91Ó°ÊÓ

A winemaker has placed a large order for the no. 9 corks described in Applied Example 6.13 (p. 285 ) and is concerned about the number of corks that might have smaller diameters. During the corking process, the corks are squeezed down to 16 to \(17 \mathrm{mm}\) in diameter for insertion into bottles with an \(18 \mathrm{mm}\) opening. The cork then expands to make the seal. The winemaker wants the corks to be as tight as possible and is therefore concerned about any that might be undersized. The diameter of each cork is measured in several places, and an average diameter is reported for each cork. The cork manufacturer has assured the winemaker that each cork has an average diameter within the specs and that all average diameters have a normal distribution with a mean of \(24.0 \mathrm{mm}\). a. Why does it make sense for the diameter of the cork to be assigned the average of several different diameter measurements? A random sample of 18 corks is taken from the batch to be shipped and the diameters (in millimeters) obtained: $$\begin{array}{llllllll}\hline 23.93 & 23.91 & 23.82 & 24.02 & 23.93 & 24.17 & 23.93 & 23.84 & 24.13 \\\24.01 & 23.83 & 23.74 & 23.73 & 24.10 & 23.86 & 23.90 & 24.32 & 23.83 \\\\\hline\end{array}$$ b. The average diameter spec is "24 \(\mathrm{mm}+0.6 \mathrm{mm} /\) \(-0.4 \mathrm{mm} . "\) Does it appear this order meets the spec on an individual cork basis? Explain. c. Does the sample in part a show sufficient reason to doubt the truthfulness of the claim, that the mean average diameter is \(24.0 \mathrm{mm},\) at the 0.02 level of significance? A different sample of 18 corks was randomly selected and the diameters (in millimeters) obtained: $$\begin{array}{lllllllll}\hline 23.90 & 23.98 & 24.28 & 24.22 & 24.07 & 23.87 & 24.05 & 24.06 & 23.82 \\\24.03 & 23.87 & 24.08 & 23.98 & 24.21 & 24.08 & 24.06 & 23.87 & 23.95 \\\\\hline\end{array}$$ d. Does the preceding sample show sufficient reason to doubt the truthfulness of the claim, that the mean average diameter is \(24.0 \mathrm{mm},\) at the 0.02 level of significance? e. What effect did the two different sample means have on the calculated test statistic in parts c and d? Explain. f. What effect did the two different sample standard deviations have on the calculated test statistic in parts c and d? Explain.

Step by step solution

01

Calculating the First Sample Mean and Standard Deviation

The first step is to calculate the mean and the standard deviation of the first sample using the given measurements. The mean is computed by adding all the measurements of the diameters and dividing by the number of measurements. On the other hand, the standard deviation is calculated by taking the square root of the variance (which is the average of the squared differences from the mean).

02

Checking Against Specification of Each Cork

To answer part 'b' which involves checking if the first order meets the specification on an individual cork basis, one has to compare each measurement with the given specification \(24 mm + 0.6 mm\) and \(24 mm - 0.4 mm\). If all the provided measurements fall within this range, it can be concluded that the order meets the specification.

03

Testing against Claimed Mean

This step involves testing the manufacturer's claim that the mean diameter is \(24.0 mm\) at a 0.02 level of significance. Here the student uses a z-test to test the hypothesis. The z-test is used when the sample size is greater than 30, or the population standard deviation is known. The Z-score is calculated using the formula \[Z = \frac{(\bar{x}-\mu)}{(\sigma/\sqrt{n})}\] where, \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the standard deviation and n is the sample size. By comparing the calculated Z-score with the critical value from a z-table for a \(0.02\) level of significance, it can be determined if there is sufficient evidence to reject the null hypothesis (that the claim is truthful). It can be considered valid if the calculated value is less than the critical value as it falls within the acceptance region.

04

Repeating The Process with Second Sample

Part 'd' requires the same process as in step 3 but this time using the second provided sample of cork measurements. The mean and standard deviation of the second sample are calculated and a new Z-score is computed. Using a similar process of hypothesis testing, the conclusions about the manufacturer's claim is checked.

05

Understanding the Effect of Different Means and Standard Deviations

The final parts 'e' and 'f' require understanding of how different sample means and standard deviations affect the calculated test statistics. The sample mean is a point estimate of the population mean. Therefore, different sample means would lead to different estimates of the parameter of interest thus changing the test statistic. Similarly, the standard deviation gives an understanding of the variability in the sample. If the variability is large (i.e., a larger standard deviation), it will result in a smaller z-score and vice versa. Therefore, the different standard deviations calculated for the two samples will result in different test statistics.

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

When dealing with hypothesis testing, the z-test plays a crucial role, especially when you need to compare the sample mean to the known population mean. This is useful when you're analyzing data to see if a claim about a population holds true. For example, the winemaker is verifying whether the mean cork diameter is indeed 24 mm as claimed.

We often use the z-test when either the sample size is large (more than 30) or the population standard deviation is known. The z-test helps us determine how many standard deviations an element is from the mean. Here’s how it works:

• First, calculate the sample mean.
• Next, determine the population standard deviation if known.
• Compute the z-score using the formula: \[ Z = \frac{(\bar{x} - \mu)}{(\sigma/\sqrt{n})} \]where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the standard deviation, and \(n\) is the sample size.
• Finally, compare the z-score to a critical value from the z-table to decide whether to accept or reject the null hypothesis.

This process systematically determines if there's enough evidence to doubt the manufacturer's claim about the corks.

standard deviation

Standard deviation is a measure that informs us about the spread or variability of the data around the mean. In the context of the winemaker's problem, understanding standard deviation is crucial because it shows how much the cork diameters deviate from the average.

The steps to calculate standard deviation are as follows:

• First, find the mean (average) of your dataset.
• Subtract the mean from each data point to find the deviation of each point.
• Square each deviation to eliminate negative values.
• Calculate the average of these squared deviations. This is called variance.
• The standard deviation is the square root of the variance.

A small standard deviation means that the values tend to be close to the mean, whereas a larger standard deviation indicates more spread out data. This is vital for ensuring that most corks are tightly within specification limits.

sample mean

The sample mean is a crucial concept in statistics as it serves as an estimate of the population mean. In hypothesis testing, we often use the sample mean to make inferences or test claims about the population from which the sample was drawn. Here’s how you determine the sample mean for the cork diameter problem:

• Add up all the diameter measurements you’ve recorded.
• Divide the total sum by the number of corks measured.This gives you the sample mean, represented as \(\bar{x}\).

Using the sample mean helps in assessing if the actual mean is quite different from what was claimed, especially when checked against the specified limits. This ensures that the corks being used will function effectively in sealing wine bottles.

significance level

The significance level, denoted by \(\alpha\), is a threshold we use to decide whether to reject the null hypothesis in hypothesis testing. It represents the probability of making a Type I error, or rejecting a true null hypothesis. For example, a \(0.02\) significance level indicates that there is a 2% risk of concluding that the cork diameter does not meet specifications when it actually does.

In the winemaker's scenario, this significance level is essential for making a data-supported decision about the corks. Here’s what you need to know about significance levels:

• A smaller significance level (e.g., \(0.01\)) indicates that you require stronger evidence to reject the null hypothesis.
• You choose the significance level based on how much risk you’re willing to accept.
• Compare the calculated p-value or test statistic against this threshold to make your decision.

Being clear about the significance level helps ensure that decisions are made with a defined certainty, minimizing the risk and ensuring quality.