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When using the \(F\) distribution to test two variances, is it essential that each of the two populations be normally distributed? Would it be all right if the populations had distributions that were mound-shaped and more or less symmetrical?

Short Answer

Expert verified
Yes, normal distribution is essential, but mound-shaped and symmetrical populations are often acceptable.

Step by step solution

01

Understand the F-Distribution

The F-distribution is applied in hypothesis testing, particularly for comparing variances of two populations. It's a ratio of two chi-square distributions and relies heavily on specific distribution characteristics of the data.
02

Identify Normal Distribution Requirement

For the F-test to be valid, the populations being tested should ideally be normally distributed. This is crucial because the F-test assumes normality in its derivation and validity.
03

Evaluate Mound-Shaped and Symmetrical Condition

If populations are not perfectly normally distributed but are mound-shaped (unimodal) and symmetrical, the F-test can still be approximately valid. These characteristics suggest a distribution not too far from normal, often allowing the test to still function adequately.
04

Considerations for Real-World Data

In practice, slight deviations from normality might not drastically affect the F-test's results, especially if sample sizes are large. However, the closer the data is to a normal distribution, the more reliable the results.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions about the properties of a population based on sample data. It involves setting up two competing statements, known as the null hypothesis
  • Null Hypothesis ( $H_0$ ): A statement that there is no effect or no difference.
  • Alternative Hypothesis ( $H_1$ ): Contradicts the null hypothesis, suggesting there is an effect or a difference.
The process begins with assuming that the null hypothesis is true and then employing statistical tests to see if there is enough evidence to reject it in favor of the alternative hypothesis.

The F-Test is a form of hypothesis testing that compares the variances of two populations. This test helps us decide whether observed differences in variance are due to chance or if they reflect true differences in the populations. It’s important to have a clear question in mind and collect data carefully to ensure the test results are reliable.
Normal Distribution
The normal distribution is a fundamental concept in statistics, serving as the backbone for many statistical methods, including the F-Test. A population or data sample that is normally distributed will have a distinct bell-shaped curve when graphed. This distribution is
  • Symmetrical: The left and right halves of the graph are mirror images.
  • Unimodal: Only one significant peak exists in the data.
  • Mean, median, and mode are equal: All three central tendency measures coincide at the center of the graph.
For the F-Test, which compares variances, the assumption of normality is key. This assumption ensures that the calculations performed are valid and lead to trustworthy conclusions. If the distribution of the population is known to be normally distributed, the conditions under which the F-Test is applied are met, elevating the confidence in the decision made from the hypothesis test.
Comparing Variances
When we talk about comparing variances, we are interested in understanding if the spread or variability in two different populations is similar or significantly different.

Here's where the F-Distribution comes in: It employs a measure called the F-ratio, which is the ratio of two sample variances.
  • If the F-ratio is approximately equal to 1, the variances are likely the same.
  • If the F-ratio is significantly greater than 1 or less than 1, we may have reason to believe the variances are different.
Before conducting an F-Test, ensure the populations are approximately normally distributed. However, in real-world scenarios where data isn't perfectly normal, the test often still holds if the data is mound-shaped and symmetrical. This property offers some flexibility when dealing with real data, though caution is advised as non-normality can sometimes affect results, especially with smaller sample sizes.

Ultimately, comparing variances is crucial in fields like quality control, research, and any area where understanding variation is essential.

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Most popular questions from this chapter

A set of solar batteries is used in a research satellite. The satellite can run on only one battery, but it runs best if more than one battery is used. The variance \(\sigma^{2}\) of lifetimes of these batteries affects the useful lifetime of the satellite before it goes dead. If the variance is too small, all the batteries will tend to die at once. Why? If the variance is too large, the batteries are simply not dependable. Why? Engineers have determined that a variance of \(\sigma^{2}=23\) months (squared) is most desirable for these batteries. A random sample of 22 batteries gave a sample variance of \(14.3\) months (squared). (i) Using a \(0.05\) level of significance, test the claim that \(\sigma^{2}=23\) against the claim that \(\sigma^{2}\) is different from 23 . (ii) Find a \(90 \%\) confidence interval for the population variance \(\sigma^{2}\). (iii) Find a \(90 \%\) confidence interval for the population standard deviation \(\sigma .\)

A new kind of typhoid shot is being developed by a medical research team. The old typhoid shot was known to protect the population for a mean time of 36 months, with a standard deviation of 3 months. To test the time variability of the new shot, a random sample of 23 people were given the new shot. Regular blood tests showed that the sample standard deviation of protection times was \(1.9\) months. Using a \(0.05\) level of significance, test the claim that the new typhoid shot has a smaller variance of protection times. Find a \(90 \%\) confidence interval for the population standard deviation.

Investing: Mutual Funds How reliable are mutual funds that invest in bonds? Again, this depends on the bond fund you buy (see reference in Problem 9). A random sample of annual percentage returns for mutual funds holding shortterm U.S. government bonds is shown below. \(\begin{array}{lllllll}4.6 & 4.7 & 1.9 & 9.3 & -0.8 & 4.1 & 10.5\end{array}\) \(\begin{array}{llllll}4.2 & 3.5 & 3.9 & 9.8 & -1.2 & 7.3\end{array}\) Use a calculator to verify that \(s^{2}=13.59\) for the preceding data. A random sample of annual percentage returns for mutual funds holding intermediate-term corporate bonds is shown below. \(\begin{array}{rrrrrrrr}-0.8 & 3.6 & 20.2 & 7.8 & -0.4 & 18.8 & -3.4 & 10.5 \\\ 8.0 & -0.9 & 2.6 & -6.5 & 14.9 & 8.2 & 18.8 & 14.2\end{array}\) Use a calculator to verify that \(s^{2} \approx 72.06\) for returns from mutual funds holding intermediate-term corporate bonds. Use \(\alpha=0.05\) to test the claim that the population variance for annual percentage returns of mutual funds holding short-term government bonds is different from the population variance for mutual funds holding intermediate- term corporate bonds. How could your test conclusion relate to the question of reliability of returns for each type of mutual fund?

A sociologist studying New York City ethnic groups wants to determine if there is a difference in income for immigrants from four different countries during their first year in the city. She obtained the data in the following table from a random sample of immigrants from these countries (incomes in thousands of dollars). Use a \(0.05\) level of significance to test the claim that there is no difference in the earnings of immigrants from the four different countries. $$ \begin{array}{rrcr} \text { Country I } & \text { Country II } & \text { Country III } & \text { Country IV } \\ 12.7 & 8.3 & 20.3 & 17.2 \\ 9.2 & 17.2 & 16.6 & 8.8 \\ 10.9 & 19.1 & 22.7 & 14.7 \\ 8.9 & 10.3 & 25.2 & 21.3 \\ 16.4 & & 19.9 & 19.8 \end{array} $$

The following problem is based on information from an article by N. Keyfitz in the American Journal of Sociology (Vol. 53, pp. \(470-480\) ). Let \(x=\) age in years of a rural Quebec woman at the time of her first marriage. In the year 1941 , the population variance of \(x\) was approximately \(\sigma^{2}=5.1 .\) Suppose a recent study of age at first marriage for a random sample of 41 women in rural Quebec gave a sample variance \(s^{2}=3.3 .\) Use a \(5 \%\) level of significance to test the claim that the current variance is less than \(5.1 .\) Find a \(90 \%\) confidence interval for the population variance.

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