Question

The stability of measurements on a manufactured product is important in maintaining product quality. A manufacturer of lithium batteries, such as the ones used for digital cameras, suspected that one of the production lines was producing batteries with a wide variation in length of life. To test this theory, he randomly selected \(n\) \(=50\) batteries from the suspect line and \(n=50\) from a line that was judged to be "in control." He then measured the length of time (in hours) until depletion to \(0.85 \mathrm{~V}\) with a 5-Ohm load for both samples. The sample means and variances for the two samples were as follows: $$ \begin{array}{ll} \text { Suspect Line } & \text { Line "in Control" } \\ \hline \bar{x}_{1}=9.40 & \bar{x}_{2}=9.25 \\ s_{1}=.25 & s_{2}=.12 \end{array} $$ a. Do the data provide sufficient evidence to indicate that batteries produced by the "suspect line" have a larger variance in length of life than those produced by the line that is assumed to be in control? Test using \(\alpha=.05\). b. Find the approximate \(p\) -value for the test and interpret its value. c. Construct a \(90 \%\) confidence interval for the variance ratio.

Short answer

In this exercise, we were asked to determine whether there is a significant difference in the variance of battery life between two different production lines. Using the F-test, we calculated the F-statistic to be 2.0833, which was larger than the critical value of 1.68. This led us to reject the null hypothesis in favor of the alternative hypothesis, suggesting that the "suspect line" has a larger variance in battery life than the "in control" line. Additionally, we calculated the p-value to be approximately 0.035, which is less than the significance level of 0.05, further supporting our previous conclusion. Finally, we constructed a 90% confidence interval for the variance ratio to be approximately (0.9536, 4.5477).

Step-by-step solution

State the null and alternative hypotheses

The null hypothesis (H0) and alternative hypothesis (H1) are as follows: H0: There is no difference in the variance of the battery life between the two production lines (σ1^2 = σ2^2). H1: The variance in battery life is larger for batteries produced by the "suspect line" (σ1^2 > σ2^2).

Perform the F-test

In this step, we will perform the F-test to test the hypothesis. The F-test statistic is calculated as follows: $$ F = \dfrac{s_1^2}{s_2^2} $$ where \(s_1^2\) is the sample variance from the "suspect line" and \(s_2^2\) is the sample variance from the "in control" line. Using the given values, we have: $$ F = \dfrac{0.25}{0.12} \approx 2.0833 $$

Find the critical value

The critical value of the F-distribution can be found by looking up an F-distribution table using the degrees of freedom (n1 - 1, n2 - 1) and the significance level α. In this case, we have: Degrees of freedom: (50 - 1, 50 - 1) = (49, 49) Significance level: α = 0.05 From the F-distribution table, the critical value is approximately 1.68.

Compare the test statistic to the critical value

Now, we will compare the F-statistic we calculated in Step 2 to the critical value we found in Step 3. $$ F = 2.0833 > 1.68 $$ Since the F-statistic is greater than the critical value, we reject the null hypothesis in favor of the alternative hypothesis.

Calculate the p-value

To calculate the p-value, we use the F-distribution with 49 degrees of freedom in the numerator and 49 degrees of freedom in the denominator. We find the probability that an F value will be greater than the F-statistic calculated in Step 2. Using a statistical software or an online calculator, the p-value is approximately 0.035.

Interpret the p-value

The p-value represents the probability of observing a value as extreme or more extreme than the F-statistic if the null hypothesis were true. Since the p-value is less than the significance level (0.035 < 0.05), we reject the null hypothesis, which means there is evidence that the "suspect line" has a larger variance in battery life than the "in control" line.

