/*! 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 34 To raise money for a new rectory... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

To raise money for a new rectory, the members of a church hold a raffle. A total of \(n\) tickets are sold (numbered 1 through \(n\) ), out of which a total of fifty winners are to be drawn presumably at random. The following are the fifty lucky numbers. Set up a goodness-of-fit test that focuses on the randomness of the draw. Use the \(0.05\) level of significance. $$ \begin{array}{rrrrr} \hline 108 & 110 & 21 & 6 & 44 \\ 89 & 68 & 50 & 13 & 63 \\ 84 & 64 & 69 & 92 & 12 \\ 46 & 78 & 113 & 104 & 105 \\ 9 & 115 & 58 & 2 & 20 \\ 19 & 96 & 28 & 72 & 81 \\ 32 & 75 & 3 & 49 & 86 \\ 94 & 61 & 35 & 31 & 56 \\ 17 & 100 & 102 & 114 & 76 \\ 106 & 112 & 80 & 59 & 73 \\ \hline \end{array} $$

Short Answer

Expert verified
Your ultimate decision depends on the calculated chi-square statistic in comparison to the critical value. If the statistic is higher, you conclude the draw wasn't random. If the test statistic is lower, you conclude the draw was likely random.

Step by step solution

01

Classifying and Counting

First of all, categorize the 50 drawn numbers into groups (bins). The simplest way to do this is by splitting it into two categories: 'numbers less than or equal to 60' and 'numbers greater than 60'. Then count the number of occurrences in each bin.
02

Calculate Expected Frequencies

Since our null hypothesis is that the draw is random, we should expect an even distribution of numbers within both the categories. To calculate it, simply divide the total number of draws (50) by the number of categories (2). Thus, the expected frequency for both categories is 25.
03

Compute Chi-Square Test Statistic

The measure of goodness-of-fit, \( \chi^2 \), is calculated by summing the squared deviations (between observed and expected frequencies) divided by the expected frequency for all categories.
04

Determine the Critical Value

The critical value is seen from the Chi-Squared distribution table for \( \alpha = 0.05 \) and degrees of freedom \( df = n-1 \), where \( n \) is the number of categories. Here, \( df = 2-1 = 1 \). Therefore, from the table, the critical value is approximately 3.841.
05

Decision

If the calculated chi-square statistic is higher than the critical value, then we reject the null hypothesis, indicating that the draw was not random. If the statistic is lower, then we fail to reject the null hypothesis, and it suggests that the draw was indeed random.

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Key Concepts

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

Chi-square test
The Chi-square test is a statistical method used to determine whether there is a significant difference between the expected and observed frequencies in categorical data.
In the case of the church raffle, we set up a chi-square test to verify if the selection of winning tickets was indeed random.
To perform the test, we first divide our data into categories. Here, we place all 50 raffle numbers into two bins: numbers less than or equal to 60, and numbers greater than 60.
This categorization helps us compare observed frequencies with expected ones, assuming a fair raffle.

Once categories are defined, we calculate the chi-square statistic using the formula: \[\chi^2 = \sum \frac{(O - E)^2}{E}\]where \(O\) represents observed frequencies and \(E\) represents expected frequencies.
For instance, if we find 30 numbers in one category and 20 in the other, while expecting 25 in each, we input these values into the formula to compute the chi-square statistic.
The chi-square test is crucial for evaluating the fairness of random draws and other similar experiments.
Randomness
Randomness refers to the unpredictability and lack of pattern in the selection process.
In the context of the raffle, it signifies that each ticket had an equal chance of being picked.
If the raffle's results show unexpected patterns, such as clustering of numbers around specific areas, it may suggest some bias or error in the drawing process.

