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Determine the number of degrees of freedom for the twosample \(t\) test or \(\mathrm{CI}\) in each of the following situations: a. \(m=10, n=10, s_{1}=5.0, s_{2}=6.0\) b. \(m=10, n=15, s_{1}=5.0, s_{2}=6.0\) c. \(m=10, n=15, s_{1}=2.0, s_{2}=6.0\) d. \(m=12, n=24, s_{1}=5.0, s_{2}=6.0\)

Short Answer

Expert verified
a: 17.44, b: 21.72, c: 18.27, d: 26.07

Step by step solution

01

Understanding Degrees of Freedom

In a two-sample t-test, the degrees of freedom (df) can be calculated in several ways depending on assumptions about the population variances. When variances are assumed to be unequal, we use the Welch-Satterthwaite approximation for df. This is calculated using a specific formula.
02

Formula for Degrees of Freedom

The formula for calculating the degrees of freedom under Welch's approximation is: \[df = \frac{\left(\frac{s_1^2}{m} + \frac{s_2^2}{n}\right)^2}{\frac{\left(\frac{s_1^2}{m}\right)^2}{m - 1} + \frac{\left(\frac{s_2^2}{n}\right)^2}{n - 1}}\]where \(m\) and \(n\) are the sample sizes, and \(s_1\) and \(s_2\) are the sample standard deviations.
03

Calculate Degrees of Freedom for Situation (a)

For (a) with \(m = 10, n = 10, s_1 = 5.0, s_2 = 6.0\):First, calculate the numerator:\[\frac{s_1^2}{m} = \frac{5.0^2}{10} = 2.5, \quad \frac{s_2^2}{n} = \frac{6.0^2}{10} = 3.6\]\[Numerator = \left(2.5 + 3.6\right)^2 = 6.1^2 = 37.21\]Then, calculate the denominator:\[\frac{\left(\frac{5.0^2}{10}\right)^2}{9} = \frac{2.5^2}{9} \approx 0.6944, \quad \frac{\left(\frac{6.0^2}{10}\right)^2}{9} = \frac{3.6^2}{9} \approx 1.44\]\[Denominator = 0.6944 + 1.44 = 2.1344\]Finally, compute the degrees of freedom:\[df = \frac{37.21}{2.1344} \approx 17.44\]
04

Calculate Degrees of Freedom for Situation (b)

For (b) with \(m = 10, n = 15, s_1 = 5.0, s_2 = 6.0\):First, calculate the numerator:\[\frac{s_1^2}{m} = \frac{5.0^2}{10} = 2.5, \quad \frac{s_2^2}{n} = \frac{6.0^2}{15} = 2.4\]\[Numerator = \left(2.5 + 2.4\right)^2 = 4.9^2 = 24.01\]Then, calculate the denominator:\[\frac{\left(\frac{5.0^2}{10}\right)^2}{9} = 0.6944, \quad \frac{\left(\frac{6.0^2}{15}\right)^2}{14} = \frac{2.4^2}{14} \approx 0.4114\]\[Denominator = 0.6944 + 0.4114 = 1.1058\]Finally, compute the degrees of freedom:\[df = \frac{24.01}{1.1058} \approx 21.72\]
05

Calculate Degrees of Freedom for Situation (c)

For (c) with \(m = 10, n = 15, s_1 = 2.0, s_2 = 6.0\):First, calculate the numerator:\[\frac{s_1^2}{m} = \frac{2.0^2}{10} = 0.4, \quad \frac{s_2^2}{n} = \frac{6.0^2}{15} = 2.4\]\[Numerator = \left(0.4 + 2.4\right)^2 = 2.8^2 = 7.84\]Then, calculate the denominator:\[\frac{\left(\frac{2.0^2}{10}\right)^2}{9} = \frac{0.4^2}{9} = 0.0178, \quad \frac{\left(\frac{6.0^2}{15}\right)^2}{14} = 0.4114\]\[Denominator = 0.0178 + 0.4114 = 0.4292\]Finally, compute the degrees of freedom:\[df = \frac{7.84}{0.4292} \approx 18.27\]
06

Calculate Degrees of Freedom for Situation (d)

