/*! 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 106 A manufacturer turns out a produ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A manufacturer turns out a product item that is labeled either "defective" or "not defective." In order to estimate the proportion defective, a random sample of 100 items is taken from production and 10 are found to be defective. Following the implementation of a quality improvement program, the experiment was conducted again. A new sample of 100 is taken and this time only 6 are found defective. (a) Give a \(95 \%\) confidence interval on \(p \mid-p_{2},\) where \(p_{1}\) is the population proportion defective before improvement, and \(p_{2}\) is the proportion defective after improvement. (b) Is there information in the confidence interval found in (a) that would suggest that \(p_{1}>p_{2} ?\) Explain.

Short Answer

Expert verified
The 95% confidence interval for the difference between the proportions of defect rate before and after the improvement is -0.031 to 0.111. This shows there isn't strong statistical evidence to suggest the proportion of defects has decreased as zero is included in the interval.

Step by step solution

01

Calculate Sample Proportions

First, calculate the sample proportions. Before improvement, \(p_{1}\) was given as 10 out of 100, or \(0.10\). After improvement, \(p_{2}\) was given as 6 out of 100, or \(0.06\).
02

Find Standard Errors

The standard error (SE) of a proportion is given by the formula \(\sqrt{p(1 - p)/n}\). For \(p_{1}\), the SE is \(\sqrt{0.10(1-0.10)/100}\), or \(0.03\), and for \(p_{2}\), the SE is \(\sqrt{0.06(1-0.06)/100}\), or \(0.02\).
03

Calculate Difference of the Proportions

The difference (\(d\)) of the proportions is \(p_{1} - p_{2}\) which equals \(0.10 - 0.06 = 0.04\).
04

Calculate Confidence Interval

A 95% confidence interval is calculated as: \(d ± Z*SE_d\), where \(SE_d\), the standard error of the difference between two proportions, is given by \(\sqrt{SE_{p1}^2 + SE_{p2}^2}\). Therefore, \(SE_d = \sqrt{0.03^2 + 0.02^2} = 0.036\). Using a Z score of 1.96 for a 95% confidence interval, the confidence interval is: \(0.04 ± 1.96*0.036\), which is \((0.04 - 0.071, 0.04 + 0.071)\) or \(-0.031, 0.111\).
05

Interpret Confidence Interval

The confidence interval contains both positive and negative values. This means the true difference could be positive (implying the proportion defective went down), zero (implying no change), or even negative (implying the proportion defective increased). This result does not suggest a clear improvement since zero is part of the 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.

Sample Proportion
When you have a large set of items, such as products produced in a factory, you sometimes want to estimate the true proportion of a certain characteristic, like how many of them are defective. A sample proportion is a way to estimate this. It is calculated as the number of defective items in your sample divided by the total number of items in the sample.

For example, if 10 out of 100 items are defective, the sample proportion is 0.10. This number helps us understand the fraction of defective items in our production without having to check every single item. It provides a snapshot or an estimate of what the proportion might be in the entire batch, factory, or production line.
Standard Error
Standard error (SE) is a measure that helps us understand how accurate our sample proportion is likely to be. It tells us how much the proportion might vary from one sample to another. Think of it like a small adjustment around our estimate that shows how precise our guess is.

To calculate the standard error of a sample proportion, use the formula \( \sqrt{\frac{p(1-p)}{n}} \), where \(p\) is the sample proportion and \(n\) is the sample size.
  • For example, if \( p = 0.10 \) and \( n = 100 \), the standard error is \( \sqrt{\frac{0.10(1-0.10)}{100}} = 0.03 \).
  • This means the estimate of the sample proportion is likely to vary by 0.03 either way if we were to take multiple samples.
Understanding standard error is crucial, especially when making predictions from a small group to a larger population.
Difference of Proportions
In statistics, comparing two different scenarios or groups often involves calculating the difference of proportions. This helps us see how two different sample proportions, taken from the same population before and after an event or change, compare with each other.

To find this difference, simply subtract one sample proportion from the other. For example, if the proportion before an improvement program is 0.10 and after is 0.06, the difference of proportions is 0.10 - 0.06 = 0.04.

This difference tells us that the defective rate decreased by 0.04, which is equivalent to 4% of the total sample size. It's an essential step in assessing the impact of interventions like quality improvements.
Quality Improvement
Quality improvement programs aim to enhance the processes or outputs of an organization, often by reducing errors or defects. In our scenario, the manufacturer attempted to produce better quality items with fewer defects.

Measuring the effectiveness of these improvements can involve looking at sample proportions before and after the change. If after implementing a quality improvement program, the sample proportion of defective products decreases, it suggests that the program might be working.

However, proper statistical analysis, like calculating confidence intervals, is necessary to understand if the observed changes are just due to chance or reflect a real improvement. If the confidence interval suggests a significant decrease in the proportion of defects, it can reinforce that the improvements have been effective.

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

An efficiency expert wishes to determine the average time that it takes to drill three holes in a certain metal clamp. How large a sample will he need to be \(95 \%\) confident that his sample mean will be within 15 seconds of the true mean? Assume that it is known from previous studies that \(\sigma=40\) seconds.

Specify the moment estimators for the \(p\) and er \(^{2}\) for the lognormal distribution.

In Section 9.3 wc emphasized the notion of "most efficient estimator" by comparing the variance of two unbiased estimators \(\Theta_{1}\) and \(\Theta_{2}\). However this does not take into account bias in case one or both estimators are not unbiased. Consider the quantity $$M S E=E(\Theta-\theta)$$ where MSE denotes mean squared error. The \(M S E\) is often used to compare two estimators \(\ddot{\Theta}_{1}\) and \(\Theta_{2}\) of 6 when either or both is unbiased because (i) it is intuitively reasonable and (ii) it accounts for bias. Show that MSE can be written $$\begin{aligned}M S E=& E\left[\Theta^{*}-E(\hat{\Theta})\right]^{2}+[E(\Theta-6))^{2} \\ &=\operatorname{Var}(\Theta)+[\operatorname{Bias}(\hat{\Theta})]^{2} \end{aligned}$$

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?

A random sample of 12 graduates of a certain secretarial school typed an average of 79.3 words per minute with a standard deviation of 7.8 words per minute. Assuming a normal distribution for the number of words typed per minute, find a \(95 \%\) confidence interval for the average number of words typed by all graduates of this school.

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.