/*! 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 20 At every setting a high-speed pa... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

At every setting a high-speed packing machine delivers a product in amounts that vary from container to container with a normal distribution of standard deviation 0.12 ounce. To compare the amount delivered at the current setting to the desired amount 64.1 ounce, a quality inspector randomly selects five containers and measures the contents of each, obtaining sample mean 63.9 ounces and sample standard deviation 0.10 ounce. Test whether the data provide sufficient evidence, at the \(5 \%\) level of significance, to conclude that the mean of all containers at the current setting is less than 64.1 ounces.

Short Answer

Expert verified
The machine delivers less than 64.1 ounces on average.

Step by step solution

01

Define Hypotheses

The null hypothesis \( H_0 \) is that the mean amount delivered, \( \mu \), is equal to 64.1 ounces, i.e., \( H_0: \mu = 64.1 \). The alternative hypothesis \( H_a \) is that the mean amount delivered, \( \mu \), is less than 64.1 ounces, i.e., \( H_a: \mu < 64.1 \).
02

Determine Significance Level

The significance level is given as \( \alpha = 0.05 \). This represents the probability of rejecting the null hypothesis when it is true.
03

Calculate Test Statistic

Since the sample size is small (n=5) and the population standard deviation is known, we use the t-test:\[t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\]where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size.\[t = \frac{63.9 - 64.1}{0.10/\sqrt{5}} = \frac{-0.2}{0.04472} \approx -4.47\]
04

Determine Critical Value

For a one-tailed t-test at \( \alpha = 0.05 \) with degrees of freedom \( df = n - 1 = 4 \), use a t-distribution table:The critical t-value is approximately \(-2.132\).
05

Compare Test Statistic to Critical Value

The calculated t-statistic is \(-4.47\), and the critical value is \(-2.132\). Since \(-4.47 < -2.132\), we reject the null hypothesis \( H_0 \).
06

Conclusion

Based on the results, there is sufficient evidence at the \(5\%\) level of significance to conclude that the mean of all containers at the current setting is less than 64.1 ounces.

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.

Understanding the t-test
The t-test is a statistical test used to determine if there is a significant difference between the means of two groups. In this scenario, we are interested in testing whether the average amount delivered by the machine is less than the desired 64.1 ounces. This is a one-tailed t-test because we are specifically looking to see if the mean is less than 64.1 ounces, rather than just different.

The formula for the t-test is: \[t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\] Where:
  • \(\bar{x}\) is the sample mean,
  • \(\mu\) is the population mean under the null hypothesis,
  • \(s\) is the sample standard deviation,
  • \(n\) is the sample size.
In our example, after calculation, the test statistic was found to be approximately -4.47. This number helps determine how far the sample mean is from the population mean under the null hypothesis, measured in terms of the number of standard deviations.
The Role of Significance Level
In hypothesis testing, the significance level, represented by \( \alpha \), is a critical threshold. It indicates the probability of making a Type I error, which means rejecting the null hypothesis when it is actually true. A common choice for \(\alpha\) is 0.05, as used in our example.

Choosing a 0.05 significance level means that there is a 5% risk of concluding that a difference exists when there is no actual difference. The significance level helps set the stage for determining the critical value against which we compare our test statistic.

Understanding and selecting the significance level is a crucial step because it directly affects your confidence in the results of the hypothesis test.
Normal Distribution and Its Importance
Normal distribution is a key concept in statistics and is often called the bell curve due to its shape. It's a continuous probability distribution characterized by a symmetric, bell-shaped curve. This distribution is critical in hypothesis testing because many statistical tests, like the t-test, assume normally distributed data.

In the example provided, the assumption of normal distribution is important for calculating the t-test and determining the significance of the observed sample mean.

Why is this important?
  • It allows for the use of the t-test as a valid method.
  • It facilitates the understanding and prediction of phenomena around the mean.
  • Many natural processes, such as the weight of machine-packed products, tend to follow a normal distribution.
Sample Mean and Its Significance
The sample mean is the average value of a sample and serves as an estimate of the population mean. In statistical tests like the t-test, the sample mean is crucial because it represents the data we assess to draw conclusions about the overall population behavior.

In our scenario, the sample mean was calculated from the contents of five randomly selected containers of product, giving a value of 63.9 ounces. This sample mean is compared against the hypothesized population mean of 64.1 ounces, helping determine if there's sufficient evidence to support our alternative hypothesis.
  • A lower sample mean suggests that the current setting might deliver less product than intended.
  • Calculating and analyzing the sample mean helps infer trends or changes in a larger population.
The proximity or distance of the sample mean from the hypothesized population mean significantly influences the outcome of the hypothesis test.

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 recommended daily calorie intake for teenage girls is 2,200 calories/day. A nutritionist at a state university believes the average daily caloric intake of girls in that state to be lower. Test that hypothesis, at the \(5 \%\) level of significance, against the null hypothesis that the population average is 2,200 calories/day using the following sample data: \(n=36, x-2,150, s=203\).

The mean score on a 25-point placement exam in mathematics used for the past two years at a large state university is 14.3. The placement coordinator wishes to test whether the mean score on a revised version of the exam differs from 14.3 . She gives the revised exam to 30 entering freshmen early in the summer; the mean score is 14.6 with standard deviation 2.4 a. Perform the test at the \(10 \%\) level of significance using the critical value approach. b. Compute the observed significance of the test. c. Perform the test at the \(10 \%\) level of significance using the \(p\) -value approach. You need not repeat the first three steps, already done in part (a).

A random sample of size 12 drawn from a normal population yielded the following results: \(x-=86.2, s=0.63\). a. Test \(H 0: \mu=85.5\) vs. Ha:\mu \(\neq 85.5 @ \alpha=0.01\). b. Estimate the observed significance of the test in part (a) and state a decision based on the \(p\) -value approach to hypothesis testing.

A literary historian examines a newly discovered document possibly written by Oberon Theseus. The mean average sentence length of the surviving undisputed works of Oberon Theseus is 48.72 words. The historian counts words in sentences between five successive 101 periods in the document in question to obtain a mean average sentence length of 39.46 words with standard deviation 7.45 words. (Thus the sample size is five.) a. Determine if these data provide sufficient evidence, at the \(1 \%\) level of significance, to conclude that the mean average sentence length in the document is less than 48.72 . b. Estimate the \(p\) -value of the test. c. Based on the answers to parts \((\mathrm{a})\) and \((\mathrm{b})\), state whether or not it is likely that the document was written by Oberon Theseus.

Large Data Set 1 lists the GPAs of 1,000 students. a. Regard the data as arising from a census of all freshman at a small college at the end of their first academic year of college study, in which the GPA of every such person was measured. Compute the population mean \(\mu\). b. Regard the first 50 students in the data set as a random sample drawn from the population of part (a) and use it to test the hypothesis that the population mean is less than \(2.50,\) at the \(10 \%\) level of significance. (The null hypothesis is that \(\mu=2.50 .)\) c. Is your conclusion in part \((b)\) in agreement with the true state of nature (which by part (a) you know), or is your decision in error? If your decision is in error, is it a Type I error or a Type II error?

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.