Construct a 90% confidence interval for the variance ratio

To construct a 90% confidence interval for the variance ratio, we will use the F-distribution. The lower and upper bounds for the confidence interval are given by: $$ \dfrac{s_1^2}{s_2^2} \times \dfrac{1}{F_{\frac{1-\text{ Confidence Level }}{2}, \text{ DF1, DF2}}}, \quad \dfrac{s_1^2}{s_2^2} \times F_{\frac{1+\text{ Confidence Level }}{2}, \text{ DF1, DF2}} $$ Using the given values and the F-distribution table: Lower bound = 2.0833 × 1/2.183 = 0.9536 Upper bound = 2.0833 × 2.183 = 4.5477 Thus, the 90% confidence interval for the variance ratio is approximately (0.9536, 4.5477).

Key concepts

F-test

The F-test is a statistical methodology used to compare two population variances. It helps us to determine if the variances from two different data sets are significantly different from each other. This is crucial when evaluating the consistency of production processes, like in the manufacture of lithium batteries. In our scenario, the F-test is used to assess if the variance in battery life length from the 'suspect line' is greater than that from the 'control line'. The formula \[ F = \frac{s_1^2}{s_2^2} \]defines how to calculate the F-statistic, where \(s_1^2\) is the variance of the suspect line and \(s_2^2\) is the variance of the control line. To determine the significance, the F-statistic is compared against a critical value from the F-distribution table, which takes account of the degrees of freedom from each group.

• If the F-statistic is higher than this critical value, as in our case with \( F = 2.0833 \) being greater than \( 1.68 \), the null hypothesis is rejected, supporting the alternative hypothesis of a larger variance in the suspect line.
• On the contrary, if it were lower, we would not have sufficient evidence to infer a greater variance in the suspect line.

Variance comparison

Variance comparison is about understanding how much variability is present within data sets. Variance provides insight into how spread out the data points are. In production settings, such as battery manufacturing, high variance might indicate inconsistencies or quality control issues. It is calculated by evaluating the square of the standard deviation.

When performing variance comparison, it is vital to assess whether the difference in variances between the two groups is statistically significant. This involves comparing computed variances of the suspect line and the control line, which in our example is \(0.25\) and \(0.12\) respectively. By comparing these values through the F-test, we determine if the observed variance in battery life from the suspect line is much larger than from the control line.

• The larger variance implies more inconsistency.
• This can lead to potential reliability issues for the end product if not managed properly.

Confidence interval

A confidence interval provides a range of plausible values for an unknown parameter, here being the variance ratio. In hypothesis testing, it offers a range within which the true value of the ratio of variances is likely to fall, with a certain level of confidence. This aids in understanding the precision and reliability of the estimate.

To construct a confidence interval for our variance ratio, we use the F-distribution. The 90% confidence interval method involves calculating bounds using the F-statistic and critical values from the F-distribution. The calculations involve\[\text{Lower bound} = \dfrac{s_1^2}{s_2^2} \times \dfrac{1}{F_{0.05,49,49}}, \quad \text{Upper bound} = \dfrac{s_1^2}{s_2^2} \times F_{0.95,49,49}\]and substituting the values accordingly.

• This provides us with a range from approximately \(0.9536\) to \(4.5477\), which means we are 90% confident that the true variance ratio lies within this interval.
• This interval gives us insight into the variability comparison, helping make informed decisions about process control.

p-value interpretation

The p-value is a key aspect of hypothesis testing. It indicates the probability of obtaining test results at least as extreme as the ones observed, assuming the null hypothesis is true. It serves a critical role in deciding whether to reject the null hypothesis.

In our analysis with an F-statistic from the suspect line, the p-value calculated is 0.035. This value helps us interpret the test results.

• If the p-value is below the given significance level \(\alpha = 0.05\), as it is here, there is strong evidence against the null hypothesis, prompting us to reject it. This implies that the variance in the suspect line is indeed greater than that in the control line.
• If it had been larger than 0.05, we would not reject the null hypothesis and would conclude there's not enough evidence to suggest a discrepancy in variances.

The p-value thus quantifies the strength of evidence against the null hypothesis, guiding decisions in quality control and production adjustments.