/*! 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 67 Shulman Steel Corporation makes ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 67

Shulman Steel Corporation makes bearings that are supplied to other companies. One of the machines makes bearings that are supposed to have a diameter of 4 inches. The bearings that have a diameter of either more or less than 4 inches are considered defective and are discarded. When working properly, the machine does not produce more than \(7 \%\) of bearings that are defective. The quality control inspector selects a sample of 200 bearings each week and inspects them for the size of their diameters. Using the sample proportion, the quality control inspector tests the null hypothesis \(p \leq .07\) against the alternative hypothesis \(p>.07\), where \(p\) is the proportion of bearings that are defective. He always uses a \(2 \%\) significance level. If the null hypothesis is rejected, the machine is stopped to make any necessary adjustments. One sample of 200 bearings taken recently contained 22 defective bearings.

Short Answer

Expert verified
At an alpha level of 2%, using a right-tailed Z test, the null hypothesis - the machine does not produce more than 7% of bearings that are defective - is not rejected. The sample data does not provide enough evidence to conclude that the machine is producing more than 7% of defective bearings.

Step by step solution

01

Calculate the Sample Proportion

The sample proportion (\(p'\)) is found by dividing the number of defective bearings by the total number of bearings. Given that there are 22 defective bearings out of 200 bearings, \(p'\) can be calculated as \(p' = \frac{22}{200} = 0.11.\)
02

Setup the Null and Alternative Hypothesis

We test the null hypothesis \(p \leq .07\) against the alternative hypothesis \(p>.07\).
03

Conduct Hypothesis Test Using the Sample Proportion

Under the null hypothesis, the sample proportion has a normal distribution with mean \(0.07\) and standard deviation \(\sqrt{\(p(1-p)/n}\) where \(p\) is proportion under the null hypothesis and \(n\) is sample size. Here, \(p = 0.07\) and \(n = 200\), so standard deviation = \(\sqrt {\(0.07 * 0.93 / 200}\) = 0.01814. The Z score for the sample proportion is then calculated as \((p' - p)/standard deviation\) = \((0.11 - 0.07) / 0.01814\) = 2.205 (rounded to three decimals).If the Z score is higher than the value of Z for the given significance level, null hypothesis is rejected.At a 2% significance level for a right tailed test, the critical value (Z score) from standard normal distribution tables is 2.33.The calculated Z is less than 2.33, thus we will not reject the null hypothesis.
04

Reach a Conclusion on the Hypothesis Test

Since the calculated Z score is less than the critical value at the 2% significance level, we do not reject the null hypothesis. This indicates that there is not enough evidence to conclude that the machine is producing more than 7% defective bearings and thus it does not need any adjustments.

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 91影视!

Key Concepts

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

Quality Control
Quality control is a vital part of manufacturing processes, ensuring that products meet specific standards and function as intended. In the context of Shulman Steel Corporation, quality control involves monitoring the machine that produces bearings to ensure they have the correct diameter of 4 inches. Bearings that deviate from this diameter are defective and must be discarded.

The company sets a benchmark where no more than 7% of the bearings can be defective. This percentage is the threshold for maintaining good quality. By checking a sample of bearings each week, the quality control inspector can identify whether the machine produces an acceptable number of defective products. If too many are defective, the machine needs adjustment.
  • Ensures product consistency
  • Identifies when machines need adjustments
  • Maintains customer satisfaction through reliable products
Sample Proportion
The concept of sample proportion is key in assessing machine performance. Here, the sample proportion is the fraction of defective bearings within a sample of 200. It's calculated by dividing the number of defective bearings by the total number of bearings in the sample.

In this case, with 22 defective bearings out of 200, the sample proportion (\(p'\)) is \(\frac{22}{200} = 0.11\). This figure indicates the proportion of the sample which is defective and acts as an estimate of the machine's performance. Understanding sample proportions helps in determining whether the null hypothesis should be rejected or not.
  • It gives an estimate of the population proportion
  • Helps in hypothesis testing
  • Affects decision making through hypothesis test outcomes
Significance Level
The significance level in hypothesis testing is a threshold used to decide whether a hypothesis should be rejected. It represents the probability of making a Type I error, which occurs when the null hypothesis is incorrectly rejected.

