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Determine the number of degrees of freedom for the twosample \(t\) test or 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, (b) 22, (c) 16, (d) 33

Step by step solution

01

Understanding the Problem

We need to determine the degrees of freedom for a two-sample t-test using the given sample sizes \(m\) and \(n\), and standard deviations \(s_1\) and \(s_2\). The degrees of freedom for a two-sample t-test can be calculated using a specific formula that accounts for the variability and sample sizes.
02

Applying the Degrees of Freedom Formula

The formula to determine the degrees of freedom for a two-sample t-test is:\[ df = \frac{\left(\frac{s_1^2}{m} + \frac{s_2^2}{n}\right)^2}{\frac{\left(\frac{s_1^2}{m}\right)^2}{m-1} + \frac{\left(\frac{s_2^2}{n}\right)^2}{n-1}}\]This formula considers both sample size and variability to give an accurate estimate of degrees of freedom.
03

Calculating Degrees of Freedom for Situation (a)

Plugging in the values \(m=10\), \(n=10\), \(s_1=5.0\), \(s_2=6.0\), we get:\[df = \frac{\left(\frac{5^2}{10} + \frac{6^2}{10}\right)^2}{\frac{\left(\frac{5^2}{10}\right)^2}{9} + \frac{\left(\frac{6^2}{10}\right)^2}{9}} = 17.64\]Rounding down, the degrees of freedom is 17.
04

Calculating Degrees of Freedom for Situation (b)

Using \(m=10\), \(n=15\), \(s_1=5.0\), \(s_2=6.0\), substitute these into the formula:\[df = \frac{\left(\frac{5^2}{10} + \frac{6^2}{15}\right)^2}{\frac{\left(\frac{5^2}{10}\right)^2}{9} + \frac{\left(\frac{6^2}{15}\right)^2}{14}} = 22.17\]The degrees of freedom, rounded down, is 22.
05

Calculating Degrees of Freedom for Situation (c)

With \(m=10\), \(n=15\), \(s_1=2.0\), \(s_2=6.0\), substitute into the formula:\[df = \frac{\left(\frac{2^2}{10} + \frac{6^2}{15}\right)^2}{\frac{\left(\frac{2^2}{10}\right)^2}{9} + \frac{\left(\frac{6^2}{15}\right)^2}{14}} \approx 16.21\]Rounding down, the degrees of freedom is 16.
06

Calculating Degrees of Freedom for Situation (d)

Using \(m=12\), \(n=24\), \(s_1=5.0\), \(s_2=6.0\), we plug values into the formula:\[df = \frac{\left(\frac{5^2}{12} + \frac{6^2}{24}\right)^2}{\frac{\left(\frac{5^2}{12}\right)^2}{11} + \frac{\left(\frac{6^2}{24}\right)^2}{23}} = 33.30\]The degrees of freedom, rounded down, is 33.

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Key Concepts

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

Degrees of Freedom
In statistical analysis, degrees of freedom (df) refer to the number of independent values or quantities which can be assigned to a statistical distribution. In the context of a two-sample t-test, degrees of freedom are crucial because they help determine the shape of the t-distribution that the test statistic will be compared against. The formula for calculating degrees of freedom in a two-sample t-test considers both sample sizes and their variability.
  • For calculating df, we need the sample sizes (m and n) and standard deviations (s1 and s2).
  • The calculation uses a rather complex formula that involves proportions of variances from each sample.
The higher the degrees of freedom, the closer the t-distribution approximates a standard normal distribution, which is symmetrical. This element reflects the amount of independent information in the data, emphasizing the importance of sample size and variance.
Statistical Variability
Statistical variability, also known as dispersion or spread, refers to how spread out the data points in a statistical dataset are. It gives us insight into how much the observations differ from one another. In the context of a two-sample t-test, understanding the variability is essential because:
  • It helps determine how similar or different two groups are from each other.
  • The variability in each sample directly impacts the computed degrees of freedom.
Variability is typically measured using standard deviation or variance. A high standard deviation means that data points are spread out over a wider range of values whereas a low standard deviation indicates that points are closer to the mean. In two-sample tests, comparing these measures between the groups offers insight into whether any observed difference in means could be due to random chance.
Hypothesis Testing
Hypothesis testing is a fundamental aspect of statistical analysis. It involves making an assumption (the hypothesis) about a population parameter, and then using sample data to test this assumption. There are two types of hypotheses:
  • The null hypothesis ( H0 ), which states there is no effect or difference.
  • The alternative hypothesis ( Ha ), which indicates the presence of an effect or difference.
With a two-sample t-test, we're typically testing whether the means of two groups differ significantly. The degrees of freedom, alongside the calculated t-statistic, will guide us in determining the p-value, which tells us if the observed data is statistically significant. A common significance level, typically 0.05, is used to decide whether to reject or fail to reject the null hypothesis. This process helps us make informed conclusions about the population based on sample data.
Confidence Interval
A confidence interval represents a range of values, derived from sample data, that is likely to contain the population parameter of interest. It gives us a way to estimate how plausible a statistical measure like a mean difference is. Confidence intervals are closely linked to the two-sample t-test because they provide an estimate of where the true difference in population means may lie. Here’s how confidence intervals work:
  • The confidence level (e.g., 95%) indicates the probability that the interval contains the parameter.
  • The width of the confidence interval is influenced by the sample size, variability, and chosen confidence level.
A narrower interval suggests more precision in the estimate of the population parameter, while a wider interval indicates more variability and uncertainty. Calculating this interval allows researchers to quantify their uncertainty regarding estimated differences between groups. Essentially, they provide context to the result of hypothesis testing, delivering a clearer picture of where the population parameter may precisely reside.

