/*! 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 7 A consumer products company is f... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A consumer products company is formulating a new shampoo and is interested in foam height (in millimeters). Foam height is approximately normally distributed and has a standard deviation of 20 millimeters. The company wishes to test \(H_{0}: \mu=175\) millimeters versus \(H_{1}: \mu>175\) millimeters, using the results of \(n=10\) samples. a. Find the type I error probability \(\alpha\) if the critical region is \(\bar{x}>185\) b. What is the probability of type II error if the true mean foam height is 185 millimeters? c. Find \(\beta\) for the true mean of 195 millimeters.

Short Answer

Expert verified
a. \( \alpha = 0.0571 \) b. \( \beta = 0.5000 \) c. \( \beta = 0.0571 \)

Step by step solution

01

Define Parameters and Hypotheses

The null hypothesis is \( H_0: \mu = 175 \) and the alternative is \( H_1: \mu > 175 \). The sample size \( n = 10 \), and the population standard deviation \( \sigma = 20 \). The critical region is set at \( \overline{x} > 185 \).
02

Calculate Critical Value for Type I Error

Since \( \overline{x} \) follows a normal distribution with mean \( \mu = 175 \) under the null hypothesis and standard deviation \( \sigma_\overline{x} = \frac{20}{\sqrt{10}} = 6.32 \), we calculate the type I error probability \( \alpha \) for \( \overline{x} > 185 \). The z-score is \( z = \frac{185 - 175}{6.32} \approx 1.58 \). Using z-tables or a calculator, find \( P(Z > 1.58) \approx 0.0571 \).
03

Calculate Type II Error for True Mean 185

The probability of a type II error \( \beta \) is the probability of failing to reject the null hypothesis when the true mean \( \mu = 185 \). Calculate \( \beta = P(\overline{x} \leq 185 | \mu = 185) \). Here, \( \bar{x} \) is normally distributed with mean \( 185 \) and \( \sigma_\overline{x} = 6.32 \). Thus, \( \beta = P(Z \leq 0) = 0.5 \).
04

Calculate Type II Error for True Mean 195

For \( \mu = 195 \), express \( \beta = P(\overline{x} \leq 185 | \mu = 195) \). With \( \bar{x} \) normally distributed (mean \( 195 \), \( \sigma_\overline{x} = 6.32 \)), find \( z = \frac{185 - 195}{6.32} \approx -1.58 \). Then calculate \( \beta = P(Z \leq -1.58) \approx 0.0571 \).

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.

Type I Error
In hypothesis testing, a Type I error occurs when the null hypothesis is rejected even though it is actually true. This is a false positive error. In other words, we're saying there is an effect or a difference when, in fact, there isn't one.

A Type I error is typically denoted by the Greek letter \( \alpha \). The significance level \( \alpha \) is the probability of making a Type I error, and it is a crucial concept in determining the reliability of a hypothesis test.
  • It sets the threshold for how extreme the data must be before we decide to reject the null hypothesis.
  • Common significance levels are 0.05, 0.01, and 0.10, though other values can be used based on the researcher's needs.
In the context of the problem, the null hypothesis \( H_0: \mu = 175 \) means we expect the average foam height to be 175 mm. A Type I error in this situation would mean claiming that the average foam height is greater than 175 mm based on sample data, even when it actually isn't.
In the provided solution, the Type I error rate was calculated as \( 0.0571 \), which means there is about a 5.71% chance of incorrectly rejecting the true null hypothesis given the critical region \( \bar{x} > 185 \).
Type II Error
A Type II error happens when we fail to reject the null hypothesis, even though the alternative hypothesis is true. This error is also called a false negative because we're missing an effect or difference that actually exists.

The probability of making a Type II error is denoted by \( \beta \). Unlike Type I error, which is set by the researcher, Type II error is usually more complex because it depends on several factors, such as sample size, variance, and the actual effect size.
  • A lower \( \beta \) value indicates more sensitivity to detecting true effects.
  • Reducing \( \beta \) usually requires larger sample sizes or more precise measurements.
In our shampoo experiment, the solution calculated \( \beta = 0.5 \) for a true mean foam height of 185 mm, meaning there's a 50% chance of failing to detect that the true mean is greater than 175 when it actually is 185.
For a true mean of 195 mm, \( \beta \) was found to be approximately \( 0.0571 \). Here, the probability of missing the greater mean is considerably lower, indicating better test sensitivity at this higher means.
Normal Distribution
The concept of normal distribution is fundamental in statistics, especially in hypothesis testing. It represents a continuous probability distribution characterized by its symmetric bell-shaped curve, where most of the observations cluster around the mean.

