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Use the information given in Exercises \(8-11\) to bound the \(p\) -value of the \(F\) statistic for a one-tailed test with the indicated degrees of freedom. \(F=6.16, d f_{1}=4, d f_{2}=13\)

Short Answer

Expert verified
Answer: The p-value lies in the interval (0, 0.05).

Step by step solution

01

Understand the F-distribution and p-value

The F-distribution is a continuous probability distribution that is used to test the equality of variance among different groups. The F-statistic is a ratio of variances, and the F-distribution is determined by two degrees of freedom: the numerator's degree of freedom (\(df_1\)) and the denominator's degree of freedom (\(df_2\)). A p-value represents the probability of obtaining an F-value as extreme or more extreme than the one obtained from the test, assuming the null hypothesis is true.
02

Use the F-distribution table to find the critical F-value

To find the critical F-value, we need to refer to an F-distribution table. The table provides critical F-values for various degrees of freedom and significance levels. Since we are performing a one-tailed test, we will look for a significance level of \(\alpha = 0.05\). Our given \(df_1 = 4\) and \(df_2 = 13\). Consulting an F-distribution table, we find the critical F-value to be \(F_{critical} = 3.259\).
03

Calculate the p-value

As we have the F-distribution table, we are limited to a range of confidence levels. We can provide the p-value within an interval. In this case, we have the critical F-value for \(\alpha = 0.05\), so we know that the p-value is lower than 0.05. Our obtained F-value (\(6.16\)) is greater than the critical F-value (\(3.259\)) found in the F-distribution table. Since the obtained F-value is in the extreme right tail of the F-distribution, this suggests that the p-value is lower than 0.05. In conclusion, the p-value for this exercise lies in the interval \((0, 0.05)\).

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Key Concepts

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

p-value calculation
Understanding the p-value is crucial for interpreting the statistical significance of test results. In hypothesis testing, the p-value helps determine whether to reject the null hypothesis. It quantifies the probability of observing a test statistic as extreme as the one obtained, given that the null hypothesis is true.

For the F-distribution, which arises when comparing variances, calculating the p-value means finding the probability that the F-statistic will be as large as or larger than the observed value. Unfortunately, precise p-value calculation often cannot be done directly, as F-distribution tables typically only provide critical values for certain confidence levels, like 0.05 or 0.01.

When the calculated F-statistic is greater than the critical value from the table, the p-value falls below the corresponding significance level. In our exercise, with an F-value of 6.16 and critical value for a 0.05 significance level being 3.259, we infer that the p-value is less than 0.05. This indicates strong evidence against the null hypothesis in a one-tailed test.
one-tailed test
In hypothesis testing, a one-tailed test, or a one-sided test, is used when we want to determine if there is an increase or decrease but not both in a certain population parameter. Contrast this with a two-tailed test, which checks for any difference, regardless of the direction.

A one-tailed test is useful when the research hypothesis predicts a specific direction of the treatment or experiment’s effect. For instance, if we expect that a new medication will not be less effective than the current standard, a one-tailed test can be used to only reject the null hypothesis if the new medication proves to be superior.

This specificity makes the one-tailed test more powerful than the two-tailed test for detecting an effect in the hypothesized direction. However, this comes at the cost of not being able to detect an effect in the opposite direction of the hypothesis.
degrees of freedom
Degrees of freedom (df) are essential in the context of statistical distributions. They represent the number of independent values that can vary in a statistical calculation without breaking any constraints. Intuitively, they can be understood as the number of 'choices' left after we estimate certain parameters.

In the F-distribution, which is used to compare variances, there are two degrees of freedom: one for the numerator (\(df_1\)) and one for the denominator (\(df_2\)). The numerator df corresponds to the variance among the group means, while the denominator df corresponds to the within-group variance.

The degrees of freedom affect the shape of the F-distribution curve, and thus, they are crucial for locating the critical F-value and calculating the p-value. More degrees of freedom typically mean a more peaked distribution, while fewer degrees of freedom result in a flatter distribution. In our exercise, the df for the numerator is 4, and the denominator is 13, which dictates the particular shape of the F-distribution we use to determine our p-value.

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Most popular questions from this chapter

Find the tabled value of \(t\left(t_{a}\right)\) corresponding to a left-tail area a and degrees of freedom given. $$ a=.10, d f=35 $$

An experiment was conducted to compare the use of iPads versus regular textbooks in teaching algebra to two classes of middle school students. \({ }^{6}\) To remove teacher-to-teacher variation, the same teacher taught both classes, and all teaching materials were provided by the same author and publisher. Suppose that after 1 month, 10 students were selected from each class and their scores on an algebra advancement test recorded. The summarized data follow. $$ \begin{array}{lcc} \hline & \text { iPad } & \text { Textbook } \\ \hline \text { Mean } & 86.4 & 79.7 \\ \text { Standard Deviation } & 8.95 & 10.7 \\ \text { Sample Size } & 10 & 10 \\ \hline \end{array} $$ a. Use the summary data to test for a significant difference in advancement scores for the two groups using $$ \alpha=.05 . $$ b. Find a \(95 \%\) confidence interval for the difference in mean scores for the two groups. c. In light of parts a and b, what can we say about using iPads versus traditional textbooks in teaching algebra at the middle school level?

Find the tabled value of \(t\left(t_{a}\right)\) corresponding to a left-tail area a and degrees of freedom given. $$ a=.05, d f=8 $$

Use the information in Exercises \(7-8 .\) Calculate the observed value of the \(t\) statistic for testing the difference between the two population means using paired data. Approximate the \(p\) -value for the test and use it to state your conclusions. $$\bar{d}=5.7, s_{d}^{2}=256, n=18, H_{\mathrm{a}}: \mu_{d}>0$$

The stability of measurements on a manufactured product is important in maintaining product quality. A manufacturer of lithium batteries suspected that one of the production lines was producing batteries with a wide variation in length of life. To test this theory, he randomly selected \(n=50\) batteries from the suspect line and \(n=50\) from a line that was judged to be "in control." He then measured the length of time (in hours) until depletion for both samples. The sample means and variances for the two samples were as follows: $$\begin{array}{ll}\hline \text { Suspect Line } & \text { Line }^{\prime \prime} \text { in Control }^{\prime \prime} \\\\\hline \bar{x}_{1}=9.40 & \bar{x}_{2}=9.25 \\\s_{1}=0.25 & s_{2}=0.12\end{array}$$ a. Do the data provide sufficient evidence to indicate that batteries produced by the "suspect line" have a larger variance in length of life than those produced by the line that is assumed to be in control? Test using \(\alpha=.05 .\) b. Find the approximate \(p\) -value for the test and interpret its value. c. Construct a \(90 \%\) confidence interval for the variance ratio.

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