/*! 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 You roll a die 60 times and reco... [FREE SOLUTION] | 91Ó°ÊÓ

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You roll a die 60 times and record the sample proportion of 5 's, and you want to test whether the die is biased to give more 5 's than a fair die would ordinarily give. To find the p-value for your sample data, you create a randomization distribution of proportions of 5 's in many simulated samples of size 60 with a fair die. (a) State the null and alternative hypotheses. (b) Where will the center of the distribution be? Why? (c) Give an example of a sample proportion for which the number of 5 's obtained is less than what you would expect in a fair die. (d) Will your answer to part (c) lie on the left or the right of the center of the randomization distribution? (e) To find the p-value for your answer to part (c), would you look at the left, right, or both tails? (f) For your answer in part (c), can you say anything about the size of the p-value?

Short Answer

Expert verified
The null hypothesis is that the die is fair and the probability of getting 5 is \(\frac{1}{6}\). The alternative hypothesis is that the die is biased with a probability of getting 5 higher than \(\frac{1}{6}\). The center of the distribution is at 0.167. An unusual sample proportion could be 0.10, which would lie to the left of the center. The p-value is calculated from the right tail, and its size will technically be 1.

Step by step solution

01

- Formulating Null and Alternative Hypotheses

The Null Hypothesis, \(H_0\), suggests that there is no significant difference between the observed and expected results. This means that the die is fair and the probability of getting a 5 is \(\frac{1}{6}\). The Alternative Hypothesis, \(H_1\), suggests that there is a significant difference between the observed and expected results. This means that the die is biased and the probability of rolling a 5 is greater than \(\frac{1}{6}\).
02

- Identifying The Center of Distribution and Reasoning

The center of the distribution for a fair die, under the Null Hypothesis, will be at the expected proportion of 5's, which is \(\frac{1}{6}\) or approximately 0.167 because a fair die has an equal chance of rolling each number.
03

- An Example of an Unusual Sample Proportion

An example of a sample proportion for which the number of 5's obtained is less than what would be expected from a fair die could be 0.10. This is because under the null hypothesis we would expect the proportion to be around 0.167, therefore 0.10 would be less than expected.
04

- Position of the Unusual Sample Proportion in Distribution

The unusual sample proportion from step 3 (0.10) would lie to the left of the center of the randomization distribution (0.167). This is because a proportion less than the expected value would lie to the left of the expected value in the distribution.
05

- Determining P-value Position for the Sample Proportion

To find the p-value for the sample proportion calculated in step 3, one would look at the right tail. This is because the p-value is determined by the probability of getting a result as extreme or more extreme than the observed result under the null hypothesis.
06

- Brief Comment on Size of the P-value for Given Sample Proportion

As the observed proportion (0.10) lies to the left and the test is for greater proportion in the alternative hypothesis, the p-value will technically be 1. This is because we are looking all values as extreme or more extreme than our observed result, and since our test is one-sided to the right, all values fall into this category.

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Key Concepts

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

Null Hypothesis
In any statistical analysis, the null hypothesis (\(H_0\)) acts as the starting assumption. It is a general statement or default position that there is no relationship between two measured phenomena or no association among groups. In the die bias test, the null hypothesis would state that the die is fair; thus, the probability of rolling a 5 is \frac{1}{6}, which is what one would expect from a fair die with six equal sides. Understanding the null hypothesis is critical because it forms the basis of the statistical test and determines what the alternative hypothesis will be, which is the next concept we'll explore.
Alternative Hypothesis
Contrary to the null hypothesis, the alternative hypothesis (\(H_1\text{ or }\H_a\)) signifies what a researcher aims to prove. It proposes that there is a statistically significant effect or relationship between variables or that a single variable differs from a specified value. Regarding the die test, the alternative hypothesis posits that the die is biased, leading to a greater than \frac{1}{6} chance of rolling a 5. The alternative hypothesis is what you test against the null hypothesis, and if evidence supports it enough, you may reject the null hypothesis.
Randomization Distribution
The randomization distribution is a critical component in the resampling or simulation-based inference methods used in statistics. It refers to the distribution of a test statistic that is created by calculating the statistic for numerous re-sampled or simulated datasets. For the die bias test, the randomization distribution would illustrate the expected proportions of rolling a 5 in a large number of simulated samples of 60 rolls each, assuming that the die is fair. It is expected to center around the null hypothesis value of 0.167. Mapping out the entire distribution is core to understanding the behavior under the null hypothesis and subsequently finding the p-value.
P-value
The p-value is a foundational concept in hypothesis testing that helps you decide whether to support or reject your null hypothesis. It is the probability of obtaining the observed sample results, or more extreme, assuming the null hypothesis is true. A low p-value (< 0.05 is a common threshold) indicates that your sample findings are unlikely under the null hypothesis, which may lead you to reject it in favor of the alternative hypothesis. In the context of the die bias test, you would calculate the p-value by looking at the right tail (since we're testing for more 5s) of the randomization distribution. However, if your sample proportion is less than the expected (like the example of 0.10), your p-value will lean towards 1, suggesting no evidence against the null hypothesis in this one-sided test.
Sample Proportion
The sample proportion is the fraction of items in a sample that have a particular attribute, and it's a statistic used to estimate a population proportion. In our exercise, if you roll the die 60 times, the sample proportion of 5's is the number of 5's rolled divided by 60. Sample proportions are affected by random sampling variations, and using them to draw conclusions about the fairness of the die involves comparing the observed proportion to the expected proportion under the null hypothesis. As in the example given, a sample proportion of 0.10 would be significantly less than the expected proportion of 0.167 for a fair die, affecting our interpretations and the resulting p-value.
Statistical Significance
Statistical significance is a term used to state that a result is not likely to occur by chance. In hypothesis testing, a result is considered statistically significant if the p-value is below a predetermined threshold, typically 0.05. This significance level indicates that there is less than a 5% probability that the observed result is due to random variation alone. In our die bias test, if the p-value is very low when testing an observed increase in the number of 5's, this would suggest a statistically significant result, providing evidence to reject the null hypothesis and support the claim of a biased die. However, a high p-value, as in the unusual sample proportion, does not provide evidence against the null hypothesis and would not be considered statistically significant.

