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A soft-drink dispensing machine is said to be out of control if the variance of the contents exceeds 1.15 deciliters. If a random sample of 25 drinks from this machine has a variance of 2.03 deciliters, does this indicate at the 0.05 level of significance that the machinc is out of control? Assume that the contents are approximately normally distributed

Short Answer

Expert verified
There is significant evidence at the 0.05 level to conclude that the machine is out of control. This is because the variance of the soft-drink contents from the machine significantly exceeds 1.15 deciliters.

Step by step solution

01

State the Hypotheses

The null hypothesis (\(H_0\)) is that the variance is equal to 1.15 deciliters. The alternative hypothesis (\(H_a\)) is that the variance is greater than 1.15 deciliters (\(\sigma^2 > 1.15\)).
02

Compute the Test Statistic

We use the Chi-square test statistic formula, \(\chi^2 = (n-1)*s^2/\sigma^2\). Where n is the sample size (25), \(s^2\) is the sample variance (2.03), and \(\sigma^2\) is the variance under null hypothesis (1.15). So, \(\chi^2 = (25-1)*2.03/1.15 = 44.84\)
03

Determine the Rejection Region

The rejection region is \(\chi^2 > \chi^2_{1-\alpha, n-1}\), where \(\alpha\) is the significance level (0.05) and n is the sample size (25). The critical value for chi-square with 24 degrees of freedom (\(n-1\)) at 0.05 significance level (from the Chi-square distribution table) is 36.415.
04

Make the Statistical Decision

Since our calculated chi-square test statistic (44.84) is higher than the critical value (36.415), we reject the null hypothesis. This means there is enough statistical evidence at the 0.05 level of significance to conclude that the variance of the machine is greater than 1.15 deciliters.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a process used to determine if there is enough evidence to reject a null hypothesis in favor of an alternative hypothesis. In this context, the null hypothesis (\(H_0\)) posits that the variance of the soft-drink machine's output is 1.15 deciliters. The alternative hypothesis (\(H_a\)) suggests that the variance is greater than 1.15 deciliters.

To test these hypotheses, a Chi-square test is employed. This statistical method assesses the variance within a sample compared to what is expected under the null hypothesis. The outcome aids in making decisions about whether the observed data significantly deviates from the assumed population variance.

Key points to remember:
  • The null hypothesis is a statement of no effect or status quo.
  • The alternative hypothesis represents a change or difference from what is assumed in the null hypothesis.
  • Hypothesis testing involves calculating a test statistic, comparing it to a critical value, and making a decision to either reject or fail to reject the null hypothesis.
Variance Analysis
Variance analysis in hypothesis testing assesses whether the variation between measured and expected values is statistically significant. Here, we analyze if the variance of drinks dispensed by a machine is significantly higher than a predetermined value of 1.15 deciliters.

The sample variance calculated from the drinks (2.03 deciliters) is compared to the expected variance using the Chi-square test statistic. This is achieved using:\[ \chi^2 = \frac{(n-1) \times s^2}{\sigma^2} \] where:
  • \( n \) is the sample size (25).
  • \( s^2 \) is the sample variance (2.03).
  • \( \sigma^2 \) is the variance under the null hypothesis (1.15).
By calculating the test statistic, one can interpret whether the apparent discrepancy in variance is merely due to random sample fluctuations or indicates a real difference. This process is essential for quality control and verifying that the machines are operating correctly.
Significance Level
The significance level, often denoted as \( \alpha \), is a threshold used in hypothesis testing to decide when to reject the null hypothesis. It represents the probability of rejecting the null hypothesis when it is actually true, which is known as a Type I error.

In our exercise, the significance level is set at 0.05. This implies a 5% risk of concluding that the machine's variance exceeds the allowed limit when, in fact, it does not.

A crucial step in hypothesis testing is to compare the calculated test statistic with a critical value derived from statistical distributions. For our problem:
  • A Chi-square distribution with 24 degrees of freedom (since \( n-1 \) = 24) is used.
  • The critical value at a 0.05 significance level is 36.415.
When the test statistic (44.84) exceeds this critical value, it indicates that the variance is unlikely to be that high by chance alone, leading to the rejection of the null hypothesis. Setting the significance level involves balancing the risks of Type I and Type II errors (failing to reject a false null hypothesis) and understanding the consequences of these errors in the given context.

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