/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 An automotive company is conside... [FREE SOLUTION] | 91Ó°ÊÓ

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An automotive company is considering two types of batteries for its automobile. Sample information on the life of the battery is being used. Twenty batteries of type \(A\) and twenty batteries of type \(B\) are being used. The summary statistics are \(X A=32.91\), \(x_{B}=30.47, s_{A}=1.57,\) and \(s_{B}-1.74 .\) Assume the data on each battery are normally distributed and assume \(\sigma_{A}=\sigma_{B}\) (a) Find a \(95 \%\) confidence interval on \(\mu_{A}-\mu_{B}\). (b) Draw some conclusion from (a) that provides some insight into whether \(A\) or \(B\) should be adopted.

Short Answer

Expert verified
The 95% confidence interval for the difference of the mean lives of the two batteries is from 1.690772 to 3.189228. It suggests that battery A has a significantly longer lifespan than battery B on average.

Step by step solution

01

Calculation of sample mean difference

Firstly, need to subtract the sample mean of battery B from the sample mean of battery A. So, \(D_{x} = X_{A} - X_{B} = 32.91 - 30.47 = 2.44\).
02

Calculation of pooled variance

Since it's given that the standard deviations of both populations are equal, calculate the pooled variance. Begin by calculating variances for each group, \(s^2_{A} = (s_{A})^2 = (1.57)^2 = 2.4649\) and \(s^2_{B} = (s_{B})^2 = (1.74)^2 = 3.0276\). The pooled variance is then calculated as \((s^2_{P}) = (s^2_{A} + s^2_{B}) / 2 = (2.4649 + 3.0276) / 2 = 2.74625\).
03

Calculation of standard error

Next is to calculate the standard error of the difference using the pooled variance and sample sizes. The standard error is given by \(\sqrt{s^2_{P} / n_{A} + s^2_{P} / n_{B}} = \sqrt{2.74625/20 + 2.74625/20} = \sqrt{0.1373125} = 0.370558\).
04

Calculation of confidence interval

Now we calculate the confidence interval for the difference in means at the 95% confidence level. From the t-distribution table, the t-score for df=38 (degrees of freedom = n-1 = 20-1 = 19, for each group and combined for both groups gives 38) with 95% confidence level is approximately 2.024. Therefore, the confidence interval will be \(D_{x} \pm t*s_{D_{x}}\) where \(D_{x}\) is the sample mean difference and \(s_{D_{x}}\) is the standard error of the difference. Plugging in the values we find, the confidence interval becomes \(2.44 \pm 2.024 * 0.370558 = 2.44 \pm 0.749228\) which gives two values (1.690772, 3.189228).
05

Conclusion

The 95% confidence interval for the difference of the mean lives of the two batteries is from 1.690772 to 3.189228. This indicates that we are 95% confident that the mean life of battery A is between 1.690772 and 3.189228 units greater than the mean life of battery B. Since all values in this interval are positive, it suggests that battery A has a significantly longer lifespan than battery B on average.

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Key Concepts

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

Battery Comparison
When comparing two different types of automotive batteries, understanding how their lifespan differs on average is key. In this scenario, we aim to determine which battery, Type A or Type B, has a longer average life. By gathering sample data from twenty batteries of each type, we're able to calculate summary statistics like the mean ( X ext{ and }x_A=32.91 ext{ and } X_B=30.47 ext{. ) To evaluate which battery is superior regarding lifespan, examining the mean difference provides insights into performance. The initial calculation involves finding the sample mean difference, also known as D_x . This is done by subtracting the mean of Battery B from that of Battery A ( 32.91 - 30.47 = 2.44 ). Through this process, it becomes evident whether there's a significant advantage in choosing one battery over the other based on mean lifespan alone, making it crucial to proceed with further statistical analysis.
Pooled Variance
Pooled variance is a technique used to estimate a common variance for two or more groups. It’s particularly useful when you assume equal population variances, which is true in this exercise for Battery Types A and B.To calculate pooled variance, we first determine the individual variances of each battery group. The sample variance for Battery A is s_A^2 = (1.57)^2 = 2.4649, and for Battery B, s_B^2 = (1.74)^2 = 3.0276. By combining these, the pooled variance is calculated as (s^2_P). The formula is:\[s^2_P = rac{(s^2_A + s^2_B)}{2} = rac{(2.4649 + 3.0276)}{2} = 2.74625\]The resulting pooled variance captures both groups' data variability more robustly than using individual variances alone. It serves as a foundation for further steps like standard error and confidence intervals, providing a shared estimate of variance.
Standard Error Calculation
After determining the pooled variance, the standard error of the difference in means provides insight into the variability expected when comparing battery life spans. The standard error reflects how much the sample means (mean life of the batteries) would vary if we were to take multiple samples.To compute the standard error:1. Apply the pooled variance (s^2_P = 2.74625) calculated earlier.2. Use the formula for standard error, given by:\[s_{D_x} = \sqrt{\frac{s^2_P}{n_A} + \frac{s^2_P}{n_B}}\]Inserting the values yields:\[s_{D_x} = \sqrt{\frac{2.74625}{20} + \frac{2.74625}{20}} = \sqrt{0.1373125} = 0.370558\]This result, (0.370558), quantifies the uncertainty in our measurement of the average lifespan difference, guiding the construction of confidence intervals.
t-Distribution
The t-distribution is essential when working with sample data, especially when the sample size is small and the population standard deviation is unknown. In comparing battery life, it helps form conclusions about means using sample data.Since we are interested in a 95% confidence interval for the mean difference between battery types, we use the t-distribution. Based on the degrees of freedom (df = 38)—the sum of sample sizes (20+20-2)—a t-score is consulted to find how values are distributed.For a 95% confidence level, this t-score is approximately 2.024. Subsequently, the confidence interval formula is:\[D_x \pm t \times s_{D_x}\]In this example, it translates to:\[2.44 \pm 2.024 \times 0.370558 = 2.44 \pm 0.749228\]This gives a confidence interval of (1.690772, 3.189228), meaning we are 95% sure the mean life of Battery A is between 1.690772 and 3.189228 units greater than Battery B. As the interval contains only positive numbers, Battery A is revealed to statistically have a longer mean lifespan.

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Most popular questions from this chapter

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