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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
For part a, the probability is the sum of the probabilities of 0 success and 1 success with p=0.4. For part b, she has committed a Type I error. For part c, the probability is 1 minus the sum of probabilities of 0 success and 1 success with p=0.2. For part d, suggestions can be increasing the sample size, increasing the number of tests or decreasing the error margin.

Step by step solution

01

Calculation of Binomial Probability For Part a

First you need to apply the formula for binomial probability, which is \(P(x) = C(n, x) \times (p)^x \times (1-p)^{n-x}\), where P(x) is the probability of x successes in n trials, C(n, x) is the combination of n items taken x at a time, and p is the probability of success. The question is looking for the probability of at most one success, which means you need to calculate P(0) and P(1), and then add them. For P(0), simply substitute the values of n=10, x=0 and p=0.4 into the formula. For P(1), substitute the values of n=10, x=1 and p=0.4 into the formula.
02

Interpretation of Error For Part b

This is a question of understanding Type I and Type II errors in hypothesis testing. If the new clay is really no better than the old one, but she thinks it is, then she has committed a Type I error - rejecting the null hypothesis when it is true.
03

Calculation of Binomial Probability For Part c

This part involves changing the probability of success (p) to 0.2 and then calculating the probability of at least 2 successes, which requires calculating 1 - P(0) - P(1). The calculation of probabilities is similar to step 1, but the probability of success is now 0.2 instead of 0.4.
04

Suggesting Improvements For Part d

Improving the power of the test can be achieved in a number of ways. Some suggestions can be: increasing the sample size (fire more pieces), increase the number of tests (repeat the firing process several times to get more data), or decrease the margin of error (be more strict with the criteria for calling a piece a 'success').

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Key Concepts

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

Understanding Type I and Type II Errors
When an artist or a researcher conducts a test, sometimes they make decisions that are incorrect due to the variability inherent in data. These incorrect decisions are known as Type I and Type II errors in the realm of hypothesis testing.

Type I error occurs when one wrongly rejects a true null hypothesis. Imagine if we're testing if a coin is fair and we conclude it's biased when it's actually not; that’s a Type I error. In the context of our artist's scenario, a Type I error would be her deciding the new clay is better (rejecting the null hypothesis), while in reality, it's not.

Conversely, a Type II error happens when one fails to reject a false null hypothesis. If the coin is indeed biased, but our test fails to show this, we're facing a Type II error. If the new clay actually reduces breakage but the artist’s test fails to detect this, she has made a Type II error.

Each type of error has different consequences and understanding both is paramount when designing an experiment, as it shapes analysis and decision making.
The Basics of Hypothesis Testing
Hypothesis testing is a core concept in statistics used to determine whether there is enough evidence in a sample of data to infer that a certain condition holds for the entire population. This process begins with proposing two hypotheses: the null hypothesis (\( H_0 \)), which is usually a statement of 'no effect' or 'no difference', and the alternative hypothesis (\( H_1 \) or \( H_a \)), which suggests a new theory or effect we wish to test.

For our pottery artist, the null hypothesis might be 'the new clay does not reduce the breakage rate', while the alternative is 'the new clay reduces breakage rate'. She then collects data through her experiment of firing 10 pieces, and uses it to decide which hypothesis to support. The decision involves calculating probabilities and considering the chance of making Type I or Type II errors.
What is Statistical Power?
In statistical hypothesis testing, the power of a test is the probability that the test correctly rejects the null hypothesis when the alternative hypothesis is true. In layman's terms, it's the test’s ability to detect an effect if there really is one. High power is desirable and means the test is reliable.

In the artist's experiment, the power relates to the test's ability to detect a reduction in the breakage rate when using the new clay, assuming the new clay is, in fact, superior. If the test has low power and fails to detect this improvement, the artist risks continuing to use inferior clay due to a Type II error. To avoid this, she needs to increase the power of her test which can often be achieved by increasing the sample size or the sensitivity of the test criteria.
Calculating Probability of Success
In the context of binomial probability, as seen in the artist example, the 'probability of success' refers to the likelihood of a single outcome we define as 'success' occurring in a trial. Success does not necessarily mean something positive; it is simply the outcome of interest. In the artist’s case, 'success' could be defined as a piece of pottery not breaking when fired.

