/*! 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 22 Rutter Nursery Company packages ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Rutter Nursery Company packages its pine bark mulch in 50 -pound bags. From a long history, the production department reports that the distribution of the bag weights follows the normal distribution, and the standard deviation of the packaging process is 3 pounds per bag. At the end of each day, Jeff Rutter, the production manager, weighs 10 bags and computes the mean weight of the sample. Below are the weights of 10 bags from today's production. \(\begin{array}{lllllllll}45.6 & 47.7 & 47.6 & 46.3 & 46.2 & 47.4 & 49.2 & 55.8 & 47.5 & 48.5\end{array}\) a. Can Mr. Rutter conclude that the mean weight of the bags is less than 50 pounds? Use the .01 significance level. b. In a brief report, tell why Mr. Rutter can use the \(z\) distribution as the test statistic. c. Compute the \(p\) -value.

Short Answer

Expert verified
No, Mr. Rutter cannot conclude the bags weigh less than 50 pounds at the 0.01 level. The p-value must be less than 0.01 to reject the null hypothesis.

Step by step solution

01

Calculate the Sample Mean

First, compute the mean of the sample weights. Add all the given weights together and divide by the number of bags, which is 10. Specifically, calculate \( \text{Mean} = \frac{45.6 + 47.7 + 47.6 + 46.3 + 46.2 + 47.4 + 49.2 + 55.8 + 47.5 + 48.5}{10} \).
02

Calculate the Standard Error of the Mean

The standard error (SE) of the mean is calculated using the formula \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma = 3 \) pounds (population standard deviation) and \( n = 10 \) (sample size).
03

Set Up the Null and Alternative Hypotheses

Define the hypotheses for the test. Null hypothesis \( H_0: \mu = 50 \) pounds. Alternative hypothesis \( H_a: \mu < 50 \) pounds. This is a one-tailed test.
04

Calculate the Test Statistic (z-score)

Use the formula for the z-score: \( z = \frac{\bar{x} - \mu}{SE} \), where \( \bar{x} \) is the sample mean, \( \mu = 50 \) pounds (claimed mean), and \( SE \) is the standard error found in Step 2.
05

Find the Critical Value and Make a Decision

Look up the critical value for \( z \) at a 0.01 significance level for a one-tailed test. If the calculated \( z \)-score from Step 4 is less than this critical value, we reject the null hypothesis, \( H_0 \).
06

Compute the p-value

The p-value is found using the z-score from Step 4. You can use a standard normal distribution table or calculator to find the probability associated with the calculated \( z \)-score. If the p-value is less than 0.01, reject \( H_0 \).
07

Explain Why the z-Distribution is Used

Mr. Rutter can use the \( z \)-distribution because the sample size is relatively small, but the population standard deviation is known and the population is normally distributed.

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.

Normal Distribution
In statistics, **normal distribution** is a fundamental concept that's often used in the context of hypothesis testing. It's sometimes called a Gaussian distribution or "bell curve" because of its distinct shape. In this specific context, Rutter Nursery's production weights follow a normal distribution, meaning most bag weights are centered around the mean with symmetry. Characteristics of a normal distribution include:
  • **Symmetry**: Half of the data falls to the right of the mean and half falls to the left, creating a symmetric bell-shaped curve.
  • **Mean, Median, and Mode are Equal**: The curve has its peak at the center.
  • **Defined Scale**: The spread of the curve is determined by its standard deviation.
Understanding this distribution helps Mr. Rutter in analyzing the sample data and understanding whether the bag weights deviate from the 50-pound target.
Z-Distribution
When assessing the bag weights, Mr. Rutter uses **z-distribution** for statistical testing. A z-distribution is derived from the normal distribution and is used to determine the probability of a sample mean in hypothesis testing. Key aspects include:
  • **Standardized**: Data is converted into z-scores, measuring how many standard deviations an element is from the mean. This helps compare different data sets with an identical distribution.
  • **Usefulness**: Since Mr. Rutter is measuring a sample mean and the population standard deviation is known, this makes it appropriate for using the z-score.
  • **Critical Values**: Using z-tables, Mr. Rutter can find threshold values for making decisions about the null hypothesis, a useful step in ensuring product consistency.
This method allows Mr. Rutter to interpret and decide whether production meets target specifications or adjustments need to be made.
Sample Mean
In any study like Mr. Rutter's, the **sample mean** is a crucial concept. The sample mean is simply the average value of sample data. It's calculated by summing up all sample observations and then dividing by the number of observations. In this scenario, Mr. Rutter takes the weights of 10 bags and calculates their average to assess the production process. Important points about the sample mean include:
  • **Representation**: It helps to represent the typical weight of the bag if the population mean is unknown or can't be measured.
  • **Influence in Testing**: This value is central to calculating the z-score and affects hypothesis testing, acting as a potential indicator if production is meeting its claimed weight standard.
  • **Estimation of Population Mean**: While it's an estimate, a well-chosen sample mean provides vital insights into the overall production quality and consistency.
Having a reliable sample mean assists managers like Mr. Rutter in making informed decisions about whether production needs any change.
P-value
The **p-value** is a critical component of hypothesis testing. It helps determine the strength of the evidence against a null hypothesis. Mr. Rutter calculates the p-value to know if the true mean weight of his products is indeed less than the claimed 50 pounds. Here's how p-values work:
  • **Decision-Making**: The p-value is the probability of obtaining a result at least as extreme as the one observed during the test, assuming the null hypothesis is true.
  • **Interpretation**: If the p-value is less than the significance level (0.01 in this case), it indicates strong evidence against the null hypothesis, prompting Mr. Rutter to reject it.
  • **Significance Level**: The set significance level (in this case 0.01) is a benchmark for deciding how unlikely a result must be for it to be considered statistically significant.
This tool aids in quantitatively assessing whether the production process needs review or if results fall within expected variations.

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 management of White Industries is considering a new method of assembling its golf cart. The present method requires a mean time of 42.3 minutes to assemble a cart. The mean assembly time for a random sample of 24 carts, using the new method, was 40.6 minutes, and the standard deviation of the sample was 2.7 minutes. Using the .10 level of significance, can we conclude that the assembly time using the new method is faster?

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?

The publisher of Celebrity Living claims that the mean sales for personality magazines that feature people such as Megan Fox or Jennifer Lawrence are 1.5 million copies per week. A sample of 10 comparable titles shows a mean weekly sales last week of 1.3 million copies with a standard deviation of 0.9 million copies. Do these data contradict the publisher's claim? Use the .01 significance level.

For a recent year, the mean fare to fly from Charlotte, North Carolina, to Chicago, Illinois, on a discount ticket was \(\$ 267 .\) A random sample of 13 round-trip discount fares on this route last month shows: \(\begin{array}{llllllllll}\$ 321 & \$ 286 & \$ 290 & \$ 330 & \$ 310 & \$ 250 & \$ 270 & \$ 280 & \$ 299 & \$ 265 & \$ 291 & \$ 275 & \$ 281\end{array}\) At the .01 significance level, can we conclude that the mean fare has increased? What is the \(p\) -value?

A recent study revealed the typical American coffee drinker consumes an average of 3.1 cups per day. A sample of 12 senior citizens revealed they consumed the following amounts of coffee, reported in cups, yesterday. \(\begin{array}{lllllllllll}3.1 & 3.3 & 3.5 & 2.6 & 2.6 & 4.3 & 4.4 & 3.8 & 3.1 & 4.1 & 3.1 & 3.2\end{array}\) At the .05 significance level, do these sample data suggest there is a difference between the national average and the sample mean from senior citizens?

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.