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Julia Jackson operates a franchised restaurant that specializes in soft ice cream cones and sundaes. Recently she received a letter from corporate headquarters warning her that her shop is in danger of losing its franchise because the average sales per customer have dropped "substantially below the average for the rest of the corporation." The statement may be true, but Julia is convinced that such a statement is completely invalid to justify threatening a closing. The variation in sales at her restaurant is bound to be larger than most, primarily because she serves more children, elderly, and single adults rather than large families who run up big bills at the other restaurants. Therefore, her average ticket is likely to be smaller and exhibit greater variability. To prove her point, Julia obtained the sales records from the whole company and found that the standard deviation was \(2.45\)dollar per sales ticket. She then conducted a study of the last 71 sales tickets at her store and found a standard deviation of \(2.95\)dollar per ticket. Is the variability in sales at Julia's franchise, at the 0.05 level of significance, greater than the variability for the company?

Step by step solution

01

Define the hypotheses

Set the null hypothesis H0: The sales variability in Julia's franchise is less or equal to the overall corporation sales variability. Set the alternative hypothesis H1: The sales variability in Julia's franchise is greater than the overall corporation sales variability. That is, \H0: \(\sigma^2_{julia} \leq \sigma^2_{corp}\) \H1: \(\sigma^2_{julia} > \sigma^2_{corp}\)

02

Calculate the test statistic

For comparing two variances, we use the F-test. The formula for the F statistic is the ratio of the two sample variances. In this case, we are testing if the variance of Julia's store is significantly higher than that of the corporation, so the ratio would be \(\frac{s_{julia}^2}{s_{corp}^2}\), where \(s_{julia}^2\) is the variance of sales in Julia's store and \(s_{corp}^2\) is the corporation's variance. Given in the problem, standard deviation in Julia's franchise is \(2.95\) dollars and that for the corporation is \(2.45\) dollars. So, \the test statistic F = \[\frac{(2.95)^2}{(2.45)^2} = 1.44979591837]

03

Determine the critical value

The critical value for the F distribution can be found using the F-distribution table. To use the F-distribution table, we need two degrees of freedom, that being the sample size subtracted by one. For Julia's sample the degrees of freedom is \(df1 = 71 - 1 = 70\), and for the corporation, being akin to an infinite population, assumed to be a large number say \(df2 = 1000\). With a 0.05 level of significance for a one-tailed test, the critical value is found to be near 1.285.

04

Make decision

As the test statistic (F = 1.45) is greater than the critical value (1.285), the null hypothesis is rejected at the 0.05 level of significance. Therefore, Julia's claim is valid. The variability (variance) in sales at her franchise is significantly higher than the corporation's variability.

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Key Concepts

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

F-test

The F-test is a statistical procedure used to compare two populations' variances to see if they are significantly different. It's most commonly applied when comparing the variability of two groups, such as in ANOVA (analysis of variance), to determine if the means of several groups are equal.

It operates under the premise that if the null hypothesis is true, the ratio of the variances (or standard deviations squared) should be close to 1. If the ratio is far from 1, it suggests that the variances are different, and thus, so too might be the populations. In the example within our exercise, Julia employs an F-test to address a specific question regarding sales variability in comparison to the corporation's variability.

Standard Deviation

Standard deviation is a measure that quantifies the amount of variation or dispersion of a set of values. It is a very commonly used statistic that can tell you how spread out the numbers in a data set are. If the values are close to the mean, then the standard deviation is small, whereas a high standard deviation indicates that the values are spread out over a wider range.

In the context of the exercise, Julia calculates the standard deviation of her sales tickets to argue that her franchise's sales variability, which she believes is affected by her unique customer demographic, is different from that of the overall corporation.

Degrees of Freedom

Degrees of freedom (df) in statistics refers to the number of values in a calculation that are free to vary. When estimating population parameters based on sample statistics, degrees of freedom are an integral part of calculating these estimations and understanding their distributions.

The number of degrees of freedom is typically equal to the number of values minus the number of parameters that need to be estimated. For the F-test, we need degrees of freedom for both the numerator and the denominator. In Julia's case, the degree of freedom for her store df1 is 70 (71-1), and it's assumed as a large number for the corporation, df2 (for instance, 1000), to reflect the larger population's variance.

Null Hypothesis

The null hypothesis (often symbolized as H0) is a statement in hypothesis testing that there is no effect or no difference, and it serves as a starting assumption. In the exercise, the null hypothesis is that Julia's franchise does not have greater variability in sales compared to the corporation as a whole. In statistical hypothesis testing, the objective is to determine whether there is enough evidence from a sample to reject the null hypothesis for the entire population.

Alternative Hypothesis

Conversely, the alternative hypothesis (H1 or Ha) suggests that there is an effect, or there is a difference. It's what you want to prove to be true. In Julia's scenario, her alternative hypothesis is that her franchise's sales variability is indeed greater than that of the broader corporation. Rejecting the null hypothesis lends support to the alternative hypothesis.

Level of Significance

The level of significance, commonly denoted by \(\alpha\), is the probability of rejecting the null hypothesis when it is actually true; it's a measure of how much evidence we require before we reject the null hypothesis. A common level of significance is 0.05, meaning there's a 5% risk of concluding there is an effect (rejecting the null hypothesis) when there isn’t one. In the given exercise, Julia uses a 0.05 level of significance to determine whether the difference in variability between her franchise's sales and that of the corporation's is statistically significant.