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F.DIST

Calculates the probability density function (PDF) or the cumulative distribution function (CDF) of the F-distribution.

Syntax:

F.DIST(x, deg_freedomOne, deg_freedomTwo, cumulative)

x is required, and is the value you want to use to evaluate the F-distribution.

deg_freedomOne is required, and is the degrees of freedom for the numerator.

deg_freedomTwo is required, and is the degrees of freedom for the denominator.

cumulative is required, and is a logical value:

TRUE: Returns the cumulative distribution function (the probability of observing an F-statistic less than or equal to x).

FALSE: Returns the probability density function (the height of the F-distribution curve at a specific value of x).

Example:

If x contains 4.5, deg_freedomOne contains 2, deg_freedomTwo contains 27 and cumulative contains TRUE:

F.DIST(4.5, 2, 27, TRUE)

returns 0.979425606

This example calculates the probability of observing an F-statistic of 4.5 or lower, given 2 and 27 degrees of freedom.

Deg_freedomOne:

Deg_freedomTwo:

Cumulative:

Result:

0.979425606

Application:

Comparing the Consistency of Two Production Lines

A car parts manufacturer has two assembly lines, Line A and Line B, that produce the same engine component. The company wants to test if the variability in the diameter of the components from Line A is different from the variability of components from Line B. The ideal diameter is a precise measurement, and any significant variance could lead to quality issues.

The quality control manager takes a random sample of components from each line and measures their diameters. The data is as follows (measurements in mm):

Line A Sample Data:

Sample Size ( $n_{1}$ ): 15
Sample Variance ( $s_{1}^{2}$ ): 0.000045

Line B Sample Data:

Sample Size ( $n_{2}$ ): 12
Sample Variance ( $s_{2}^{2}$ ): 0.000078

Hypothesis Test:

The manager performs an F-test for equality of variances.

Null Hypothesis ( $H_{0}$ ): The population variances are equal ( $σ_{1}^{2} = σ_{2}^{2}$ ). There is no difference in the consistency of the two lines.
Alternative Hypothesis ( $H_{a}$ ): The population variances are not equal ( $σ_{1}^{2} \neq = σ_{2}^{2}$ ). The consistency of the two lines is different.

Calculating the F-Statistic:

The F-statistic is the ratio of the two sample variances. By convention, the larger variance is placed in the numerator.

$F = \frac{s _{B}^{2}}{s _{A}^{2}} = \frac{0.000078}{0.000045} = 1.733$

Using the F.DIST Function:

To determine if this F-statistic is significant, the manager needs to find the probability of observing an F-statistic this large (or larger) under the null hypothesis. This is where the F.DIST function is used.

The F.DIST function has the following syntax:

F.DIST(x, deg_freedom1, deg_freedom2, cumulative)

x: The F-statistic calculated from the sample data. In this case, 1.733.
deg_freedom1: The degrees of freedom for the numerator's variance. For a sample, degrees of freedom is n−1. For Line B, this is 12−1=11.
deg_freedom2: The degrees of freedom for the denominator's variance. For a sample, this is 15−1=14.
cumulative: A logical value. TRUE returns the cumulative distribution function (CDF), which is the probability that a random F-value is less than or equal to x. FALSE returns the probability density function. For hypothesis testing, we use TRUE.

Table Representation:

This information can be organized in a table to easily apply the F.DIST function.

	Parameter	Value
	A	B
1	x (F-statistic)	1.733
2	Numerator Degrees of Freedom	11
3	Denominator Degrees of Freedom	14
4	Cumulative	TRUE

Applying the Function:

Using a spreadsheet program, the formula would look like this:

F.DIST(1.733, 11, 14, TRUE)

This function returns the cumulative probability, which is the area under the F-distribution curve to the left of our calculated F-statistic. Let's assume the result is approximately 0.835.

Interpreting the Result:

The result of 0.835 means that there is an 83.5% chance of getting an F-statistic of 1.733 or smaller, assuming the null hypothesis (equal variances) is true.

For a two-tailed test (since our alternative hypothesis is "not equal"), we are interested in the probability of getting an F-statistic as extreme as ours in either direction. The p-value for a two-tailed test is calculated as:

$p -value = 2 \times min (F.DIST, 1 - F.DIST)$

$p -value = 2 \times (1 - 0.835) = 2 \times 0.165 = 0.33$

The p-value is 0.33. If the company set a significance level (α) of 0.05, we would compare our p-value to this level. Since 0.33 > 0.05, we do not have sufficient evidence to reject the null hypothesis.

Conclusion:

Based on the F-test, there is no statistically significant difference in the variability of the component diameters produced by Line A and Line B. The quality control manager can conclude that both lines are operating with a similar level of consistency.

Result for F.DIST(1.733, 11, 14, TRUE):

0.835

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