In hypothesis testing, normal distribution helps quantify probabilities and critical values.
  • A standard normal distribution has a mean of zero and a standard deviation of one.
  • To convert any normal distribution to a standard normal distribution, use the z-score formula \( z = \frac{x - \mu}{\sigma} \).
In our exercise, we presume the foam height follows a normal distribution, which is crucial for setting up the hypothesis test effectively.
The given population standard deviation is 20 mm, which helps define the distribution's spread. By using the central limit theorem, it's assumed that the sampling distribution of the sample mean will be approximately normal because we have a reasonably large sample size (10 samples in this scenario).
This assumption enables the use of normal distribution tables or calculators to find the critical region and calculate error probabilities like \( \alpha \) and \( \beta \). Understanding the interplay between normal distribution and these error calculations empowers us to draw meaningful conclusions from the sample data.

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 mean bond strength of a cement product must be at least 10000 psi. The process by which this material is manufactured must show equivalence to this standard. If the process can manufacture cement for which the mean bond strength is at least 9750 psi, it will be considered equivalent to the standard. A random sample of six observations is available, and the sample mean and standard deviation of bond strength are 9360 psi and 42.6 psi, respectively. a. State the appropriate hypotheses that must be tested to demonstrate equivalence. b. What are your conclusions using \(\alpha=0.05 ?\)

An article in Growth: A Journal Devoted to Problems of Normal and Abnormal Growth ["Comparison of Measured and Estimated Fat-Free Weight, Fat, Potassium and Nitrogen of Growing Guinea Pigs" (1982, Vol. \(46(4),\) pp. \(306-321\) ) ] reported the results of a study that measured the body weight (in grams) for guinea pigs at birth. $$ \begin{array}{rrrrr} 421.0 & 452.6 & 456.1 & 494.6 & 373.8 \\ 90.5 & 110.7 & 96.4 & 81.7 & 102.4 \\ 241.0 & 296.0 & 317.0 & 290.9 & 256.5 \\ 447.8 & 687.6 & 705.7 & 879.0 & 88.8 \\ 296.0 & 273.0 & 268.0 & 227.5 & 279.3 \\ 258.5 & 296.0 & & & \end{array} $$ a. Test the hypothesis that mean body weight is 300 grams. Use \(\alpha=0.05\) b. What is the smallest level of significance at which you would be willing to reject the null hypothesis? c. Explain how you could answer the question in part (a) with a two-sided confidence interval on mean body weight.

Cloud seeding has been studied for many decades as a weather modification procedure (for an interesting study of this subject, see the article in Technometrics, "A Bayesian Analysis of a Multiplicative Treatment Effect in Weather Modification," \(1975,\) Vol. \(17,\) pp. \(161-166) .\) The rainfall in acre-feet from 20 clouds that were selected at random and seeded with silver nitrate follows: \(18.0,30.7,19.8,27.1,22.3,18.8,31.8,23.4,\) 21.2,27.9,31.9,27.1,25.0,24.7,26.9,21.8,29.2,34.8,26.7 and 31.6 a. Can you support a claim that mean rainfall from seeded clouds exceeds 25 acre-feet? Use \(\alpha=0.01 .\) Find the \(P\) -value. b. Check that rainfall is normally distributed. c. Compute the power of the test if the true mean rainfall is 27 acre-feet. d. What sample size would be required to detect a true mean rainfall of 27.5 acre-feet if you wanted the power of the test to be at least \(0.9 ?\) e. Explain how the question in part (a) could be answered by constructing a one-sided confidence bound on the mean diameter.

Supercavitation is a propulsion technology for undersea vehicles that can greatly increase their speed. It occurs above approximately 50 meters per second when pressure drops sufficiently to allow the water to dissociate into water vapor, forming a gas bubble behind the vehicle. When the gas bubble completely encloses the vehicle, supercavitation is said to occur. Eight tests were conducted on a scale model of an undersea vehicle in a towing basin with the average observed speed \(\bar{x}=102.2\) meters per second. Assume that speed is normally distributed with known standard deviation \(\sigma=4\) meters per second a. Test the hypothesis \(H_{0}: \mu=100\) versus \(H_{1}: \mu<100\) using \(\alpha=0.05\) b. What is the \(P\) -value for the test in part (a)? c. Compute the power of the test if the true mean speed is as low as 95 meters per second. d. What sample size would be required to detect a true mean speed as low as 95 meters per second if you wanted the power of the test to be at least \(0.85 ?\) e. Explain how the question in part (a) could be answered by constructing a one-sided confidence bound on the mean speed.

The impurity level (in ppm) is routinely measured in an intermediate chemical product. The following data were observed in a recent test: $$ \begin{array}{l} 2.4,2.5,1.7,1.6,1.9,2.6,1.3,1.9,2.0,2.5,2.6,2.3,2.0, \\ 1.8,1.3,1.7,2.0,1.9,2.3,1.9,2.4,1.6 \end{array} $$ Can you claim that the median impurity level is less than \(2.5 \mathrm{ppm} ?\) a. State and test the appropriate hypothesis using the sign test with \(\alpha=0.05 .\) What is the \(P\) -value for this test? b. Use the normal approximation for the sign test to test \(H_{0}: \tilde{\mu}=2.5\) versus \(H_{1}: \tilde{\mu}<2.5 .\) What is the \(P\) -value for this test?

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.