/*! 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 78 Bottles of a popular cola are su... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 78

Bottles of a popular cola are supposed to contain 300 milliliters (ml) of cola. There is some variation from bottle to bottle because the filling machinery is not perfectly precise. An inspector measures the contents of six randomly selected bottles from a single day's production. The results are $$ \begin{array}{llllll} 299.4 & 297.7 & 301.0 & 298.9 & 300.2 & 297.0 \end{array} $$ Do these data provide convincing evidence that the mean amount of cola in all the bottles filled that day differs from the target value of \(300 \mathrm{ml} ?\)

Short Answer

Expert verified
No, the sample mean does not significantly differ from 300 ml.

Step by step solution

01

Calculate the Sample Mean

First, calculate the average amount of cola in the six bottles. To find the sample mean, sum up all the measured values and divide by the number of bottles. \[ \text{Sample Mean} = \frac{299.4 + 297.7 + 301.0 + 298.9 + 300.2 + 297.0}{6} \] Perform the calculations to get the sample mean.
02

Compute the Sample Standard Deviation

Find the standard deviation of the sample to understand the variability of the bottle measurements. Use the formula: \[ s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2} \] where \( n \) is the number of samples and \( \bar{x} \) is the sample mean calculated in Step 1. Compute \( s \) using the data.
03

State the Hypotheses

Formulate the null and alternative hypotheses. - Null Hypothesis \( H_0: \mu = 300 \) ml (the population mean is 300 ml).- Alternative Hypothesis \( H_a: \mu eq 300 \) ml (the population mean is not 300 ml).
04

Conduct a t-Test

Perform a t-test to determine if the sample mean significantly differs from the hypothesized population mean (300 ml). Use the formula: \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \] Substitute the sample mean, the population mean (300), the sample standard deviation, and the number of samples to find the t-value.
05

Determine the Critical t-Value

Decide on a significance level \( \alpha \), commonly 0.05, and find the critical t-value from the t-distribution table with \( n-1 \) degrees of freedom (which is 5 in this case). This helps in determining the rejection region for the null hypothesis.
06

Compare t-Value with Critical Value

Compare the calculated t-value from Step 4 with the critical t-value from Step 5. If the calculated t-value falls in the rejection region, we reject the null hypothesis; otherwise, we fail to reject it.

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.

Sample Mean
The sample mean is a statistical measure that helps us summarize a data set. It's often the first step in analyzing data like the cola bottle volumes. In our given exercise, the sample mean tells us the average amount of cola present in the six bottles we measured.

To find the sample mean, sum up all individual measurements and divide by the total number of measurements. It's mathematically expressed as:
  • Sample Mean = \( \frac{\text{Sum of all measured values}}{\text{Total number of bottles}} \)
The sample mean offers a quick overview of the data set, helping us see if it's close to the claimed mean value of 300 ml. In this exercise, after calculating this step, we can learn whether these bottles, on average, have less, more, or exactly 300 ml of cola.
Standard Deviation
Standard deviation represents the amount of variation or dispersion in a set of values. In the context of the cola bottles, it tells us how much the amount of cola varies from bottle to bottle.

To compute standard deviation, follow these steps:
  • Calculate the mean (average) of the dataset.
  • Subtract the mean from each data point and square the result.
  • Mean these squared differences.
  • Take the square root of that mean.
For our exercise, the greater the standard deviation, the more variability in cola amounts across bottles. Low standard deviation implies less variation, and the bottles are more consistently filled. Calculating this value helps us understand how precise the cola filling process is.
t-Test
The t-test is a statistical test used to compare the sample mean to a known value - here it's the supposed mean bottle volume of 300 ml. It helps decide whether any observed difference is significant or just due to chance.

To conduct a t-test, this formula is used:
  • \( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \)
where:
  • \( \bar{x} \) = sample mean
  • \( \mu \) = hypothesized population mean (300 ml)
  • \( s \) = sample standard deviation
  • \( n \) = sample size
Performing this test allows us to determine if the difference in bottle fill quantity is statistically significant. If the t-value falls beyond the critical t-value of our set significance level, it suggests our findings are significant.
Null Hypothesis
A cornerstone of hypothesis testing, the null hypothesis posits there is no effect or difference. It's essentially a statement we aim to test or disprove. In our bottle example, the null hypothesis is that the average amount of cola in the bottles is indeed 300 ml.

This is expressed as:
  • \( H_0: \mu = 300 \) ml
Here \( \mu \) represents the overall population mean. During the hypothesis testing, we assume the null hypothesis to be true until evidence suggests otherwise. Rejecting the null hypothesis implies that the bottles' average volume differs significantly from 300 ml.
Alternative Hypothesis
The alternative hypothesis is the counterpart to the null hypothesis and suggests there is a significant effect or difference. It's what we consider true if the null hypothesis is rejected.

In the cola bottle scenario, our alternative hypothesis claims that the average amount of cola is different from the expected 300 ml. We express this as:
  • \( H_a: \mu eq 300 \) ml
This two-sided hypothesis does not specify if the average bottle volume is higher or lower鈥攐nly that it's different. By proving the alternative hypothesis, we show that the measured differences in bottle fill volumes are unlikely caused by random chance alone.

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 most important condition for sound conclusions from statistical inference is that (a) the data come from a well-designed random sample or randomized experiment. (b) the population distribution be exactly Normal. (c) the data contain no outliers. (d) the sample size be no more than \(10 \%\) of the population size. (e) the sample size be at least 30 .

A state's Division of Motor Vehicles (DMV) claims that \(60 \%\) of teens pass their driving test on the first attempt. An investigative reporter examines an SRS of the DMV records for 125 teens; 86 of them passed the test on their first try. Is there convincing evidence at the \(\alpha=\) 0.05 significance level that the DMV's claim is incorrect?

Which of the following is not a condition for performing a significance test about a population proportion \(p ?\) (a) The data should come from a random sample or randomized experiment. (b) Both \(n p_{0}\) and \(n\left(1-p_{0}\right)\) should be at least 10 . (c) If you are sampling without replacement from a finite population, then you should sample no more than \(10 \%\) of the population. (d) The population distribution should be approximately Normal, unless the sample size is large. (e) All of the above are conditions for performing a significance test about a population proportion.

You read that a statistical test at significance level \(\alpha=0.05\) has power 0.78 . What are the probabilities of Type I and Type II errors for this test?

In a recent year, \(73 \%\) of firstyear college students responding to a national survey identified "being very well-off financially" as an important personal goal. A state university finds that 132 of an SRS of 200 of its first-year students say that this goal is important. Is there convincing evidence at the \(\alpha=0.05\) significance level that the proportion of all first-year students at this university who think being very well-off is important differs from the national value, \(73 \% ?\)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

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.