/*! 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 37 Too little or too much wine? Win... [FREE SOLUTION] | 91Ó°ÊÓ

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Too little or too much wine? Wine-pouring vending machines, previously available in Europe and international airports, have become popular in the past few years in the United States. They are even approved to dispense wine in some Walmart stores. The available pouring options are a 5 -ounce glass, a 2.5 -ounce half-glass, and a 1-ounce taste. When the machine is in statistical control (see Exercise 7.43 ), the amount dispensed for a full glass is 5.1 ounces. Four observations are taken each day to plot a daily mean over time on a control chart to check for irregularities. The most recent day's observations were \(5.05,5.15,4.95,\) and \(5.11 .\) Could the difference between the sample mean and the target value be due to random variation, or can you conclude that the true mean is now different from \(5.1 ?\) Answer by showing the five steps of a significance test, making a decision using a 0.05 significance level.

Short Answer

Expert verified
The sample mean difference could be due to random variation; we do not reject the null hypothesis.

Step by step solution

01

Formulate Hypotheses

In the first step, we need to set up our null and alternative hypotheses.**Null Hypothesis \( H_0 \):** The true mean amount of wine dispensed is \( 5.1 \) ounces, which means any difference is due to random variation. \( \mu = 5.1 \).**Alternative Hypothesis \( H_a \):** The true mean amount of wine dispensed is not \( 5.1 \) ounces. \( \mu eq 5.1 \).
02

Collect Data and Calculate Sample Mean and Standard Deviation

We are given four observations: \(5.05, 5.15, 4.95, \) and \( 5.11 \). Calculate the sample mean (\( \bar{x} \)) and standard deviation (\( s \)).\[ \bar{x} = \frac{5.05 + 5.15 + 4.95 + 5.11}{4} = 5.065 \text{ ounces} \]First, calculate the deviations from the mean, then square them, average these, and finally take the square root to get \( s \).
03

Determine the Test Statistic

Use the sample mean and sample standard deviation to calculate the test statistic under the assumption that the null hypothesis is true. The formula for the t-statistic for a sample is:\[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]Here, \( \bar{x} = 5.065 \), \( \mu = 5.1 \), and \( n = 4 \) so you'll need the standard deviation \( s \) from Step 2.
04

Define the Significance Level and Find Critical Values

For a significance level of \( \alpha = 0.05 \), and since this is a two-sided test, find the critical values of the t-distribution with \( n-1 = 3 \) degrees of freedom. Use a t-distribution table to find these values, which will be used to compare against our t-statistic from Step 3.
05

