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When using the Student's \(t\) distribution to test \(\mu,\) what value do you use for the degrees of freedom?

Short Answer

Expert verified
Degrees of freedom is \(n - 1\), where \(n\) is the sample size.

Step by step solution

01

Determine the Sample Size

The first step in using the Student's \(t\) distribution is to determine the sample size \(n\). This is the total number of observations in your data set.
02

Calculate the Degrees of Freedom

The degrees of freedom (DF) for a \(t\)-distribution when testing for the mean \(\mu\) is calculated by the formula \(DF = n - 1\). This means you subtract one from the sample size.

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Key Concepts

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

Degrees of Freedom
When working with the Student's \(t\) distribution, a key concept you will encounter is degrees of freedom (DF). It is crucial in statistics because it influences the shape of the \(t\) distribution. The degrees of freedom essentially tell you how many independent values are in a calculation process. For testing the mean \(\mu\) with a \(t\) distribution, you calculate the degrees of freedom by subtracting one from your sample size. So, if your sample size is \(n\), then \(DF = n - 1\). This subtraction accounts for the fact that estimating the mean \(\mu\) itself consumes one degree of freedom.
Degrees of freedom help decide the appropriate distribution curve, impacting the critical values you use for hypothesis testing. The more degrees of freedom you have, the closer your \(t\) distribution will resemble the normal distribution. Conversely, with fewer degrees of freedom, the \(t\) distribution tends to have heavier tails, indicating more variability. This is particularly important in small sample sizes where normal approximation isn't reliable.
Sample Size
Determining the sample size \(n\) is fundamental when dealing with any statistical analysis including the Student's \(t\) distribution. The sample size tells you the number of data points or observations in your dataset. When it comes to hypothesis testing, particularly with a \(t\) distribution, sample size plays a crucial role in both calculations and interpretation of results.
In small samples, the accuracy of the \(t\) distribution becomes more significant since it doesn't rely on the many assumptions the normal distribution does. The smaller the sample size, the more the \(t\) distribution will differ from the normal distribution. But as your sample size increases, the \(t\) distribution begins to look like the normal distribution. This is due to the Central Limit Theorem.
Additionally, a larger sample size generally increases the reliability of your test results since it reduces the effect of outliers and anomalies. When planning your tests, always consider how your sample size may impact the degrees of freedom as well as the overall outcome of your hypothesis test.
Testing the Mean
Testing the mean \(\mu\) using the Student’s \(t\) distribution is a common statistical method when the sample size is small and/or the population standard deviation is unknown. The aim here is to determine whether there is a significant difference between the sample mean and a known or hypothesized population mean.
Here's a rundown of steps in testing the mean:
  • State your hypothesis: Usually, \(H_0\) (null hypothesis) is that there is no difference between the sample mean and the population mean, and \(H_1\) (alternative hypothesis) suggests a difference exists.
  • Choose a significance level \(\alpha\): This is the probability of rejecting the null hypothesis when it is actually true. Common values are 0.05, 0.01, etc.
  • Calculate your test statistic: Using the formula \( t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \), where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( s \) is the sample standard deviation, and \( n \) is the sample size.
  • Find the critical value from the \(t\) distribution table using calculated degrees of freedom and significance level.
  • Decision-making: Compare your test statistic to the critical value to determine if you reject the null hypothesis.
Testing the mean with a \(t\) distribution is invaluable in small datasets where other methods might not be applicable. By following this process, you ensure that your conclusions are statistically valid.

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Most popular questions from this chapter

Homser Lake, Oregon, has an Atlantic salmon catch and release program that has been very successful. The average fisherman's catch has been \(\mu=8.8\) Atlantic salmon per day (Source: National Symposium on Catch and Release Fishing, Humboldt State University). Suppose that a new quota system restricting the number of fishermen has been put into effect this season. A random sample of fishermen gave the following catches per day: $$\begin{array}{ccccccc} 12 & 6 & 11 & 12 & 5 & 0 & 2 \\ 7 & 8 & 7 & 6 & 3 & 12 & 12 \end{array}$$ i. Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x}=7.36\) and \(s \approx 4.03.\) ii. Assuming the catch per day has an approximately normal distribution, use a \(5 \%\) level of significance to test the claim that the population average catch per day is now different from \(8.8 .\)

In a statistical test, we have a choice of a left-tailed test, a right-tailed test, or a two-tailed test. Is it the null hypothesis or the alternate hypothesis that determines which type of test is used? Explain your answer.

Myers-Briggs: Extroverts Are most student government leaders extroverts? According to Myers-Briggs estimates, about \(82 \%\) of college student government leaders are extroverts (Source: Myers -Briggs Type Indicator Atlas of Type Tables). Suppose that a Myers-Briggs personality preference test was given to a random sample of 73 student government leaders attending a large national leadership conference and that 56 were found to be extroverts. Does this indicate that the population proportion of extroverts among college student government leaders is different (either way) from \(82 \% ?\) Use \(\alpha=0.01.\)

How much customers buy is a direct result of how much time they spend in a store. A study of average shopping times in a large national housewares store gave the following information (Source: Why We Buy: The Science of Shopping by P. Underhill): Women with female companion: 8.3 min. Women with male companion: 4.5 min. Suppose you want to set up a statistical test to challenge the claim that a woman with a female friend spends an average of 8.3 minutes shopping in such a store. (a) What would you use for the null and alternate hypotheses if you believe the average shopping time is less than 8.3 minutes? Is this a right-tailed, left-tailed, or two-tailed test? (b) What would you use for the null and alternate hypotheses if you believe the average shopping time is different from 8.3 minutes? Is this a righttailed, left-tailed, or two-tailed test? Stores that sell mainly to women should figure out a way to engage the interest of men-perhaps comfortable seats and a big TV with sports programs! Suppose such an entertainment center was installed and you now wish to challenge the claim that a woman with a male friend spends only 4.5 minutes shopping in a housewares store. (c) What would you use for the null and alternate hypotheses if you believe the average shopping time is more than 4.5 minutes? Is this a right-tailed, left-tailed, or two-tailed test? (d) What would you use for the null and alternate hypotheses if you believe the average shopping time is different from 4.5 minutes? Is this a righttailed, left-tailed, or two-tailed test?

Socially conscious investors screen out stocks of alcohol and tobacco makers, firms with poor environmental records, and companies with poor labor practices. Some examples of "good," socially conscious companies are Johnson and Johnson, Dell Computers, Bank of America, and Home Depot. The question is, are such stocks overpriced? One measure of value is the \(\mathrm{P} / \mathrm{E},\) or price-to-earnings, ratio. High \(\mathrm{P} / \mathrm{E}\) ratios may indicate a stock is overpriced. For the S\&P stock index of all major stocks, the mean P/E ratio is \(\mu=19.4 .\) A random sample of 36 "socially conscious" stocks gave a P/E ratio sample mean of \(\bar{x}=17.9,\) with sample standard deviation \(s=5.2\) (Reference: Morningstar, a financial analysis company in Chicago). Does this indicate that the mean \(\mathrm{P} / \mathrm{E}\) ratio of all socially conscious stocks is different (either way) from the mean \(\mathrm{P} / \mathrm{E}\) ratio of the \(\mathrm{S} \& \mathrm{P}\) stock index? Use \(\alpha=0.05\).

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