/*! 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 42 Pottery An artist experimenting ... [FREE SOLUTION] | 91Ó°ÊÓ

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Pottery An artist experimenting with clay to create pottery with a special texture has been experiencing difficulty with these special pieces. About \(40 \%\) break in the kiln during firing. Hoping to solve this problem, she buys some more expensive clay from another supplier. She plans to make and fire 10 pieces and will decide to use the new clay if at most one of them breaks. a) Suppose the new, expensive clay really is no better than her usual clay. What's the probability that this test convinces her to use it anyway? (Hint: Use a Binomial model.) b) If she decides to switch to the new clay and it is no better, what kind of error did she commit? c) If the new clay really can reduce breakage to only \(20 \%,\) what's the probability that her test will not detect the improvement? d) How can she improve the power of her test? Offer at least two suggestions.

Short Answer

Expert verified
a) 0.047 probability; b) Type I error; c) 0.625 probability; d) Increase sample size or adjust threshold.

Step by step solution

01

Define the Problem

We need to determine the probability of misleading the artist (in part a), understand the type of error (part b), calculate the missed detection (part c), and suggest improvements (part d).
02

Part a - Use Binomial Model Calculation

The probability that the new clay is no better is given by the binomial model with parameters: number of pieces, \( n = 10 \), and the probability of failure, \( p = 0.4 \). We are looking for \(\Pr(X \leq 1)\), where \(X\) represents the number of broken pieces. Calculate \(\Pr(X = 0) + \Pr(X = 1)\). \[\Pr(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]Calculate for \(X = 0\) and \(X = 1\):\[\Pr(X = 0) = \binom{10}{0} (0.4)^0 (0.6)^{10} \approx 0.006 \]\[\Pr(X = 1) = \binom{10}{1} (0.4)^1 (0.6)^9 \approx 0.041\]Thus, \(\Pr(X \leq 1) = 0.006 + 0.041 = 0.047\).
03

Part b - Identify the Error Type

If the clay is no better and the artist decides to switch, she has committed a Type I error, which is a false positive—rejecting a true null hypothesis.
04

Part c - Calculate Probability of Missing Improvement

If the new clay reduces breakage to 20%, and she doesn’t detect it, calculate \(\Pr(X > 1)\) with \(n = 10\) and \(p = 0.2\), using the complement rule: \[\Pr(X \leq 1) = \Pr(X = 0) + \Pr(X = 1)\]\[\Pr(X = 0) = \binom{10}{0} (0.2)^0 (0.8)^{10} = 0.107\]\[\Pr(X = 1) = \binom{10}{1} (0.2)^1 (0.8)^9 = 0.268\]Thus, \(\Pr(X \leq 1) = 0.107 + 0.268 = 0.375\) and \(\Pr(X > 1) = 1 - 0.375 = 0.625\).
05

Part d - Suggest Improvements to Power

To improve the test's power, the artist could (1) increase the sample size (more pieces made and fired) to better detect differences or (2) use a lower threshold for accepting the new clay (e.g., at most two break).

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Key Concepts

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

Type I Error
When we talk about Type I Error in statistics, we're referring to a situation where you think you've found a change or difference, but actually, there isn't one. It's like having a false alarm. In the context of the pottery exercise, if the artist uses the new clay thinking it is better, when in fact it isn't, she has made a Type I Error.

This happens when the artist inadvertently rejects the true null hypothesis. The null hypothesis in this case would be: "The new expensive clay is no better than the usual clay." If she decides on using the new clay based on a faulty test result where there seems to be an improvement, she has committed this error.
  • It's often symbolized as α, which is the probability of committing a Type I Error.
  • Controlling for Type I Error is critical because it affects the validity of the test results.
Bearing this in mind, the probability we calculated in Step 2 of \(\Pr(X \leq 1) = 0.047\), gives us a ballpark likelihood of this error occurring. This means there's a 4.7% chance that she'll incorrectly conclude that the expensive clay is better when it’s not, leading to a false positive decision.
Power of a Test
The power of a test is about its ability to discover the truth. Specifically, it refers to the probability that the test will correctly reject a false null hypothesis. Think of it as the test’s sensitivity in detecting an actual improvement when one truly exists.