Make a Decision

Compare the calculated t-statistic from Step 3 with the critical values determined in Step 4. If the t-statistic is beyond the critical values (either greater than the upper critical value or less than the lower critical value), reject the null hypothesis \( H_0 \). Otherwise, do not reject \( H_0 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hypothesis Testing
Hypothesis testing is an essential method in statistics that helps us make decisions based on data. It involves comparing a sample statistic to a population parameter to determine the likelihood that a particular claim is true. The process starts with stating two hypotheses: the null hypothesis (\( H_0\)) and the alternative hypothesis (\( H_a\)).
- The null hypothesis asserts that there is no effect or no difference, meaning any observed variation is due to chance.
- The alternative hypothesis suggests that there is a significant effect or difference.
In the wine vending machine example, the null hypothesis asserts that the mean amount of wine dispensed is 5.1 ounces, while the alternative hypothesis suggests it's different. A significance level is set, often at 5% (\( \alpha = 0.05\)), to determine how much risk of being wrong one is willing to accept. By conducting a test and comparing the test statistic against critical values, one can decide whether to reject or not reject \( H_0\), based on the likelihood of observing the data if \( H_0\) were true.
Control Chart
A control chart is a graphical tool used to monitor if a process is in statistical control over time. It displays the data points of process measurements against upper and lower limits, which are usually set at \( \pm 3\sigma\)-\( \text{limits}\) above and below the process mean.
These charts are primarily used in quality control processes to identify special cause variations, which might indicate a shift in the process.
Some important aspects of control charts include:
- A central line that represents the average process measurement.
- Upper and lower control limits that indicate the threshold for acceptable variation.
- Data points plotted in time order to observe any trends or patterns.
In the wine-pouring machine case, a control chart helps track daily pour volumes to quickly see if any irregularity occurs beyond random variation, thereby maintaining the consistency of the pours.
Statistical Process Control
Statistical Process Control (SPC) involves using statistical methods to measure and control the quality of processes. This methodology relies heavily on control charts to detect and prevent issues before they result in product defects. The aim of SPC is to ensure a process runs efficiently and produces outputs adhering to specifications.
Key elements of SPC include:
- Continuous monitoring of the process through control charts.
- Identifying both common cause (natural) and special cause (unusual) variations.
- Implementing solutions when process deviations are detected to maintain quality.
In contexts like the wine vending machine, SPC ensures that every glass poured meets quality standards, alerting operators about when adjustments are needed to maintain the desired pour volume.
t-Statistic
The t-statistic is a pivotal concept in hypothesis testing, especially with small sample sizes. It compares the sample mean to a known population mean to evaluate if they are statistically different.
The formula for the t-statistic is
\[t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\]
where:
  • \(\bar{x}\) is the sample mean.
  • \(\mu\) is the population mean.
  • \(s\) is the sample standard deviation.
  • \(n\) is the sample size.
The t-statistic helps in determining whether to reject \( H_0\) by comparing it to critical values from the t-distribution, which considers the degrees of freedom (n-1 for a single sample).
In our wine vending machine example, the t-statistic is used to test if the mean volume of wine poured is significantly different from the expected 5.1 ounces, given the sample data.

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Most popular questions from this chapter

For a test of \(\mathrm{H}_{0}: p=0.50\), the sample proportion is 0.35 based on a sample size of 100 . a. Show that the test statistic is \(z=-3.0\). b. Find the \(P\) -value for \(H_{a}: p<0.50\). c. Does the \(P\) -value in part b give much evidence against \(\mathrm{H}_{0} ?\) Explain.

\(z\) test statistic To test \(\mathrm{H}_{0}: p=0.50\) that a population proportion equals 0.50 , the test statistic is a \(z\) -score that measures the number of standard errors between the sample proportion and the \(\mathrm{H}_{0}\) value of \(0.50 .\) If \(z=3.6,\) do the data support the null hypothesis, or do they give strong evidence against it? Explain.

The question about the opinion on the increased use of fracking from the November 2014 survey mentioned in Example 6 was also included in an earlier survey in September 2013. Using this earlier survey, let's again focus on those who oppose the increased use of fracking. a. Define the parameter of interest and set up hypotheses to test that those who oppose fracking in 2013 are in the minority. b. Of the 1506 respondents in the 2013 survey, 740 indicated that they oppose the increased use of fracking. Find and interpret the test statistic. c. Report the P-value. Indicate your decision, in the context of this survey, using a 0.05 significance level. d. Check whether the sample size was large enough to conduct the inference in part \(c .\) Indicate what the assumptions are for your inferences to apply to the entire U.S. population. e. Find the P-value for the two-sided alternative that the proportion opposing is different from 0.50 .

When the 583 female workers in the 2012 GSS were asked how many hours they worked in the previous week, the mean was 37.0 hours, with a standard deviation of 15.1 hours. Does this suggest that the population mean work week for females is significantly different from 40 hours? Answer by: a. Identifying the relevant variable and parameter. b. Stating null and alternative hypotheses. c. Reporting and interpreting the P-value for the test statistic value. d. Explaining how to make a decision for the significance level of 0.01

Example 8 tested a therapy for anorexia, using hypotheses \(\mathrm{H}_{0}: \mu=0\) and \(\mathrm{H}_{a}: \mu \neq 0\) about the population mean weight change \(\mu .\) In the words of that example, what would be (a) a Type I error and (b) a Type II error?

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