/*! 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 90 A manufacturer of interocular le... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 90

A manufacturer of interocular lenses is qualifying a new grinding machine and will qualify the machine if there is evidence that the percentage of polished lenses that contain surface defects does not exceed \(2 \%\). A random sample of 250 lenses contains six defective lenses. (a) Formulate and test an appropriate set of hypotheses to determine if the machine can be qualified. Use \(\alpha=0.05\). Find the \(P\) -value. (b) Explain how the question in part (a) could be answered with a confidence interval.

Short Answer

Expert verified
The machine is not qualified as the P-value (0.7661) is greater than 0.05, and the confidence interval includes 0.02.

Step by step solution

01

Define Hypotheses

We start by defining our null hypothesis (\(H_0\)) and alternative hypothesis (\(H_a\)).\[H_0: p = 0.02\]\[H_a: p < 0.02\]Here, \(p\) is the true percentage of lenses with surface defects. We are testing whether the defect rate is less than \(2\%\).
02

Identify the Sample Proportion

Calculate the proportion of defective lenses in the sample.Number of defective lenses, \(X = 6\).Sample size, \(n = 250\).Sample proportion \(\hat{p} = \frac{X}{n} = \frac{6}{250} = 0.024\).
03

Determine the Test Statistic

Use the standard normal (z) test for proportions to compute the test statistic.The formula for the test statistic is:\[z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]Substitute the values:\[p_0 = 0.02, \hat{p} = 0.024, n = 250 \]\[z = \frac{0.024 - 0.02}{\sqrt{\frac{0.02 \times 0.98}{250}}} \approx 0.727 \]
04

Find the P-value

The P-value is computed from the normal distribution using the z-test statistic.For \(z = 0.727\), we look up the P-value from the standard normal distribution table or use a calculator.The P-value is approximately 0.7661, indicating a high probability level.
05

Decision Making

Given \(\alpha = 0.05\), compare the P-value to determine whether to reject the null hypothesis.Since the P-value (0.7661) is greater than \(\alpha\), we fail to reject the null hypothesis.There is not enough evidence to conclude that the defect rate is less than \(2\%\).
06

Confidence Interval Approach

Compute a confidence interval for the true proportion to test the hypothesis.Use the formula:\[\hat{p} \pm z_{\frac{\alpha}{2}} \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]For a \(95\%\) confidence level, the critical value \(z_{0.025}\) is approximately 1.96.\[CI: 0.024 \pm 1.96 \times \sqrt{\frac{0.024 \times 0.976}{250}} \approx (0.0067, 0.0413)\]The interval includes \(0.02\), so there's insufficient evidence to suggest the defect rate is below \(2\%\).

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.

Confidence Interval
When conducting a hypothesis test, the confidence interval provides a range of values that most likely contain the population proportion. In this exercise, we calculated a confidence interval for the percentage of defective lenses produced by the machine.The formula used is:\[ \hat{p} \pm z_{\frac{\alpha}{2}} \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]Here, \(\hat{p}\) represents the sample proportion, \(z_{\frac{\alpha}{2}}\) is the critical value from the standard normal distribution for the desired confidence level, and \(n\) is the sample size.### Why Confidence Intervals Are Useful
  • They help in assessing the reliability of our estimate for the population proportion.
  • They offer an alternative way to test hypotheses.
  • If the hypothesized proportion lies within the confidence interval, we have not found enough evidence against the null hypothesis.
In our scenario, the 95% confidence interval for the defect rate was approximately (0.0067, 0.0413), which includes 0.02. Therefore, this supports our decision not to reject the null hypothesis.
Proportion Test
A proportion test is used when we want to know if a sample proportion is significantly different from a known proportion. In this exercise, we conducted a proportion test to determine if the percentage of defective lenses was less than 2%.### Steps in the Proportion Test1. **State the Hypotheses** - Null hypothesis \(H_0: p = 0.02\) suggests the defect rate is 2%. - Alternative hypothesis \(H_a: p < 0.02\) argues it is below 2%. 2. **Compute the Test Statistic** - The test statistic \(z\) is calculated using the formula: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \] - Substituting our values, the calculated \(z\)-score was about 0.727.This \(z\)-value tells us how many standard deviations the sample proportion is from the hypothesized proportion of 2%. A higher \(z\)-value indicates a greater difference from the hypothesized value. However, our \(z\)-value of 0.727 was not far enough from 0 to support the alternative hypothesis.
P-value
The P-value plays a crucial role in hypothesis testing, indicating the probability of obtaining test results at least as extreme as the observed results, assuming the null hypothesis is true. In this situation, our calculated P-value was approximately 0.7661.### Interpreting the P-value
  • A small P-value (\(< \alpha\)) usually implies strong evidence against the null hypothesis, suggesting it should be rejected.
  • Conversely, a large P-value (> \(\alpha\)) indicates weak evidence against the null hypothesis, meaning we fail to reject it.
Given \(\alpha = 0.05\), our P-value of 0.7661 was much greater, leading us to fail to reject the null hypothesis. This implies there is insufficient statistical evidence to conclude that the defect rate of lenses is less than 2% based on our sample.