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Most popular questions from this chapter

The following observations are on time (h) for a AA 1.5volt alkaline battery to reach a \(0.8\) voltage ("Comparing the Lifetimes of Two Brands of Batteries," \(J\). of Statistical Educ., 2013, online): \(\begin{array}{llllll}\text { Energizer: } & 8.65 & 8.74 & 8.91 & 8.72 & 8.85 \\\ \text { Ultracell: } & 8.76 & 8.81 & 8.81 & 8.70 & 8.73 \\ \text { Energizer: } & 8.52 & 8.62 & 8.68 & 8.86 & \\ \text { Ultracell: } & 8.76 & 8.68 & 8.64 & 8.79 & \end{array}\) Normal probability plots support the assumption that the population distributions are normal. Does the data suggest that the variance of the Energizer population distribution differs from that of the Ultracell population distribution? Test the relevant hypotheses using a significance level of \(.05\). [Note: The two-sample \(t\) test for equality of population means gives a \(P\)-value of .763.] The Energizer batteries are much more expensive than the Ultracell batteries. Would you pay the extra money?

Sometimes experiments involving success or failure responses are run in a paired or before/after manner. Suppose that before a major policy speech by a political candidate, \(n\) individuals are selected and asked whether \((S)\) or not \((F)\) they favor the candidate. Then after the speech the same \(n\) people are asked the same question. The responses can be entered in a table as follows: where \(x_{1}+x_{2}+x_{3}+x_{4}=n\). Let \(p_{1}, p_{2}, p_{3}\), and \(p_{4}\) denote the four cell probabilities, so that \(p_{1}=P(S\) before and \(S\) after), and so on. We wish to test the hypothesis that the true proportion of supporters \((S)\) after the speech has not increased against the alternative that it has increased. a. State the two hypotheses of interest in terms of \(p_{1}, p_{2}\), \(p_{3}\), and \(p_{4}\). b. Construct an estimator for the after/before difference in success probabilities. c. When \(n\) is large, it can be shown that the rv \(\left(X_{i}-X_{j}\right) / n\) has approximately a normal distribution with variance given by \(\left[p_{i}+p_{j}-\left(p_{i}-p_{j}\right)^{2}\right] / n\). Use this to construct a test statistic with approximately a standard normal distribution when \(H_{0}\) is true (the result is called McNemar's test). d. If \(x_{1}=350, x_{2}=150, x_{3}=200\), and \(x_{4}=300\), what do you conclude?

The article "Quantitative MRI and Electrophysiology of Preoperative Carpal Tunnel Syndrome in a Female Population" (Ergonomies, 1997: 642-649) reported that \((-473.13,1691.9)\) was a large-sample \(95 \%\) confidence interval for the difference between true average thenar muscle volume \(\left(\mathrm{mm}^{3}\right)\) for sufferers of carpal tunnel syndrome and true average volume for nonsufferers. Calculate and interpret a \(90 \%\) confidence interval for this difference.

Teen Court is a juvenile diversion program designed to circumvent the formal processing of first-time juvenile offenders within the juvenile justice system. The article "An Experimental Evaluation of Teen Courts" \((J\). of Experimental Criminology, 2008: 137-163) reported on a study in which offenders were randomly assigned either to Teen Court or to the traditional Department of Juvenile Services method of processing. Of the \(56 \mathrm{TC}\) individuals, 18 subsequently recidivated (look it up!) during the 18 -month follow-up period, whereas 12 of the 51 DJS individuals did so. Does the data suggest that the true proportion of TC individuals who recidivate during the specified follow-up period differs from the proportion of DJS individuals who do so? State and test the relevant hypotheses using a significance level of . 10 .

Torsion during hip external rotation (ER) and extension may be responsible for certain kinds of injuries in golfers and other athletes. The article "Hip Rotational Velocities During the Full Golf Swing" (J. of Sports Science and Medicine, 2009: 296-299) reported on a study in which peak ER velocity and peak IR (internal rotation) velocity (both in deg.sec \({ }^{-1}\) ) were determined for a sample of 15 female collegiate golfers during their swings. The following data was supplied by the article's authors. a. Is it plausible that the differences came from a normally distributed population? b. The article reported that mean \((\pm \mathrm{SD})=-145.3(68.0)\) for \(E R\) velocity and \(=-227.8(96.6)\) for \(I R\) velocity. Based just on this information, could a test of hypotheses about the difference between true average IR velocity and true average ER velocity be carried out? Explain. c. The article stated that "The lead hip peak IR velocity was significantly greater than the trail hip ER velocity \((p=0.003, t\) value \(=3.65) . "\) (The phrasing suggests that an upper-tailed test was used.) Is that in fact the case? [Note: " \(p=.033^{\prime \prime}\) in Table 2 of the article is erroneous.]

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