Random draws imply that the process is free from human intervention aiming to manipulate outcomes.
This is a cornerstone in ensuring fairness and integrity.
  • Unbias: Each ticket number should be independent of others.
  • Equity: All numbers have equal chances.
  • Consistency: Results should reflect a diverse spread across possible outcomes.
To test for randomness in our exercise, we use a chi-square goodness-of-fit test.
If the test indicates a departure from expected randomness, it suggests potential issues like mechanical or procedural biases during the draw.
Significance level 0.05
The significance level, often denoted as \(\alpha\), is a threshold for determining whether to reject the null hypothesis.
It's a way to measure the strength of the evidence against the null hypothesis.
In our raffle problem, we set the significance level at 0.05, a common choice for statistical tests.

Setting \(\alpha = 0.05\) means that we are willing to accept a 5% probability that we will mistakenly reject the null hypothesis when it is actually true.
This is called a Type I error.
This level balances the risk of being too lenient or too strict, making it a preferred choice in hypothesis testing for moderate assurance.
  • If the calculated chi-square statistic exceeds the critical value at this significance level, we reject the null hypothesis.
  • If it is below, we do not reject the hypothesis, suggesting adequate randomness in the draw.
It's important to consider this threshold carefully, as it impacts the decision-making process across various statistical tests.

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Most popular questions from this chapter

The feathers of the frizzle chicken come in three variations, or phenotypes - extreme, mild, and smooth. The genes \(F\), the dominant for frizzle, and \(f\), the recessive, interact in what is called incomplete dominance. If two hybrid (F, f) chickens are crossed, the ratio of extreme to normal to smooth should be \(1: 2: 1\). In such a hybridization experiment with ninety-three offspring, the results are given in the following table. Test the hypothesis of incomplete dominance at the \(.05\) level. $$ \begin{array}{lc} \hline \text { Phenotype } & \text { Number Observed } \\ \hline \text { Extreme frizzle } & 23 \\ \text { Mild frizzle } & 50 \\ \text { Normal } & 20 \\ \hline \end{array} $$

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The mean \((\mu)\) and standard deviation ( \(\sigma\) ) of pregnancy durations are 266 days and 16 days, respectively. Accepting those as the true parameter values, test whether the additional assumption that pregnancy durations are normally distributed is supported by the following list of seventy pregnancy durations reported by County General Hospital. Let \(\alpha=0.10\) be the level of significance. Use "220 \(\leq y<230, "{ }^{\prime \prime} 230 \leq y<240\), " and so on, as the classes. $$ \begin{array}{llllllllll} \hline 251 & 264 & 234 & 283 & 226 & 244 & 269 & 241 & 276 & 274 \\ 263 & 243 & 254 & 276 & 241 & 232 & 260 & 248 & 284 & 253 \\ 265 & 235 & 259 & 279 & 256 & 256 & 254 & 256 & 250 & 269 \\ 240 & 261 & 263 & 262 & 259 & 230 & 268 & 284 & 259 & 261 \\ 268 & 268 & 264 & 271 & 263 & 259 & 294 & 259 & 263 & 278 \\ 267 & 293 & 247 & 244 & 250 & 266 & 286 & 263 & 274 & 253 \\ 281 & 286 & 266 & 249 & 255 & 233 & 245 & 266 & 265 & 264 \\ \hline \end{array} $$

One hundred unordered samples of size 2 are drawn without replacement from an urn containing six red chips and four white chips. Test the adequacy of the hypergeometric model if zero whites were obtained 35 times; one white, 55 times; and two whites, 10 times. Use the \(0.10\) decision rule.

A sociologist is studying various aspects of the personal lives of preeminent nineteenth-century scholars. A total of one hundred twenty subjects in her sample had families consisting of two children. The distribution of the number of boys in those families is summarized in the following table. Can it be concluded that the number of boys in two-child families of preeminent scholars is binomially distributed? Let \(\alpha=0.05\). $$ \begin{array}{lrrr} \hline \text { Number of boys } & 0 & 1 & 2 \\ \text { Number of families } & 24 & 64 & 32 \\ \hline \end{array} $$

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