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What is the critical F-value when the sample size for the numerator is four and the sample size for the denominator is seven? Use a one-tailed test and the . 01 significance level.

Short Answer

Expert verified
The critical F-value is approximately 8.94.

Step by step solution

01

Identify Degrees of Freedom

For an F-test, we need to determine the degrees of freedom for both the numerator and the denominator. Given a sample size for the numerator (n1 = 4), the degrees of freedom for the numerator is calculated as df1 = n1 - 1. Therefore, df1 = 4 - 1 = 3. For the sample size of the denominator (n2 = 7), the degrees of freedom for the denominator is df2 = n2 - 1. So, df2 = 7 - 1 = 6.
02

Look Up Critical F-value

Using the found degrees of freedom, look up the critical F-value from the F-distribution table. We need an F-table or statistical software for a 0.01 significance level (one-tailed test) with df1 = 3 and df2 = 6.
03

Confirm and Conclude

Upon consulting the F-distribution table, find that the critical F-value for the one-tailed test with df1 = 3, df2 = 6 at the 0.01 significance level is approximately 8.94.

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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 are fundamental to statistical tests like the F-test. They represent the number of values in a calculation that are free to vary. For the F-distribution, you have two sets of degrees of freedom: one for the numerator (\(df_1\)) and one for the denominator (\(df_2\)).
To calculate \(df_1\), the formula is based on the sample size used in the numerator. Simply subtract one from the sample size (\(n_1\)), giving you \(df_1 = n_1 - 1\). Similarly, for the denominator, \(df_2\) is calculated using \(df_2 = n_2 - 1\).
In the context of this exercise, with a sample size of 4 for the numerator, \(df_1\) becomes 3. For the denominator with a sample size of 7, \(df_2\) becomes 6. Understanding degrees of freedom is crucial as they directly impact the shape of the F-distribution and the critical F-value.
Critical F-value
The critical F-value is an important threshold used in F-tests to determine if the variance between two samples is significantly different. This value is influenced by the degrees of freedom for both the numerator and the denominator, as well as the chosen significance level.
You can find the critical F-value using an F-distribution table or statistical software by looking up your specific degrees of freedom combination and the significance level.
For a significance level of 0.01 and the degrees of freedom \(df_1 = 3\) and \(df_2 = 6\), you locate the intersection in the table to find the critical F-value. In this exercise, it is approximately 8.94. This means, for our one-tailed test, if the calculated F-ratio is above 8.94, we have enough evidence to reject the null hypothesis.
One-tailed Test
A one-tailed test in hypothesis testing is used when we want to assess if there is a significant difference in one specific direction. For an F-test, this might mean you are interested in knowing if one sample has a larger variance than the other.
The choice between a one-tailed and two-tailed test depends on your hypothesis; a one-tailed test will only look at the extreme values on one end of the distribution.
By using a one-tailed test at a 0.01 significance level, you are testing a hypothesis with 99% confidence that the difference is in a specific direction—either greater or less than, but not both. This requires a critical F-value that is more stringent at the tail end of the distribution, which in this exercise is indicated by an F-value close to 8.94. Conducting a one-tailed test can increase your test's power but requires a clear hypothesis about the direction of the effect.

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Most popular questions from this chapter

A study of the effect of television commercials on 12 -year-old children measured their attention span, in seconds. The commercials were for clothes, food, and toys. At the .05 significance level, is there a difference in the mean attention span of the children for the various commercials? Are there significant differences between pairs of means? Would you recommend dropping one of the three commercial types? $$ \begin{array}{|ccc|} \hline \text { Clothes } & \text { Food } & \text { Toys } \\ \hline 26 & 45 & 60 \\ 21 & 48 & 51 \\ 43 & 43 & 43 \\ 35 & 53 & 54 \\ 28 & 47 & 63 \\ 31 & 42 & 53 \\ 17 & 34 & 48 \\ 31 & 43 & 58 \\ 20 & 57 & 47 \\ & 47 & 51 \\ & 44 & 51 \\ & 54 & \\ \hline \end{array} $$

One variable that Google uses to rank pages on the Internet is page speed, the time it takes for a web page to load into your browser. A source for women's clothing is redesigning their page to improve the images that show its products and to reduce its load time. The new page is clearly faster, but initial tests indicate there is more variation in the time to load. A sample of 16 different load times showed that the standard deviation of the load time was 22 hundredths of a second for the new page and 12 hundredths of a second for the current page. At the .05 significance level, can we conclude that there is more variation in the load time of the new page?

A stockbroker at Critical Securities reported that the mean rate of return on a sample of 10 oil stocks was \(12.6 \%\) with a standard deviation of \(3.9 \%\). The mean rate of return on a sample of 8 utility stocks was \(10.9 \%\) with a standard deviation of \(3.5 \%\). At the .05 significance level, can we conclude that there is more variation in the oil stocks?

The following are four observations collected from each of three treatments. Test the hypothesis that the treatment means are equal. Use the . 05 significance level. $$ \begin{array}{|rcc|} \hline \text { Treatment } 1 & \text { Treatment } 2 & \text { Treatment } 3 \\\ \hline 8 & 3 & 3 \\ 6 & 2 & 4 \\ 10 & 4 & 5 \\ 9 & 3 & 4 \\ \hline \end{array} $$ a. State the null and the alternate hypotheses. b. What is the decision rule? c. Compute SST, SSE, and SS total. d. Complete an ANOVA table. e. State your decision reqarding the null hypothesis.

When only two treatments are involved, ANOVA and the Student's \(t\) test (Chapter 11) result in the same conclusions. Also, for computed test statistics, \(t^{2}=F\). To demonstrate this relationship, use the following example. Fourteen randomly selected students enrolled in a history course were divided into two groups, one consisting of 6 students who took the course in the normal lecture format. The other group of 8 students took the course in a distance format. At the end of the course, each group was examined with a 50 -item test. The following is a list of the number correct for each of the two groups. $$ \begin{array}{|cc|} \hline \text { Traditional Lecture } & \text { Distance } \\ \hline 37 & 50 \\ 35 & 46 \\ 41 & 49 \\ 40 & 44 \\ 35 & 41 \\ 34 & 42 \\ & 45 \\ & 43 \\ \hline \end{array} $$ a. Using analysis of variance techniques, test \(H_{0}\) that the two mean test scores are equal; \(\alpha=.05\) b. Using the \(t\) test from Chapter 11 , compute \(t\). c. Interpret the results.

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