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

The cooling system in a nuclear submarine consists of an assembly of welded pipes through which a coolant is circulated. Specifications require that weld strength must meet or exceed 150 psi. (a) Suppose that the design engineers decide to test the hypothesis \(H_{0}: \mu=150\) versus \(H_{1}: \mu>150 .\) Explain why this choice of alternative hypothesis is better than \(H_{1}: \mu<150\) (b) A random sample of 20 welds results in \(\bar{x}=153.7\) psi and \(s=11.3\) psi. What conclusions can you draw about the hypothesis in part (a)? State any necessary assumptions about the underlying distribution of the data.

Consider the computer output below. One-Sample T: Test of \(m u=100\) vs not \(=100\) $$ \begin{array}{llllclll} \text { Variable } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & 95 \% \mathrm{CI} & \mathrm{T} & \mathrm{P} \\ \mathrm{X} & 16 & 98.33 & 4.61 & ? & (?, ?) & ? & ? \end{array} $$ (a) How many degrees of freedom are there on the \(t\) -statistic? (b) Fill in the missing information. You may use bounds on the \(P\) -value. (c) What are your conclusions if \(\alpha=0.05 ?\) (d) What are your conclusions if the hypothesis is \(H_{0}: \mu=\) 100 versus \(H_{0}: \mu>100 ?\)

The proportion of residents in Phoenix favoring the building of toll roads to complete the freeway system is believed to be \(p=0.3\). If a random sample of 10 residents shows that 1 or fewer favor this proposal, we will conclude that \(p<0.3\). (a) Find the probability of type I error if the true proportion is \(p=0.3\) (b) Find the probability of committing a type II error with this procedure if \(p=0.2\) (c) What is the power of this procedure if the true proportion is \(p=0.2 ?\)

The mean pull-off force of an adhesive used in marufacturing a connector for an automotive engine application should be at least 75 pounds. This adhesive will be used unless there is strong evidence that the pull-off force does not meet this requirement. A test of an appropriate hypothesis is to be conducted with sample size \(n=10\) and \(\alpha=0.05 .\) Assume that the pull-off force is normally distributed, and \(\sigma\) is not known. (a) If the true standard deviation is \(\sigma=1\), what is the risk that the adhesive will be judged acceptable when the true mean pull-off force is only 73 pounds? Only 72 pounds? (b) What sample size is required to give a \(90 \%\) chance of detecting that the true mean is only 72 pounds when \(\sigma=1 ?\) (c) Rework parts (a) and (b) assuming that \(\sigma=2\). How much impact does increasing the value of \(\sigma\) have on the answers you obtain?

For the hypothesis test \(H_{0}: \mu=5\) against \(H_{1}: \mu<5\) with variance unknown and \(n=12\), approximate the \(P\) -value for each of the following test statistics. (a) \(t_{0}=2.05\) (b) \(t_{0}=-1.84\) (c) \(t_{0}=0.4\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

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.