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Employees of a certain state university system can choose from among four different health plans. Each plan differs somewhat from the others in terms of hospitalization coverage. Four random samples of recently hospitalized individuals were selected, each sample consisting of people covered by a different health plan. The length of the hospital stay (number of days) was determined for each individual selected. a. What hypotheses would you test to decide whether the mean lengths of stay are not the same for all four health plans? b. If each sample consisted of eight individuals and the value of the ANOVA \(F\) statistic was \(F=4.37\), what conclusion would be appropriate for a test with \(\alpha=0.01 ?\) c. Answer the question posed in Part (b) if the \(F\) value given there resulted from sample sizes \(n_{1}=9, n_{2}=8, n_{3}=7\), and \(n_{4}=8\).

Short Answer

Expert verified
a. Null hypothesis (\(H_{0}\)): \(\mu_{1} = \mu_{2} = \mu_{3} = \mu_{4}\) and alternative hypothesis (\(H_{a}\)): at least one mean is different. b. Since the calculated F statistic \(4.37\) is less than the critical F value \(5.57\), we fail to reject the null hypothesis. There is insufficient evidence to claim that the mean lengths of stay are different for all four health plans at the 0.01 significance level. c. The calculated F statistic \(4.37\) is still less than the critical F value \(5.72\), and we fail to reject the null hypothesis. There is insufficient evidence to claim that the mean lengths of stay are different for all four health plans at the 0.01 significance level.

Step by step solution

01

State the Hypotheses

To decide whether the mean length of stays are not the same for all four health plans, we need to set up the null and alternative hypotheses: Null hypothesis (\(H_{0}\)): The mean lengths of stay are the same for all four health plans, i.e., \(\mu_{1} = \mu_{2} = \mu_{3} = \mu_{4}\). Alternative hypothesis (\(H_{a}\)): The mean lengths of stay are not the same for all four health plans, i.e., at least one mean is different.
02

Calculate F Statistic and Degrees of Freedom for Part (b)

We are given the F statistic as \(F=4.37\) and each sample consists of 8 individuals. To find the appropriate conclusion for a test with a significance level of \(α=0.01\), we need to calculate the critical F value. The degrees of freedom for the between-groups variance (numerator) is \(k-1 = 4-1 = 3\), where \(k\) is the number of groups. The degrees of freedom for the within-groups variance (denominator) is \(N-k = (8+8+8+8)-4 = 28\), where \(N\) is the total number of observations. Now, we will find the critical F value using the F-distribution table for \(α=0.01\), \(df_{1} = 3\), and \(df_{2} = 28\).
03

Determine the Conclusion for Part (b)

The critical F value for \(α=0.01\), \(df_{1} = 3\), and \(df_{2} = 28\) is approximately \(5.57\). Since the calculated F statistic \(4.37\) is less than the critical F value \(5.57\), we fail to reject the null hypothesis. Therefore, we conclude that there is insufficient evidence to claim that the mean lengths of stay are different for all four health plans at the 0.01 significance level.
04

Calculate F Statistic and Degrees of Freedom for Part (c)

For Part (c), we are given the same F statistic \(F=4.37\), but the sample sizes are different: \(n_{1}=9, n_{2}=8, n_{3}=7,\) and \( \)n_{4}=8$. As before, we need to calculate the critical F value. The degrees of freedom for the between-groups variance (numerator) remains \(k - 1 = 3\). The degrees of freedom for the within-groups variance (denominator) is now \(N-k = (9+8+7+8)-4 = 26\). Now, we will find the critical F value using the F-distribution table for \(α=0.01\), \(df_{1} = 3\), and \(df_{2} = 26\).
05

Determine the Conclusion for Part (c)

The critical F value for \(α=0.01\), \(df_{1} = 3\), and \(df_{2} = 26\) is approximately \(5.72\). Since the calculated F statistic \(4.37\) is still less than the critical F value \(5.72\), we fail to reject the null hypothesis. As a result, our conclusion remains the same: there is insufficient evidence to claim that the mean lengths of stay are different for all four health plans at the 0.01 significance level.

