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A soft drink distributor is considering two new models of dispensing machines. Both the Harvard Company machine and the Fizzit machine can be adjusted to fill the cups to a certain mean amount. However, the variation in the amount dispensed from cup to cup is a primary concern. Ten cups dispensed from the Harvard machine showed a variance of \(0.065,\) whereas 15 cups dispensed from the Fizzit machine showed a variance of \(0.033 .\) The factory representative from the Harvard Company maintains that his machine had no more variability than the Fizzit machine. Assume the amount dispensed is normally distributed. At the 0.05 level of significance, does the sample refute the representative's assertion? a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

Step by step solution

01

Set up the Hypotheses

The null hypothesis (H0) is that the variance of the Harvard machine is equal to that of the Fizzit machine, and the alternative hypothesis (H1) is that the variances are not equal. So, \(H0: \sigma^2_H = \sigma^2_F\), \(H1: \sigma^2_H \neq \sigma^2_F\)

02

Calculation of the F statistic

Calculate the F statistic, which is the ratio of two variances, i.e., Harvard's variance to Fizzit's variance. We find that the F statistic is given by \(F = \sigma^2_H/ \sigma^2_F = 0.065 / 0.033 = 1.9696\)

03

Calculation of the critical value

Calculate the critical value for a two-tailed test at 0.05 level of significance for \(8\) and \(14\) degrees of freedom (calculated as sample size minus 1 for each sample). The critical values using the F-distribution table are \(2.90\) and \(0.34\)

04

Decision making

Since the calculated F value does not fall in the critical region, we do not reject the null hypothesis for the classical approach. Also, find the p-value associated with the calculated F. Since there isn't a direct way to calculate the p-value from the F-distribution, online calculators or statistical software may be used. If the computed p-value is larger than \(0.05\), we do not reject the null hypothesis using the p-value approach.

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Key Concepts

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

Null Hypothesis

In any statistical testing, the null hypothesis, often denoted as \( H_0 \), stands as the default or status quo assumption. It typically asserts that there is no effect or no difference between groups or variables being studied. For instance, when assessing the variance of dispensing machines, as in the problem we are discussing, the null hypothesis asserts that the variances of the two different machines are equal. This can be mathematically represented as \( H_0: \sigma^2_H = \sigma^2_F \), where \( \sigma^2_H \) and \( \sigma^2_F \) are the population variances of the Harvard and Fizzit machines, respectively. Establishing a null hypothesis provides a baseline for testing whether the gathered evidence from the samples is strong enough to reject this initial assumption.

Alternative Hypothesis

The alternative hypothesis, denoted as \( H_1 \) or \( H_a \), represents a statement that contradicts the null hypothesis. It is what the researcher seeks to prove or at least gather evidence in support of. In the context of our variance test, the alternative hypothesis posits that there is an actual difference in variance between the Harvard and Fizzit machines, formulated as \( H_1: \sigma^2_H eq \sigma^2_F \). In statistical testing, we evaluate the evidence to see if we can reject the null hypothesis in favor of the alternative, thereby concluding that there is a statistically significant difference.

F-distribution

The F-distribution is a continuous probability distribution that arises commonly when comparing the variances of two independent samples, as we are in our dispenser machine example. This distribution is asymmetric and depends on degrees of freedom, which are derived from sample sizes. When conducting a variance ratio test, the test statistic follows an F-distribution under the null hypothesis. By calculating the F-statistic as the ratio of sample variances and comparing it to a value from the F-distribution, we can make decisions about the likelihood of observing such a ratio under the null hypothesis. It essentially helps determine if the variances are significantly different.

P-value approach

The p-value approach in hypothesis testing provides a measure of the evidence against the null hypothesis given by the data. It is the probability of obtaining a test statistic at least as extreme as the one that was actually observed, assuming that the null hypothesis is true. A small p-value indicates strong evidence against the null hypothesis, suggesting that we should reject it in favor of the alternative. If the p-value is larger than the chosen significance level, typically 0.05 or 5%, we do not reject the null hypothesis. This approach is widely used because it provides a clear criterion for decision making and is readily calculable with the aid of statistical software or online calculators, as in the step-by-step solution for our exercise.

Classical Approach

The classical approach, also known as the critical value approach, is another method of hypothesis testing. It involves comparing the calculated test statistic, in this case, the F statistic, against a critical value that defines the borders of the acceptance and rejection region of the test. These critical values are derived from the probability distribution of the test statistic and are dependent on the chosen significance level and the degrees of freedom associated with the data. If the test statistic falls into the rejection region, the null hypothesis is rejected; otherwise, it is not rejected. The classical approach does not provide a probability-like the p-value but offers a direct way to make the decision based on pre-determined critical values.