In the exercise, a 2% significance level (\(\alpha = 0.02\)) is used, indicating a very strict criterion. This means that there is only a 2% risk of concluding that the machine is producing more than 7% defective bearings, when it is not. By setting a low significance level, the quality control process ensures that adjustments to the machine are only made when absolutely necessary, thereby preventing unnecessary downtime.
  • Represents risk of incorrectly rejecting the null hypothesis
  • Influences the choice of critical value in hypothesis testing
  • Dictates the stringency of the test
Standard Deviation
Standard deviation is a measure of variance or spread in a set of data. In the hypothesis testing process, it helps us understand the distribution of sample proportions under the null hypothesis.

The formula for standard deviation of a sample proportion is: \[\text{Standard deviation} = \sqrt{\frac{p(1-p)}{n}} \], where \(p\) is the hypothesized proportion (0.07 in this case) and \(n\) is the sample size (200 here). Therefore, the standard deviation is \(\sqrt{\frac{0.07 \cdot 0.93}{200}} \approx 0.01814\).

By determining the standard deviation, we evaluate how much the sample proportions could vary. This value is crucial for calculating the Z score, which helps determine whether the observed sample proportion is significantly different from what's expected under the null hypothesis.
  • Measures variability in data sets
  • Essential for hypothesis testing calculations
  • Influences interpretation of test results

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

Consider \(H_{0}: \mu=55\) versus \(H_{1}: \mu \neq 55\). a. What type of error would you make if the null hypothesis is actually false and you fail to reject it? b. What type of error would you make if the null hypothesis is actually true and you reject it?

According to a Bureau of Labor Statistics release of March 25, 2015, financial analysts earned an average of \(\$ 110,510\) in 2014 . Suppose that the 2014 earnings of all financial analysts had a mean of \(\$ 110,510\). A recent sample of 400 financial analysts showed that they earn an average of \(\$ 114,630\) a year. Assume that the standard deviation of the annual earnings of all financial analysts is \(\$ 30,570\). a. Using the critical-value approach, can you conclude that the current average annual earnings of financial analysts is higher than \(\$ 110,510 ?\) Use \(\alpha=.01\). b. What is the Type I error in part a? Explain. What is the probability of making this error in part a? c. Will your conclusion of part a change if the probability of making a Type I error is zero? d. Calculate the \(p\) -value for the test of part a. What is your conclusion if \(\alpha=.01 ?\)

Explain which of the following is a two-tailed test, a left-tailed test, or a right-tailed test. a. \(H_{0}: \mu=12, H_{1}: \mu<12\) b. \(H_{0}: \mu \leq 85, H_{1}: \mu>85\) c. \(H_{0}: \mu=33, H_{1}: \mu \neq 33\) Show the rejection and nonrejection regions for each of these cases by drawing a sampling distribution curve for the sample mean, assuming that it is normally distributed.

The XO Group Inc., released the results of its annual Real Weddings Study on March 27, 2014 (www.theknot.com). According to this study, the average cost of a wedding in America was \(\$ 29,858\) in \(2013 .\) A recent sample of 100 American couples who got married this year produced a mean wedding cost of \(\$ 32,084\) with a standard deviation of \(\$ 9275 .\) Using a \(2.5 \%\) significance level and the criticalvalue approach, can you conclude that the current mean cost of a wedding in America is higher than \(\$ 29,858\) ? Find the range for the \(p\) -value for this test. What will your conclusion be using this \(p\) -value range and \(\alpha=.05 ?\)

Acme Bicycle Company makes derailleurs for mountain bikes. Usually no more than \(4 \%\) of these parts are defective, but occasionally the machines that make them get out of adjustment and the rate of defectives exceeds \(4 \%\). To guard against this, the chief quality control inspector takes a random sample of 130 derailleurs each week and checks each one for defects. If too many of these parts are defective, the machines are shut down and adjusted. To decide how many parts must be defective to shut down the machines, the company's statistician has set up the hypothesis test $$ H_{0}: p \leq .04 \text { versus } H_{1}: p>.04 $$ where \(p\) is the proportion of defectives among all derailleurs being made currently. Rejection of \(H_{0}\) would call for shutting down the machines. For the inspector's convenience, the statistician would like the rejection region to have the form, "Reject \(H_{0}\) if the number of defective parts is \(C\) or more." Find the value of \(C\) that will make the significance level (approximately) \(.05\).
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
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.