Question

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

Answer 1

To determine whether the machine is out of order based on the sample variance, we can perform a hypothesis test. Here's how we can approach it:

Null Hypothesis (H0): The machine is not out of order (variance ≤ 1.15 deciliters). Alternative Hypothesis (H1): The machine is out of order (variance > 1.15 deciliters).

Significance level (α) = 0.05

Test statistic: We can use the chi-square distribution to test the hypothesis. The test statistic follows a chi-square distribution with (n-1) degrees of freedom, where n is the sample size. In this case, n = 25 - 1 = 24.

Critical chi-square value: To determine the critical chi-square value at the 0.05 level of significance with 24 degrees of freedom, we consult a chi-square distribution table or use statistical software. The critical value is found to be approximately 36.42.

Calculation: The test statistic is calculated using the formula:

chi-square = ((n - 1) * sample variance) / population variance

In this case, population variance = 1.15 deciliters.

chi-square = (24 * 2.03) / 1.15 = 42.12

Comparison: Since the test statistic (42.12) is greater than the critical chi-square value (36.42), we reject the null hypothesis.

Conclusion: At the 0.05 level of significance, there is sufficient evidence to conclude that the machine is out of order because the sample variance (2.03 deciliters) exceeds the threshold of 1.15 deciliters.

Therefore, based on the given information, we can say that the machine is out of order.