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Most popular questions from this chapter

Match the four \(\mathrm{p}\) -values with the appropriate conclusion: (a) The evidence against the null hypothesis is significant, but only at the \(10 \%\) level. (b) The evidence against the null and in favor of the alternative is very strong. (c) There is not enough evidence to reject the null hypothesis, even at the \(10 \%\) level. (d) The result is significant at a \(5 \%\) level but not at a \(1 \%\) level. I. 0.0875 II. 0.5457 III. 0.0217 IV. \(\quad 0.00003\)

Give null and alternative hypotheses for a population proportion, as well as sample results. Use StatKey or other technology to generate a randomization distribution and calculate a p-value. StatKey tip: Use "Test for a Single Proportion" and then "Edit Data" to enter the sample information. Hypotheses: \(H_{0}: p=0.5\) vs \(H_{a}: p \neq 0.5\) Sample data: \(\hat{p}=42 / 100=0.42\) with \(n=100\)

Interpreting a P-value In each case, indicate whether the statement is a proper interpretation of what a p-value measures. (a) The probability the null hypothesis \(H_{0}\) is true. (b) The probability that the alternative hypothesis \(H_{a}\) is true. (c) The probability of seeing data as extreme as the sample, when the null hypothesis \(H_{0}\) is true. (d) The probability of making a Type I error if the null hypothesis \(H_{0}\) is true. (e) The probability of making a Type II error if the alternative hypothesis \(H_{a}\) is true.

Hypotheses for a statistical test are given, followed by several possible confidence intervals for different samples. In each case, use the confidence interval to state a conclusion of the test for that sample and give the significance level used. Hypotheses: \(H_{0}: \rho=0\) vs \(H_{a}: \rho \neq 0\). In addition, in each case for which the results are significant, give the sign of the correlation. (a) \(95 \%\) confidence interval for \(\rho: 0.07\) to 0.15 . (b) \(90 \%\) confidence interval for \(\rho:-0.39\) to -0.78 . (c) \(99 \%\) confidence interval for \(\rho:-0.06\) to 0.03 .

Do iPads Help Kindergartners Learn: A Subtest The Auburn, Maine, school district conducted an early literacy experiment in the fall of 2011 . In September, half of the kindergarten classes were randomly assigned iPads (the intervention group) while the other half of the classes got them in December (the control group.) Kids were tested in September and December and the study measures the average difference in score gains between the control and intervention group. \(^{41}\) The experimenters tested whether the mean score for the intervention group was higher on the HRSIW subtest (Hearing and Recording Sounds in Words) than the mean score for the control group. (a) State the null and alternative hypotheses of the test and define any relevant parameters. (b) The p-value for the test is 0.02 . State the conclusion of the test in context. Are the results statistically significant at the \(5 \%\) level? (c) The effect size was about two points, which means the mean score for the intervention group was approximately two points higher than the mean score for the control group on this subtest. A school board member argues, "While these results might be statistically significant, they may not be practically significant." What does she mean by this in this context?

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