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Find the critical values \(\chi_{1-\alpha / 2}^{2}\) and \(\chi_{\alpha / 2}^{2}\) for the given level of confidence and sample size. \(98 \%\) confidence, \(n=23\)

Short Answer

Expert verified
χ_{1-\text{α/2}}^{2} = 40.289 and χ_{\text{α/2}}^{2} = 8.643.

Step by step solution

01

Determine the Degrees of Freedom

The degrees of freedom (df) for this problem is calculated using the formula: \[ \text{df} = n - 1 \] Given that the sample size (\text{n}) is 23, the degrees of freedom will be \[ \text{df} = 23 - 1 = 22 \]
02

Determine the Alpha Value

The confidence level is given as 98%. To find the alpha (\text{α}) value, use the formula: \[ \text{α} = 1 - \text{confidence level} \] Thus, \[ \text{α} = 1 - 0.98 = 0.02 \]
03

Find Critical Values

Determine the critical values \text{χ^2} for \text{1-α/2} and \text{α/2} using chi-square tables or relevant statistical tools. For \text{α/2 = 0.01} and \text{1-α/2 = 0.99} with \text{df = 22}, the critical values are: \[\text{χ_{1-α/2}^{2}} = χ_{0.99}^{2} \text{(with 22 degrees of freedom)} = 40.289 \]\[\text{χ_{\text{α/2}}^{2}} = χ_{0.01}^{2} \text{(with 22 degrees of freedom)} = 8.643 \]

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Key Concepts

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

Degrees of Freedom
Degrees of freedom (df) are a crucial concept in statistical analysis. The term 'degrees of freedom' generally refers to the number of values in a calculation that are free to vary. In the context of chi-square distribution, it helps to determine the shape of the chi-square distribution curve.
For any given dataset, you can calculate the degrees of freedom using the formula:
\( \text{df} = n - 1 \)
where \( n \) is the sample size.
For example, in a sample size of 23 (\( n = 23 \)), the degrees of freedom would be
\( 22 \) because \( 23 - 1 = 22 \).
Knowing the degrees of freedom is essential for locating the right critical values in a chi-square distribution table. It also ensures that your statistical calculations are accurate.
Always remember to subtract one from your sample size to get the degrees of freedom.
Confidence Level
The confidence level is an indicator of how certain we are that a given interval contains the true population parameter. It's expressed as a percentage, and in this instance, it is given as 98%.
To compute the alpha (\( \text{α} \)) value, you use the formula:
\(α = 1 - \text{confidence level} \)
For a 98% confidence level, we get:
\( α = 1 - 0.98 = 0.02 \)
The confidence level also plays a role when we determine the critical values of a chi-square distribution since it divides the alpha value between the two tails of the distribution.
A higher confidence level means you are more confident that the interval includes the population parameter, but it also means a wider interval. Conversely, a lower confidence level results in a narrower interval but with less confidence that it encloses the parameter.
Chi-Square Distribution
The chi-square distribution is a statistical distribution commonly used in hypothesis testing and setting confidence intervals for a population variance. With the proper usage of degrees of freedom and the alpha value, critical values can be determined from the chi-square distribution table.
For example, in the given problem, the critical values \( \text{χ_{1-α/2}^{2}} \) and \( \text{χ_{α/2}^{2}} \) need to be found for a 98% confidence level and a sample size of 23. First, calculate the degrees of freedom (\( \text{df} = 22 \)) and then find the alpha value (\( α = 0.02 \)).
The \( α/2 \) value will be:
\( \text{α/2} = 0.01 \)
The \( 1-α/2 \) value will be:
\( 1-α/2 = 0.99 \)
Now, using the chi-square table or a statistical tool, find the critical values for the given degrees of freedom.
The critical values will be:
\[ \text{χ_{1-α/2}^{2}} = 40.289 \] \[ \text{χ_{\text{α/2}}^{2}} = 8.643 \]
These values demarcate the tails of the chi-square distribution, and values within them indicate a high probability that the true population parameter lies within your confidence interval.

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Most popular questions from this chapter

The exponential probability distribution can be used to model waiting time in line or the lifetime of electronic components. Its density function is skewed right. Suppose the wait-time in a line can be modeled by the exponential distribution with \(\mu=\sigma=5\) minutes. (a) Use StatCrunch, Minitab, or some other statistical software to generate 100 random samples of size \(n=4\) from this population. (b) Construct \(95 \% t\) -intervals for each of the 100 samples found in part (a). (c) How many of the intervals do you expect to include the population mean? How many of the intervals actually contain the population mean? Explain what your results mean. (d) Repeat parts (a)-(c) for samples of size \(n=15\) and \(n=25\). Explain what your results mean.

The American Society for Microbiology (ASM) and the Soap and Detergent Association (SDA) jointly commissioned two separate studies, both of which were conducted by Harris Interactive. In one of the studies, 1001 adults were interviewed by telephone and asked about their handwashing habits. In the telephone interviews, 921 of the adults said they always wash their hands in public restrooms. In the other study, the hand-washing behavior of 6076 adults was inconspicuously observed within public restrooms in four U.S. cities and 4679 of the 6076 adults were observed washing their hands. (a) In the telephone survey, what is the variable of interest? Is it qualitative or quantitative? (b) What is the sample in the telephone survey? What is the population to which this study applies? (c) Verify that the requirements for constructing a confidence interval for the population proportion of adults who say they always wash their hands in public restrooms are satisfied. (d) Using the results from the telephone interviews, construct a \(95 \%\) confidence interval for the proportion of adults who say they always wash their hands in public restrooms. (e) In the study where hand-washing behavior was observed, what is the variable of interest? Is it qualitative or quantitative? (f) We are told that 6076 adults were inconspicuously observed, but were not told how these adults were selected. We know randomness is a key ingredient in statistical studies that allows us to generalize results from a sample to a population. Suggest some ways randomness might have been used to select the individuals in this study. (g) Verify the requirements for constructing a confidence interval for the population proportion of adults who actually washed their hands while in a public restroom. (h) Using the results from the observational study, construct a \(95 \%\) confidence interval for the proportion of adults who wash their hands in public restrooms. (i) Based on your findings in parts (a) through (h), what might you conclude about the proportion of adults who say they always wash their hands versus the proportion of adults who actually wash their hands in public restrooms? (j) Cite some sources of variability in both studies.

True or False: The value of \(t_{0.10}\) with 5 degrees of freedom is greater than the value of \(t_{0.10}\) with 10 degrees of freedom.

Indicate whether a confidence interval for a proportion or mean should be constructed to estimate the variable of interest. Justify your response. A developmental mathematics instructor wishes to estimate the typical amount of time students dedicate to studying mathematics in a week. She asks a random sample of 50 students enrolled in developmental mathematics at her school to report the amount of time spent studying mathematics in the past week.

As the number of degrees of freedom in the \(t\) -distribution increases, the spread of the distribution ________ (increases/decreases).

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