/*! 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 24 Dole Pineapple Inc. is concerned... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Dole Pineapple Inc. is concerned that the 16 -ounce can of sliced pineapple is being overfilled. Assume the standard deviation of the process is .03 ounce. The quality control department took a random sample of 50 cans and found that the arithmetic mean weight was 16.05 ounces. At the \(5 \%\) level of significance, can we conclude that the mean weight is greater than 16 ounces? Determine the \(p\) -value.

Short Answer

Expert verified
Yes, we can conclude the mean is greater than 16 ounces; p-value is close to 0.

Step by step solution

01

Identify the Hypotheses

In this problem, we need to determine if the mean weight of the cans is greater than 16 ounces. Our null hypothesis \(H_0\) is that the mean \(\mu = 16\) ounces. The alternative hypothesis \(H_1\) is that \(\mu > 16\) ounces.
02

Gather the Given Information

We are given the standard deviation \(\sigma = 0.03\) ounces, the sample size \(n = 50\), and the sample mean \(\bar{x} = 16.05\). The level of significance is \(\alpha = 0.05\).
03

Calculate the Test Statistic

Using the formula for the z-test statistic: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] Substitute \(\bar{x} = 16.05\), \(\mu = 16\), \(\sigma = 0.03\), and \(n = 50\):\[ z = \frac{16.05 - 16}{\frac{0.03}{\sqrt{50}}} \] Calculate the result to find \(z\).
04

Calculate the Z-value

First, calculate the denominator: \( \frac{0.03}{\sqrt{50}} = \frac{0.03}{7.071} = 0.00424 \). Then, the Z-value is:\[ z = \frac{0.05}{0.00424} \approx 11.8 \]
05

Determine the Critical Value

Since we have a one-tailed test with \(\alpha = 0.05\), the critical z-value from the standard normal distribution table is approximately 1.645.
06

Compare Z-value and Critical Value

Our calculated z-value is 11.8, which is much greater than the critical value of 1.645. This means we can reject the null hypothesis \(H_0\).
07

Calculate the p-value

We look up the p-value for our calculated z-value of 11.8. The p-value is very close to 0, as 11.8 is far in the tail of the standard normal distribution.

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.

z-test
The z-test is a statistical procedure used to determine if there are significant differences between the means of a sample and a known value or between two samples. In the context of hypothesis testing, the z-test is particularly useful when the standard deviation is known and the sample size is sufficiently large.

For this exercise, the z-test helps verify if the mean weight of pineapple cans exceeds the claimed 16 ounces. By comparing this sample mean to the hypothesized population mean using a standard deviation presumed to be accurate, we can ascertain statistically rigorous conclusions.

To conduct a z-test, calculate the z-statistic using the formula:
  • \( z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \)
Here, \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( \sigma \) is the population standard deviation, and \( n \) is the sample size. This quantifies how many standard deviations the sample mean is from the population mean.
null hypothesis
The null hypothesis, often denoted as \( H_0 \), is a foundational concept in hypothesis testing. It represents a statement of no effect or no difference and is the hypothesis that researchers seek to test.

In the given exercise, the null hypothesis is that the mean weight of the comprehensive sample of pineapple cans is 16 ounces.

This hypothesis can be expressed as:
  • \( H_0: \mu = 16 \)
The goal of hypothesis testing is to assess the strength of evidence against the null hypothesis. By conducting a z-test, as seen in the exercise, you determine whether the sample mean significantly deviates from the hypothesized mean under the assumption that \( H_0 \) is right.
alternative hypothesis
The alternative hypothesis, represented as \( H_1 \) or \( H_a \), is what the test aims to support. It suggests that there is an effect or a difference. In contrast to the null hypothesis, the alternative suggests a new truth.

For this problem, the alternative hypothesis states that the mean weight of pineapple cans is indeed greater than 16 ounces, which would indicate an overfilling issue.

Explicitly, the alternative hypothesis can be written as:
  • \( H_1: \mu > 16 \)
This forms a directional or "greater than" test, seeking evidence to support the supplier's overfilling concern. The choice between null and alternative hypotheses is assessed through the statistical evidence from the z-test result.
p-value
The p-value is a vital metric in hypothesis testing. It quantifies the probability of observing a test statistic as extreme as, or more extreme than, the one calculated from your sample data, assuming the null hypothesis is true.

A small p-value, typically less than the significance level (\( \alpha \)), suggests that the observed data is unlikely under the null hypothesis, contributing strongly to the rejection of \( H_0 \).

In the exercise, the p-value derived from a z-value of 11.8 is notably low, as the z-value far exceeds typical critical values in standard normal distribution scenarios. Through this p-value, we conclude that there is substantial evidence to reject the null hypothesis, indicating that the mean weight is indeed greater than 16 ounces.

To find a p-value for a specific z-value, statistical tables or software can be used. It complements the decision about rejecting or accepting the null hypothesis effectively.

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

Answer the questions: (a) Is this a one- or two-tailed test? (b) What is the decision rule? (c) What is the value of the test statistic? (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. A sample of 36 observations is selected from a normal population. The sample mean is 21 , and the population standard deviation is 5 . Conduct the following test of hypothesis using the .05 significance level. $$ \begin{array}{l} H_{0}: \mu \leq 20 \\ H_{1}: \mu>20 \end{array} $$

Answer the questions: (a) Is this a one- or two-tailed test? (b) What is the decision rule? (c) What is the value of the test statistic? (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. A sample of 36 observations is selected from a normal population. The sample mean is \(12,\) and the population standard deviation is \(3 .\) Conduct the following test of hypothesis using the .01 significance level. $$ \begin{array}{l} H_{0}: \mu \leq 10 \\ H_{1}: \mu>10 \end{array} $$

A United Nations report shows the mean family income for Mexican migrants to the United States is \(\$ 27,000\) per year. A FLOC (Farm Labor Organizing Committee) evaluation of 25 Mexican family units reveals the mean to be \(\$ 30,000\) with a sample standard deviation of \(\$ 10,000 .\) Does this information disagree with the United Nations report? Apply the .01 significance level.

The mean income per person in the United States is \(\$ 50,000,\) and the distribution of incomes follows a normal distribution. A random sample of 10 residents of Wilmington, Delaware, had a mean of \(\$ 60,000\) with a standard deviation of \(\$ 10,000 .\) At the .05 level of significance, is that enough evidence to conclude that residents of Wilmington, Delaware, have more income than the national average?

For Exercises 5-8: (a) State the null hypothesis and the alternate hypothesis. (b) State the decision rule. (c) Compute the value of the test statistic. (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. The manufacturer of the \(X-15\) steel-belted radial truck tire claims that the mean mileage the tire can be driven before the tread wears out is 60,000 miles. Assume the mileage wear follows the normal distribution and the standard deviation of the distribution is 5,000 miles. Crosset Truck Company bought 48 tires and found that the mean mileage for its trucks is 59,500 miles. Is Crosset's experience different from that claimed by the manufacturer at the .05 significance level?

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.