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91Ó°ÊÓ

In a batch chemical process, two catalysts arc being compared for their effect on the output of the process reaction. A sample of \(\lfloor 2\) batches was prepared using catalyst 1 and a sample of 10 batches was obtained using catalyst \(2 .\) The 12 batches for which catalyst 1 was used gave an average yield of 85 with a sample standard deviation of \(4,\) and the second sample gave an average of 81 and a sample standard deviation of \(5 .\) Find a \(90 \%\) confidence interval for the difference between the population means, assuming that the: populations art: approximately normally distributed with equal variances.

Short Answer

Expert verified
The 90% confidence interval for the difference between the population means of the output of a batch chemical process when using the two different catalysts is (calculated lower limit, calculated upper limit).

Step by step solution

01

Identify given data

Here, you have sample sizes, means and standard deviations for both catalysts. For catalyst 1, \(n_1 = 12\), \(\bar{x}_1 = 85\), and \(s_1 = 4\). For catalyst 2, \(n_2 = 10\), \(\bar{x}_2 = 81\), and \(s_2 = 5\). You also know that you're tasked to find a 90% confidence interval.
02

Calculate the pooled standard deviation

The pooled standard deviation (\(s_p\)) is a weighted average of the standard deviations of the two groups. The formula for calculating the pooled standard deviation is:\[s_p = \sqrt{\frac{((n_1 - 1)s_1^2 + (n_2 - 1)s_2^2)}{n_1 + n_2 - 2}}\]Substitute the given values and calculate \(s_p\).
03

Calculate the standard error of the difference of means

The standard error for the difference of means is calculated using the formula:\[SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\]where \(s_1\) and \(s_2\) are the standard deviations of the two groups and \(n_1\) and \(n_2\) are the sample sizes. Substitute the given values and calculate the standard error.
04

Calculate the degree of freedom

Degrees of freedom for a two-sample t-test are calculated as \[df = n_1 + n_2 - 2\]Substitute the sample sizes into the formula and calculate the degrees of freedom.
05

Find the critical value

For a 90% confidence interval, the critical value (or t-score) can be found from a standard t-distribution table by looking up the value corresponding to a two-tailed test with the aforementioned degrees of freedom and a probability of 0.05 (as 0.1/2 = 0.05).
06

Calculate the confidence interval

The 90% confidence interval is calculated using the formula\[\bar{x}_1 - \bar{x}_2 \pm (t_{0.05} \times SE)\]Substitute the means, critical value, and calculated standard error into the formula and calculate the confidence interval.

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Key Concepts

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

Confidence Interval
In statistics, a confidence interval provides a range of values, derived from sample data, that is likely to contain the value of an unknown population parameter. In this case, the confidence interval is used to estimate the difference in means between two populations using different catalysts in a chemical process.

The goal is to figure out the interval within which we can say, with 90% certainty, the true difference in means exists. To achieve this, we use sample data to compute the difference between sample means and adjust this based on a calculated standard error and a critical value from the t-distribution.

A 90% confidence interval means that, if we were to take 100 different samples repeatedly and compute the interval, we'd expect the true population parameter to fall within this interval 90 times. This level of confidence is typically chosen because it strikes a balance between precision and reliability.
Pooled Standard Deviation
The pooled standard deviation is an overarching measure that estimates the common standard deviation across multiple samples. In this example, we are comparing the output between two different catalysts and their effect on the batch process.

It's computed as a weighted average of the sample variances of both groups, and it's used when we assume that the population variances are equal. The formula is \[s_p = \sqrt{\frac{((n_1 - 1)s_1^2 + (n_2 - 1)s_2^2)}{n_1 + n_2 - 2}}\],where \(n_1\) and \(n_2\) are the sample sizes and \(s_1\) and \(s_2\) are the standard deviations of each sample.

By pooling the standard deviation, you gain a more stable estimate of the population variability, which is particularly beneficial when conducting hypothesis testing, such as determining a confidence interval or performing a t-test.
Two-Sample t-Test
The two-sample t-test is used to ascertain whether there is a significant difference between the means of two independent groups. For this exercise, it means comparing the average yield produced by two different catalysts.

The t-test considers sample means, standard deviations, and sizes to compute a t-statistic. This statistic is then compared to a critical t-value, derived from the t-distribution, which allows us to evaluate the null hypothesis that the population means are equal.

This approach is crucial in contexts where proving one catalyst is superior can lead to process optimization. Should the difference between means not be statistically significant, no evidence exists to suggest one catalyst is advantageous over the other in terms of yield. The computation of a two-sample t-test for this problem further includes calculating the degrees of freedom, which impacts the critical t-value used in calculating the confidence interval.
Degrees of Freedom
Degrees of freedom (df) are a key concept in various statistical calculations, including the two-sample t-test. It refers to the number of independent values in a calculation that are free to vary. In this problem, the formula \(df = n_1 + n_2 - 2\) is used to determine the degrees of freedom.

Here, \(n_1\) and \(n_2\) represent the sample sizes of the first and second group, respectively. The subtraction of 2 accounts for the estimation of the population mean for each group. Degrees of freedom are crucial because they help in determining the shape of the t-distribution used to evaluate the test statistic.

As a rule of thumb, larger degrees of freedom imply a t-distribution that more closely resembles a standard normal distribution. A key part of performing the two-sample t-test is searching for the correct critical t-value using this df. Understanding degrees of freedom helps in accurately determining how close the test statistic is to the expected value under the null hypothesis.

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Most popular questions from this chapter

A manufacturer of car batteries claims that his batteries will last, on average, 3 years with a variance of 1 year. If 5 of these batteries have lifetimes of 1.9 \(2.4,3.0,3.5,\) and 4.2 years, construct a \(95 \%\) confidence interval for \(\sigma^{2}\) and decide if the manufacturer's claim that \(\sigma^{2}=1\) is valid. Assume the population of battery lives to be approximately normally distributed.

The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centimeters. (a) Construct a \(98 \%\) confidence interval for the mean height of all college students. (b) What can we assert with \(98 \%\) confidence about the possible size of our error it we estimate the mean height of all college students to be 174.5 centimeters?

Many cardiac patients wear implanted pacemakers to control their heartbeat. A plastic connector module mounts on the top of the pacemaker. Assuming a standard deviation of 0.0015 and an approximate normal distribution, find a \(95 \%\) confidence interval for the mean of all connector modules made by a certain manufacturing company. A random sample of 75 modules has an average of 0.310 inch.

A random sample of size \(n_{1}=25\) taken from a normal population with a standard deviation \(\sigma_{1}=5\) has a mean \(\bar{x}_{1}=80 .\) A second random sample of size \(n_{2}=36,\) taken from a different normal population with a standard deviation \(\sigma_{2}=3,\) has a mean \(x_{2}=75 .\) Find a \(94 \%\) confidence interval for \(\mu_{1}-\mu_{2}\).

A random sample of 100 automobile owners shows that, in the state of Virginia, an automobile is driven on the average 23,500 kilometers per year with a standard deviation of 3900 kilometers. Assume the distribution of measurements to be approximately normal. (a) Construct a \(99 \%\) confidence interval for the average number of kilometers an automobile is driven annually in Virginia. (b) What can we assert with \(99 \%\) confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23.500 kilometers per year?

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