gets to a normal distribution.
Significance Level
The significance level (alpha, \( \alpha \)) is used to define the probability of rejecting the null hypothesis when it is true, also known as Type I error.

The significance level complements the confidence level since:
\[ \alpha = 1 - \text{Confidence Level} \]

In our example, with a 99% confidence level, the significance level is:
\[ \alpha = 1 - 0.99 = 0.01 \]

We further split this \( \alpha \) value to find the critical values on both tails of the chi-square distribution. So:
\[ \alpha / 2 = 0.005 \]

Using these values, we then refer to a chi-square distribution table to obtain our critical values: \( \chi_{0.005}^{2} \) and \( \chi_{0.995}^{2} \). These values are crucial in hypothesis testing and help us decide whether to reject the null hypothesis.

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Most popular questions from this chapter

In a survey of 1008 adult Americans, the Gallup organization asked, "When you retire, do you think you will have enough money to live comfortably or not?" Of the 1008 surveyed, 526 stated that they were worried about having enough money to live comfortably in retirement. Construct a \(90 \%\) confidence interval for the proportion of adult Americans who are worried about having enough money to live comfortably in retirement.

Suppose the following data represent the heights (in inches) of a random sample of males: 68,72,73,70,75,71 . Which of the following could be a possible bootstrap sample? (a) 68,72,72,68,70,71 (b) 75,72,73,73,68 (c) 70,71,68,73,73,71,68 (d) 72,73,75,71,73,63 (e) 68,72,73,71,72,73

The data in the next column represent the age (in weeks) at which babies first crawl based on a survey of 12 mothers conducted by Essential Baby. In Problem 34 from Section \(9.2,\) it was verified that the data are normally distributed and that \(s=10.00\) weeks. Construct and interpret a \(95 \%\) confidence interval for the population standard deviation of the age (in weeks) at which babies first crawl. $$ \begin{array}{llllll} \hline 52 & 30 & 44 & 35 & 39 & 26 \\ \hline 47 & 37 & 56 & 26 & 39 & 28 \\ \hline \end{array} $$

IQ scores are known to be approximately normally distributed with mean 100 and standard deviation \(15 .\) (a) Simulate obtaining a random sample of 12 IQ scores from this population. (b) Use the data from part (a) to construct a \(95 \%\) confidence interval for the mean IQ using Student's \(t\) -distribution. (c) Use the data from part (a) to obtain 1000 bootstrap samples. For each sample, find the mean. (d) Determine an estimate of the standard error of the mean from the 1000 bootstrap means found in part (c). Compare this result to the theoretical standard error of the mean, \(\frac{\sigma}{\sqrt{n}}\). Compare this result to the estimate of the standard error of the mean based on the sample data, \(\frac{s}{\sqrt{n}}\). (e) Construct a \(95 \%\) confidence interval for the mean IQ using the bootstrap sample from part (c).

(a) Find the \(t\) -value such that the area in the right tail is 0.02 with 19 degrees of freedom. (b) Find the \(t\) -value such that the area in the right tail is 0.10 with 32 degrees of freedom. (c) Find the \(t\) -value such that the area left of the \(t\) -value is 0.05 with 6 degrees of freedom. [Hint: Use symmetry.] (d) Find the critical \(t\) -value that corresponds to \(95 \%\) confidence. Assume 16 degrees of freedom.
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