/*! 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 38 In a batch chemical process, two... [FREE SOLUTION] | 91Ó°ÊÓ

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A clinical trial is conducted to determine if a certain type of inoculation has an effect on the incidence of a certain disease. A sample of 1000 rats was kept in a controlled environment for a period of 1 year and 500 of the rats were given the inoculation. Of the group not given the drug, there were 120 incidences of the disease, while 98 of the inoculated group contracted it. If we call \(p_{1}\) the probability of incidence of the disease in uninoculated rats and \(p_{2}\) the probability of incidence after receiving the drug, compute a \(90 \%\) confidence interval for \(p_{1}-p_{2}\).

A random sample of 25 bottles of buffered aspirin contain, on average, \(325.05 \mathrm{mg}\) of aspirin with a standard deviation of \(0.5 \mathrm{mg}\). Find the \(95 \%\) tolerance limits that will contain \(90 \%\) of the aspirin contents for this brand of buffered aspirin. Assume that the aspirin content is normally distributed.

The following measurements were recorded for the drying time, in hours, of a certain brand of latex paint: $$\begin{array}{lllll}3.4 & 2.5 & 4.8 & 2.9 & 3.6 \\\2.8 & 3.3 & 5.6 & 3.7 & 2.8 \\\4.4 & 4.0 & 5.2 & 3.0 & 4.8\end{array}$$ Assuming that the measurements represent a random sample from a normal population, find the \(99 \%\) tolerance limits that will contain \(95 \%\). of the drying times.

According to USA Today (March 17. \(\lfloor 997\) ), women made up \(33.7 \%\) of the editorial staff at local TV stations in 1990 and \(36.2 \%\) in \(1994 .\) Assume 20 new employees were: hired as editorial staff. (a) Estimate the number that would have been women in each year, respectively, (b) Compute a \(95 \%\) confidence interval to see if there is evidence that the proportion of women hired as editorial staff in 1994 was higher than the proportion hired in 1990 .

An anthropologist is interested in the proportion of individuals in two Indian tribes with double occipital hair whorls. Suppose that independent samples are taken from each of the two tribes, and it is found that 24 of 100 Indians from tribe \(A\) and 36 of 120 Indians from tribe \(B\) possess this characteristic. Construct a \(95 \%\) confidence interval for the difference \(p_{B}-p_{A}\) between the proportions of these two tribes with occipital hair whorls.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.