In the pottery example, if the new clay genuinely decreases breakage to 20%, the power of the test would indicate how likely it is that the artist identifies this improvement. The calculation from Step 4 shows a lesser power, as there's a significant \(\Pr(X > 1) = 0.625\) chance that she'll miss the improvement. Thus, a lower power results in higher chances of making a Type II Error—failing to detect a true effect.
  • Increasing the power can be achieved by enlarging the sample size, ensuring more sensitivity to detect true differences.
  • Setting a more reasonable threshold for determining success can also heighten the power, leading to better decision-making outcomes.
Therefore, enhancing the power of a test is critical to ensure real improvements are noted and not missed by chance or error.
Probability Calculation
Probability calculations are fundamental in statistics to determine the likelihood of particular outcomes. In this problem, we are especially interested in applying these calculations through the binomial model. This model helps in understanding the outcomes of processes where there are two possible results, like a pottery piece either breaking or not breaking.

The binomial model utilizes the formula:
\ \Pr(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \
where \( n \) is the number of trials, \( p \) is the probability of success (or failure, depending on how you define it), and \( k \) is the number of successful outcomes. This calculation can be used to find specific probabilities, such as the likelihood of 0 or 1 piece breaking, or more general probabilities like fewer than 2 pieces breaking.
  • For this pottery exercise, these calculations provide insights into the artist's decisions about whether to use the new clay.
  • They're essential in evaluating and contrasting the performance of different clay types based on empirical data.
Therefore, understanding and correctly utilizing probability calculations can offer solid ground for making informed choices within experimental and practical settings.

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Most popular questions from this chapter

Another P-value Have harsher penalties and ad campaigns increased seat-belt use among drivers and passengers? Observations of commuter traffic failed to find evidence of a significant change compared with three years ago. Explain what the study's P-value of 0.17 means in this context.

Hypotheses and parameters As in Exercise \(1,\) for each of the following situations, define the parameter and write the null and alternative hypotheses in terms of parameter values. a) Seat-belt compliance in Massachusetts was \(65 \%\) in \(2008 .\) The state wants to know if it has changed. b) Last year, a survey found that \(45 \%\) of the employees were willing to pay for on-site day care. The company wants to know if that has changed. c) Regular card customers have a default rate of \(6.7 \%\) A credit card bank wants to know if that rate is different for their Gold card customers.

More errors For each of the following situations, state whether a Type I, a Type II, or neither error has been made. a) A test of \(\mathrm{H}_{0}: p=0.8\) vs. \(\mathrm{H}_{\mathrm{A}}: p<0.8\) fails to reject the null hypothesis. Later it is discovered that \(p=0.9\) b) A test of \(\mathrm{H}_{0}: p=0.5\) vs. \(\mathrm{H}_{\mathrm{A}}: p \neq 0.5\) rejects the null hypothesis. Later is it discovered that \(p=0.65\) c) A test of \(\mathrm{H}_{0}: p=0.7\) vs. \(\mathrm{H}_{\mathrm{A}}: p<0.7\) fails to reject the null hypothesis. Later is it discovered that \(p=0.6\)

Alpha again Environmentalists concerned about the impact of high-frequency radio transmissions on birds found that there was no evidence of a higher mortality rate among hatch lings in nests near cell towers. They based this conclusion on a test using \(\alpha=0.05 .\) Would they have made the same decision at \(\alpha=0.10 ?\) How about \(\alpha=0.01 ?\) Explain.

Significant again? A new reading program may reduce the number of elementary school students who read below grade level. The company that developed this program supplied materials and teacher training for a large-scale test involving nearly 8500 children in several different school districts. Statistical analysis of the results showed that the percentage of students who did not meet the grade-level goal was reduced from \(15.9 \%\) to \(15.1 \%\) The hypothesis that the new reading program produced no improvement was rejected with a P-value of 0.023 a) Explain what the P-value means in this context. b) Even though this reading method has been shown to be significantly better, why might you not recommend that your local school adopt it?

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