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Key Concepts

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

Null Hypothesis
In the realm of hypothesis testing, the null hypothesis serves as a default statement asserting that no effect, difference, or relationship exists among the groups being studied. For the health plan analysis, the null hypothesis \(H_0\) posits that the mean lengths of stay in the hospital for all four health plans are equal, mathematically expressed as \(\mu_1 = \mu_2 = \mu_3 = \mu_4\). The null hypothesis provides a benchmark that the observed data are compared against to determine if there is statistical evidence to support the existence of any significant differences. It is crucial to understand that 'failing to reject' or 'rejecting' the null does not prove the hypothesis true or false; it simply reflects the strength of the evidence against it.

When establishing a null hypothesis, clarity and simplicity are key. It should be straightforward and present a precise claim that can be tested with data. By setting a defined null hypothesis, researchers can apply statistical tests to explore the validity of their predictions and gain insights into the underlying relationships within their data.
Alternative Hypothesis
The alternative hypothesis, symbolized as \(H_a\) or \(H_1\), stands in contrast to the null hypothesis. It represents the assertion that there is a significant difference, effect, or relationship present in the population, which the researcher wishes to support with empirical data. In the context of the health plan example, the alternative hypothesis posits that not all four health plans have the same mean length of hospital stay – shorthand, at least one mean differs from the others. An effective alternative hypothesis encompasses all possible scenarios that differ from the null hypothesis and sets the stage for discovering new findings through hypothesis testing. If the data provide compelling evidence against the null, one may endorse the alternative hypothesis, suggesting that further investigation into the nature of these differences is warranted.

Employing a clear alternative hypothesis enables investigators to pinpoint the nature of the differences they are seeking and directs the analysis in a scientifically meaningful manner.
F Statistic
In hypothesis testing, the F statistic plays a critical role in ANOVA (Analysis of Variance). This statistic is a ratio that compares the variance among group means to the variance within the groups, serving as a tool to test if the means of different groups are significantly different from each other. In simpler terms, a large F statistic suggests group means are not all equal, hence providing evidence against the null hypothesis. For the given exercise, an F statistic value of \(F=4.37\) was calculated based on the provided data from the four health plans.

A higher F value typically indicates a greater disparity among group means, possibly signaling that at least one group differs significantly from the others. It's the comparison between this calculated F statistic and a critical value from the F distribution that informs the researcher whether to reject the null hypothesis. Understanding the meaning of the F statistic is essential in interpreting ANOVA results and drawing the correct conclusions from the data.
Degrees of Freedom
The concept of degrees of freedom is vital when computing statistics like the F statistic. Degrees of freedom refer to the number of independent values or quantities that can be assigned to a statistical distribution while adhering to a given set of constraints. We often see degrees of freedom represented as \(df\), and in ANOVA, there are two kinds: one for the numerator (between-groups) and one for the denominator (within-groups).

In our exercise, the degrees of freedom for the between-groups variance \(df_1\) is \(k - 1\), where \(k\) is the number of groups, and for the within-groups variance \(df_2\), it is \(N - k\), with \(N\) being the total number of observations. For example, in scenario (b), \(df_1\) is 3 and \(df_2\) is 28. The correct calculation of degrees of freedom ensures the accurate application of statistical tests and is essential for deriving meaningful results from data.
Significance Level
The significance level, denoted by \(\alpha\), is a threshold that determines the likelihood of observing a test statistic as extreme as, or more so, than the one computed from the data, assuming the null hypothesis is true. In hypothesis testing, \(\alpha\) is pre-selected by the researcher and commonly set at 0.05, 0.01, or another level, depending on how stringent they wish the test to be. For our exercise, a significance level of \(\alpha = 0.01\) was chosen, indicating a 1% risk of concluding that a difference exists within the health plan groups' mean lengths of stay when there is none.

By choosing a low significance level, researchers show that they require strong evidence before rejecting the null hypothesis. This level reflects the balance between the likelihood of making a Type I error (false positive) and the desire to detect a true effect. A proper understanding of the significance level helps in assessing the rigor and reliability of hypothesis test results.
Failing to Reject the Null Hypothesis
The concept of failing to reject the null hypothesis can be perplexing; it does not mean that the null hypothesis is true, but rather that there's not enough evidence to conclude it's false. In our analysis of health plans, an F statistic of 4.37 was compared against critical values from F-distribution tables, and because it was not greater than the critical values (5.57 for part b and 5.72 for part c), we

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