For (d) with \(m = 12, n = 24, s_1 = 5.0, s_2 = 6.0\):First, calculate the numerator:\[\frac{s_1^2}{m} = \frac{5.0^2}{12} = 2.0833, \quad \frac{s_2^2}{n} = \frac{6.0^2}{24} = 1.5\]\[Numerator = \left(2.0833 + 1.5\right)^2 = 3.5833^2 = 12.841\]Then, calculate the denominator:\[\frac{\left(\frac{5.0^2}{12}\right)^2}{11} = \frac{2.0833^2}{11} = 0.3948, \quad \frac{\left(\frac{6.0^2}{24}\right)^2}{23} = \frac{1.5^2}{23} \approx 0.0978\]\[Denominator = 0.3948 + 0.0978 = 0.4926\]Finally, compute the degrees of freedom:\[df = \frac{12.841}{0.4926} \approx 26.07\]

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Key Concepts

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

Degrees of Freedom
The concept of degrees of freedom is pivotal in statistical hypothesis testing, including the two-sample t-test. It can be imagined as the number of independent values or quantities that can vary when calculating a statistical estimate. For instance, if you have a dataset with 10 numbers, the degrees of freedom would typically be 9, because one of the numbers is usually used for calculating the mean, which fixes one quantity. In a two-sample t-test, determining the degrees of freedom is crucial. Calculations for degrees of freedom can vary based on assumptions about the population variances, leading to different statistical implications.
Understanding and calculating the degrees of freedom correctly helps to decide the critical value of the t-distribution, which in turn helps to determine the significance of your test results. This calculation often involves using the sample sizes and variance estimates of the two groups involved.
Welch-Satterthwaite Approximation
The Welch-Satterthwaite approximation is a technique used to adjust the degrees of freedom for the two-sample t-test when population variances are assumed unequal. Instead of using a straightforward calculation, the Welch-Satterthwaite method provides a more accurate estimate by taking into account the variances in each group. The formula is:
\[df = \frac{\left(\frac{s_1^2}{m} + \frac{s_2^2}{n}\right)^2}{\frac{\left(\frac{s_1^2}{m}\right)^2}{m - 1} + \frac{\left(\frac{s_2^2}{n}\right)^2}{n - 1}}\]
Here, \(s_1\) and \(s_2\) represent the sample standard deviations, and \(m\) and \(n\) are the respective sample sizes for the two groups. This method gives a more flexible degree of freedom calculation, allowing for more accurate hypothesis testing, especially when the assumption of equal variances is not reasonable.
It's especially useful when dealing with real-world data, where perfect equality of variances is often not observed and allows for more accurate conclusions and improved statistical power.
Population Variance
Population variance is a measure of the dispersion or spread of a set of data points in a population. It gives an overall picture of how much the data varies from the average value. Mathematically, it's defined as the average of the squared differences from the mean. Understanding population variance is critical in hypothesis testing, as it affects the reliability and validity of statistical conclusions.
When conducting a two-sample t-test, assumptions about population variance can influence the test used and the interpretation of results. If the variances of two populations are assumed to be unequal, methods like the Welch-Satterthwaite approximation come into play, offering a solution to adjust the degrees of freedom accordingly.
Variance is expressed in squared units of the data, and to get it back to the original units, we use the square root of variance, known as the standard deviation. For example, if the variance within a data set of exam scores is high, it shows that some students scored much higher or lower than the average score, indicating wide disparity. This provides insights into the distribution and helps in making generalizations about the population being studied.

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Most popular questions from this chapter

The article "Pine Needles as Sensors of Atmospheric Pollution" (Environ. Monitoring, 1982: 273-286) reported on the use of neutron-activity analysis to determine pollutant concentration in pine needles. According to the article's authors, "These observations strongly indicated that for those elements which are determined well by the analytical procedures, the distribution of concentration is lognormal. Accordingly, in tests of significance the logarithms of concentrations will be used." The given data refers to bromine concentration in needles taken from a site near an oil-fired steam plant and from a relatively clean site. The summary values are means and standard deviations of the log-transformed observations. \begin{tabular}{lccc} Site & Sample Size & Mean Log Concentration & SD of Log Concentration \\ \hline Steam plant & 8 & \(18.0\) & \(4.9\) \\ Clean & 9 & \(11.0\) & \(4.6\) \\ \hline \end{tabular} Let \(\mu_{1}^{*}\) be the true average log concentration at the first site, and define \(\mu_{2}^{*}\) analogously for the second site. a. Use the pooled \(t\) test (based on assuming normality and equal standard deviations) to decide at significance level .05 whether the two concentration distribution means are equal. b. If \(\sigma_{1}^{*}\) and \(\sigma_{2}^{*}\) (the standard deviations of the two log concentration distributions) are not equal, would \(\mu_{1}\) and \(\mu_{2}\) (the means of the concentration distributions) be the same if \(\mu_{1}^{*}=\mu_{2}^{*}\) ? Explain your reasoning.

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