To calculate the probability of having a certain number of successes across multiple trials, we use the binomial probability formula: \[P(x) = C(n, x) \times (p)^x \times (1-p)^{n-x}\], where \(p\) is the probability of success on a single trial, \(n\) is the number of trials, and \(x\) is the number of successes. The artist uses this formula to calculate the likelihood of 0 or 1 pieces breaking, with the understanding that a lower breakage would convince her to switch to the new clay.

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Most popular questions from this chapter

A company is sued for job discrimination because only \(19 \%\) of the newly hired candidates were minorities when \(27 \%\) of all applicants were minorities. Is this strong evidence that the company's hiring practices are discriminatory? a. Is this a one-tailed or a two-tailed test? Why? b. In this context, what would a Type I error be? c. In this context, what would a Type II error be? d. In this context, what is meant by the power of the test? e. If the hypothesis is tested at the \(5 \%\) level of significance instead of \(1 \%\), how will this affect the power of the test? \(\mathrm{f}\). The lawsuit is based on the hiring of 37 employees. Is the power of the test higher than, lower than, or the same as it would be if it were based on 87 hires?

A company is willing to renew its advertising contract with a local radio station only if the station can prove that more than \(20 \%\) of the residents of the city have heard the ad and recognize the company's product. The radio station conducts a random phone survey of 400 people. a. What are the hypotheses? b. The station plans to conduct this test using a \(10 \%\) level of significance, but the company wants the significance level lowered to \(5 \%\). Why? c. What is meant by the power of this test? d. For which level of significance will the power of this test be higher? Why? e. They finally agree to use \(\alpha=0.05,\) but the company proposes that the station call 600 people instead of the 400 initially proposed. Will that make the risk of Type II error higher or lower? Explain.

The manufacturer of a metal stand for home TV sets must be sure that its product will not fail under the weight of the TV. Since some larger sets weigh nearly 300 pounds, the company's safety inspectors have set a standard of ensuring that the stands can support an average of over 500 pounds. Their inspectors regularly subject a random sample of the stands to increasing weight until they fail. They test the hypothesis \(\mathrm{H}_{0}: \mu=500\) against \(\mathrm{H}_{\mathrm{A}}: \mu>500,\) using the level of significance \(\alpha=0.01\). If the sample of stands fails to pass this safety test, the inspectors will not certify the product for sale to the general public. a. Is this an upper-tail or lower-tail test? In the context of the problem, why do you think this is important? b. Explain what will happen if the inspectors commit a Type I error. c. Explain what will happen if the inspectors commit a Type II error.

For each of the following situations, find thecritical value(s) for \(z\) or \(t\). a. \(\mathrm{H}_{0}: p=0.5\) vs. \(\mathrm{H}_{\mathrm{A}}: p \neq 0.5\) at \(\alpha=0.05\). b. \(\mathrm{H}_{0}: p=0.4 \mathrm{vs} . \mathrm{H}_{\mathrm{A}}: p>0.4\) at \(\alpha=0.05\). c. \(\mathrm{H}_{0}: \mu=10\) vs. \(\mathrm{H}_{\mathrm{A}}: \mu \neq 10\) at \(\alpha=0.05 ; n=36\). d. \(\mathrm{H}_{0}: p=0.5\) vs. \(\mathrm{H}_{\mathrm{A}}: p>0.5\) at \(\alpha=0.01 ; n=345\) e. \(\mathrm{H}_{0}: \mu=20\) vs. \(\mathrm{H}_{\mathrm{A}}: \mu<20\) at \(\alpha=0.01 ; n=1000\)

Environmentalists concerned about the impact of high-frequency radio transmissions on birds found that there was no evidence of a higher mortality rate among hatchlings 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.

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