Question

A production plant has two complex fabricating systems, both of which are maintained at 2 -week intervals. However, one system is twice as old as the other. The number of finished products fabricated daily by each of the systems is recorded for 30 working days, with the results given in the table. Do these data present sufficient evidence to conclude that the variability in daily production warrants increased maintenance of the older fabricating system? Use the \(p\) -value approach. $$ \begin{array}{ll} \text { New System } & \text { Old System } \\ \hline \bar{x}_{1}=246 & \bar{x}_{2}=240 \\ s_{1}=15.6 & s_{2}=28.2 \end{array} $$

Short answer

Answer: Yes, there is sufficient evidence to conclude that the variability in daily production is significantly greater for the older system, which warrants increased maintenance for the older fabricating system.

Step-by-step solution

Calculate the sample variances

Calculate the sample variances for both the new system (\(s_1^2\)) and old system (\(s_2^2\)) using their respective given standard deviations (\(s_1\) and \(s_2\)): $$ s_1^2 = (15.6)^2 = 243.36 $$ $$ s_2^2 = (28.2)^2 = 795.24 $$

Calculate the F-statistic

Calculate the F-statistic using the sample variances calculated in Step 1 (\(s_1^2\) and \(s_2^2\)). The F-statistic is determined by dividing the largest variance by the smallest variance: $$ F = \frac{s_2^2}{s_1^2} = \frac{795.24}{243.36} = 3.267 $$

Determine the degrees of freedom

Find the degrees of freedom for the numerator (df1) and the denominator (df2). Since both systems were observed for 30 working days, we subtract 1 from 30 to find the degrees of freedom for each system: $$ \text{df1} = 30 - 1 = 29 $$ $$ \text{df2} = 30 - 1 = 29 $$

Determine the critical value and p-value

Using an F-distribution table or calculator, find the critical value and p-value corresponding to the F-statistic (3.267) and the degrees of freedom (29, 29). Here, we will assume a significance level of 0.05: $$ \text{Critical value (at 0.05 significance level)} \approx 1.835 $$ $$ \text{p-value} \approx 0.013 $$

Compare the p-value with the significance level and draw a conclusion

With the p-value of 0.013 being less than the significance level of 0.05, we have sufficient evidence to reject the null hypothesis that the two systems have equal variances. This means that the variability in daily production is significantly greater for the older system, which warrants increased maintenance for the older fabricating system.

Key concepts

F-statistic

The F-statistic is a crucial part of comparing two sample variances to determine if they are significantly different. It is calculated by taking the ratio of the two variances, specifically dividing the larger variance by the smaller one. This value tells us how much the variance of one sample differs from another. In this exercise, the F-statistic was calculated by using the variances from the older and new fabricating systems.

• Calculation: The formula for the F-statistic is \[F = \frac{s_2^2}{s_1^2}\] where \(s_2^2\) is the variance of the old system (795.24) and \(s_1^2\) is the variance of the new system (243.36).
• Interpretation: A high F-statistic indicates a larger difference in variances, suggesting that the variability in data is not due to random chance. In contrast, an F-statistic near 1 suggests similar variances.

Understanding the F-statistic helps in determining whether maintenance decisions are warranted based on the variability observed in production systems.

Degrees of freedom

Degrees of freedom (df) refer to the number of values in a statistical calculation that are free to vary. When calculating the F-statistic, each sample variance requires us to account for the degrees of freedom.

• Calculation:For the production data, we subtract 1 from the sample size of each system (30), which gives 29 degrees of freedom for each.\[\text{df1} = n_1 - 1 = 29, \quad \text{df2} = n_2 - 1 = 29\]
• Role in Hypothesis Testing:The degrees of freedom are used to determine the critical value from F-distribution tables. Having the correct degrees of freedom ensures the results are statistically valid and the comparison between variances is sound.

Degrees of freedom are key to understanding how much information is incorporated into our statistical tests, ultimately affecting the p-value and decision-making process.

Null Hypothesis

In hypothesis testing, the null hypothesis (\(H_0\)) represents a statement of no effect or no difference, which we seek to test against an alternative hypothesis.

• Statement:In this exercise, the null hypothesis is that the variances of the two fabrication systems are equal, indicating that the differences in production variability are not significant.
• Decision Based on p-value:We'll compare our calculated p-value to a pre-defined significance level (commonly 0.05). If the p-value is lower than the significance level, we reject the null hypothesis, suggesting that there is significant evidence to indicate a difference in variances.

Rejecting the null hypothesis leads to the conclusion that increased maintenance is justified for the older system due to its higher variability.

Sample Variance

Sample variance is a measure of the dispersion or spread of a set of data points. It tells us how much the data varies from its mean and is crucial in the context of hypothesis testing. In this problem, we calculate variances for each system to understand the variability in production.

• Calculation:Sample variance \(s^2\) for each system is calculated by squaring the standard deviation \(s\). In this exercise, for the new system: \[s_1^2 = (15.6)^2 = 243.36\] For the old system:\[s_2^2 = (28.2)^2 = 795.24\]
• Importance:The variance is crucial as the F-statistic relies on these values. Larger variance means more spread in data, impacting reliability and maintenance decisions for the systems.

By examining sample variance, we gain insights into the consistency and reliability of production across different systems.