/*! 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 11 The yield of a chemical process ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The yield of a chemical process is being studied. From previous experience, yield is known to be normally distributed and \(\sigma=3\). The past five days of plant operation have resulted in the following percent yields: \(91.6,88.75,90.8,89.95,\) and \(91.3 .\) Find a \(95 \%\) two-sided confidence interval on the true mean yield.

Short Answer

Expert verified
The 95% confidence interval for the mean yield is (87.85, 93.11).

Step by step solution

01

Calculate the Sample Mean

First, we need to calculate the sample mean \( \bar{x} \) of the given yields. The yields are \(91.6, 88.75, 90.8, 89.95, \text{ and } 91.3\). Sum these values and divide by the number of yields (which is 5).\[\bar{x} = \frac{91.6 + 88.75 + 90.8 + 89.95 + 91.3}{5} = \frac{452.4}{5} = 90.48\]
02

Identify Standard Deviation and Sample Size

For a confidence interval calculation, notice that the standard deviation \( \sigma = 3\) is provided, and the sample size \( n = 5 \).
03

Determine the Confidence Interval Formula

Since we are estimating the mean with a known standard deviation and a normal distribution, we use the z-distribution. The formula for the confidence interval is:\[CI = \bar{x} \pm z_{\alpha/2} \cdot \left(\frac{\sigma}{\sqrt{n}}\right)\]Here, \( \alpha = 0.05 \) for a 95% confidence level, making \(\alpha/2 = 0.025\). The z-value corresponding to \(0.025\) in either tail of a standard normal distribution is approximately \(1.96\).
04

Calculate the Margin of Error

The margin of error is computed using:\[ME = z_{\alpha/2} \cdot \left(\frac{\sigma}{\sqrt{n}}\right)\]Substitute in the values:\[ME = 1.96 \cdot \left(\frac{3}{\sqrt{5}}\right) = 1.96 \cdot 1.3416 \approx 2.63\]
05

Compute the Confidence Interval

Now that we have the margin of error, calculate the confidence interval by adding and subtracting the margin of error from the sample mean:\[CI = 90.48 \pm 2.63\]This gives us:\[CI = (90.48 - 2.63, 90.48 + 2.63) = (87.85, 93.11)\]
06

Interpret the Result

The 95% confidence interval for the true mean yield is from 87.85 to 93.11. This means we are 95% confident that the true mean yield of the chemical process lies within this 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.

Normal Distribution
Normal distribution is one of the most important statistical concepts. It describes how data points are spread out.
This distribution is bell-shaped and symmetric around the mean. Data near the mean are more frequent than data far from the mean.
  • Mean: The average value, central peak of the distribution.
  • Standard deviation: Measures how dispersed the values are around the mean.
  • 68-95-99.7 rule: About 68% of data falls within one standard deviation from the mean, 95% within two, and 99.7% within three.
Understanding normal distribution helps in estimating probabilities and making predictions based on data. When a dataset follows a normal distribution, we can apply statistical methods, like confidence intervals, to draw meaningful conclusions.
Sample Mean Calculation
The sample mean is calculated by adding all data values together and dividing by the number of values. It is symbolized by \(\bar{x}\). This measure provides an estimate of the population mean.
In the exercise, the chemical yields were: 91.6, 88.75, 90.8, 89.95, and 91.3. Adding these values results in 452.4. Then, divide by the total number of observations (5) to get the sample mean:\[\bar{x} = \frac{452.4}{5} = 90.48\]
  • Ensures each data point is equally represented.
  • Balances individual differences, providing a central tendency.
The sample mean provides a foundation for further statistical analyses, like calculating confidence intervals.
Margin of Error
The margin of error (ME) quantifies the uncertainty in the estimate of the mean. It helps in understanding how precise our estimate is.
It is calculated by multiplying the z-value (from the z-distribution) by the standard error of the mean.From the exercise, the given standard deviation (\(\sigma\)) is 3, and the sample size (\(n\)) is 5. Using the z-value for a 95% confidence level (1.96), the margin of error is:\[ME = z_{\alpha/2} \cdot \left(\frac{\sigma}{\sqrt{n}}\right) = 1.96 \cdot \left(\frac{3}{\sqrt{5}}\right)\approx 2.63\]
  • Provides a range around the sample mean within which we expect the population mean to fall.
  • Smaller ME indicates more precise estimates.
A well-calculated margin of error gives us the assurance of our confidence interval's reliability.
Z-Distribution
Z-distribution is used in statistics for normal distribution problems, especially when calculating confidence intervals.
It is a type of normal distribution standardized to have a mean of 0 and a standard deviation of 1, often referred to as the standard normal distribution. Z-values are crucial for determining confidence intervals. They represent the number of standard deviations a data point is from the mean. In the context of a confidence interval, they help find the right value to cover the specified probability. In the solution, a 95% confidence interval requires a z-value corresponding to 0.025 in each tail of the distribution. This is found to be approximately 1.96.
  • Used when population standard deviation is known.
  • Ideal for large sample sizes or when data is normally distributed.
The z-distribution ensures we apply a consistent method to calculate and interpret confidence intervals.

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

Ishikawa et al. (Journal of Bioscience and Bioengineering, 2012) studied the adhesion of various biofilms to solid surfaces for possible use in environmental technologies. Adhesion assay is conducted by measuring absorbance at \(\mathrm{A}_{590} .\) Suppose that for the bacterial strain Acinetobacter, five measurements gave readings of 2.69,5.76,2.67,1.62 and 4.12 dyne-cm \(^{2}\). Assume that the standard deviation is known to be 0.66 dyne-cm \(^{2}\). (a) Find a \(95 \%\) confidence interval for the mean adhesion. (b) If the scientists want the confidence interval to be no wider than 0.55 dyne-cm \(^{2}\), how many observations should they take?

An article in Cancer Research ["Analyses of Litter-Matched Time-to-Response Data, with Modifications for Recovery of Interlitter Information" \((1977,\) Vol. \(37,\) pp. \(3863-\) 3868 ) ] tested the tumorigenesis of a drug. Rats were randomly selected from litters and given the drug. The times of tumor appearance were recorded as follows: $$ \begin{array}{l} 101,104,104,77,89,88,104,96,82,70,89,91,39,103,93, \\ 85,104,104,81,67,104,104,104,87,104,89,78,104,86 \\ 76,103,102,80,45,94,104,104,76,80,72,73 \end{array} $$ Calculate a \(95 \%\) confidence interval on the standard deviation of time until a tumor appearance. Check the assumption of normality of the population and comment on the assumptions for the confidence interval.

An electrical component has a time-to-failure (or lifetime) distribution that is exponential with parameter \(\lambda,\) so the mean lifetime is \(\mu=1 / \lambda .\) Suppose that a sample of \(n\) of these components is put on test, and let \(X_{i}\) be the observed lifetime of component \(i\). The test continues only until the \(r\) th unit fails, where \(r

The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 circuits is tested, revealing 13 defectives. (a) Calculate a \(95 \%\) two-sided CI on the fraction of defective circuits produced by this particular tool. (b) Calculate a \(95 \%\) upper confidence bound on the fraction of defective circuits.

The diameter of holes for a cable harness is known to have a normal distribution with \(\sigma=0.01\) inch. A random sample of size 10 yields an average diameter of 1.5045 inch. Find a \(99 \%\) two-sided confidence interval on the mean hole diameter.

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.