/*! 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 16 A gambler complained about the d... [FREE SOLUTION] | 91影视

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A gambler complained about the dice. They seemed to be loaded! The dice were taken off the table and tested one at a time. One die was rolled 300 times and the following frequencies were recorded. $$ \begin{array}{l|rrrrrr} \hline \text { Outcome } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { Observed frequency } O & 62 & 45 & 63 & 32 & 47 & 51 \\ \hline \end{array} $$ Do these data indicate that the die is unbalanced? Use a \(1 \%\) level of significance. Hint: If the die is balanced, all outcomes should have the same expected frequency.

Short Answer

Expert verified
The data do not indicate the die is unbalanced, as there is insufficient evidence at the 1% level of significance.

Step by step solution

01

Formulate Hypotheses

Determine the null and alternative hypotheses. The null hypothesis \((H_0)\) is that the die is fair, meaning each face should appear with equal probability. The alternative hypothesis \((H_a)\) is that at least one face appears with a different probability than others, indicating the die is loaded.
02

Determine Expected Frequencies

If the die is fair, each face should appear with equal frequency. We divide the total number of rolls by the number of faces: \[ E = \frac{300}{6} = 50 \]Thus, the expected frequency for each outcome (1 through 6) is 50.
03

Calculate Chi-Square Statistic

Use the formula for the chi-square statistic: \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency. Calculate for each face:\[ \chi^2 = \frac{(62-50)^2}{50} + \frac{(45-50)^2}{50} + \frac{(63-50)^2}{50} + \frac{(32-50)^2}{50} + \frac{(47-50)^2}{50} + \frac{(51-50)^2}{50} \]\[ \chi^2 = \frac{144}{50} + \frac{25}{50} + \frac{169}{50} + \frac{324}{50} + \frac{9}{50} + \frac{1}{50} \]\[ \chi^2 = 2.88 + 0.5 + 3.38 + 6.48 + 0.18 + 0.02 \]\[ \chi^2 = 13.44 \]
04

Determine Degrees of Freedom and Critical Value

The degrees of freedom \((df)\) is calculated as the number of categories minus one: \[ df = 6 - 1 = 5 \]For a \(1\%\) level of significance and \(5\) degrees of freedom, the chi-square critical value is approximately \(15.09\) (from chi-square distribution tables).
05

Compare Chi-Square Statistic to Critical Value

Compare the calculated chi-square statistic with the critical value. Since \(13.44 < 15.09\), we do not reject the null hypothesis.
06

Conclusion

Since the calculated chi-square statistic is less than the critical value, we conclude that there is not enough evidence at the \(1\%\) level of significance to indicate that the die is unbalanced.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical method that helps us make decisions about population parameters based on sample data. It's like a judge deciding whether a statement about our sample is plausible or not. In our dice example, we have two hypotheses to consider:
  • The null hypothesis \(H_0\) assumes that the die is fair, meaning each outcome from 1 to 6 should happen with equal probability.
  • The alternative hypothesis \(H_a\) suggests that the die might be loaded, which means some outcomes appear more frequently than others.
We start by assuming the null hypothesis is true. We use the chi-square test to determine if our observations provide enough evidence to reject the null hypothesis. If the calculated chi-square statistic exceeds a critical value (based on our chosen significance level), we reject \(H_0\) and suggest \(H_a\) might be true. But if it doesn鈥檛 exceed the critical value, we do not reject \(H_0\). It鈥檚 important to note this doesn鈥檛 prove the null hypothesis is true, it just indicates a lack of evidence to reject it.
Degrees of Freedom
Degrees of freedom (df) refer to the number of values in a calculation that are free to vary. Think of it like having a fixed set of scores on a test; if you know all the scores but one, you can figure out the last score because it is dependent on the total. In the context of our chi-square test for the dice, the degrees of freedom help us determine the critical value from the chi-square distribution table. The formula for degrees of freedom in a chi-square test is:\[ df = \ ext{number of categories} - 1 \]For our six-sided die, there are six possible outcomes, so:\[ df = 6 - 1 = 5 \]These degrees of freedom help us understand how to interpret the calculated chi-square statistic and whether our observed differences could happen by random chance or indicate something more significant.
Observed Frequency
Observed frequency is simply the actual counts of how often each outcome appeared in our experiment with the dice. For the gambler's complaint, the dice outcomes were documented after rolling 300 times, and we observed:
  • Outcome 1: 62
  • Outcome 2: 45
  • Outcome 3: 63
  • Outcome 4: 32
  • Outcome 5: 47
  • Outcome 6: 51
These numbers represent our real, tangible results 鈥 the actual occurrences that we recorded. The goal is to compare these numbers to what we expected (if the die were fair) to determine if there's a deviation that suggests the die might be biased. Observed frequencies are crucial in calculating the chi-square statistic, as we contrast them with expected frequencies to see if the differences are statistically significant.
Expected Frequency
Expected frequency represents the outcome numbers we would anticipate for each category if the null hypothesis were true. In our example, assuming a fair die, each side should have an equal likelihood of appearing. Since the die is tossed 300 times and there are six sides, each outcome should ideally appear fifty times. This is calculated as:\[ E = \frac{300}{6} = 50 \]This value is critical because it's what we consider 鈥榥ormal鈥 under our assumption of fairness. By comparing this expected frequency with our observed frequencies, we can assess discrepancies. This comparison is the essence of the chi-square test, where we ask, 鈥淒o these observed differences arise by chance, or is the die possibly unfair?鈥 With the chi-square statistic formula, we weigh these differences mathematically to make informed conclusions on hypothesis testing.

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Most popular questions from this chapter

The following problem is based on information taken from Academe, Bulletin of the American Association of University Professors. Let \(x\) represent the average annual salary of college and university professors (in thousands of dollars) in the United States. For all colleges and universities in the United States, the population variance of \(x\) is approximately \(\sigma^{2}=47.1\). However, a random sample of 15 colleges and universities in Kansas showed that \(x\) has a sample variance \(s^{2}=83.2 .\) Use a \(5 \%\) level of significance to test the claim that the variance for colleges and universities in Kansas is greater than 47.1. Find a \(95 \%\) confidence interval for the population variance.

In general, is the \(F\) distribution symmetrical? Can values of \(F\) be negative?

Academe, Bulletin of the American Association of University Professors (Vol. 83, No. 2 ) presents results of salary surveys (average salary) by rank of the faculty member (professor, associate, assistant, instructor) and by type of institution (public, private). List the factors and the number of levels of each factor. How many cells are there in the data table?

For a chi-square goodness-of-fit test, how are the degrees of freedom computed?

A random sample of companies in electric utilities (I), financial services (II), and food processing (III) gave the following information regarding annual profits per employee (units in thousands of dollars). $$ \begin{array}{ccc} \text { I } & \text { II } & \text { III } \\ 49.1 & 55.6 & 39.0 \\ 43.4 & 25.0 & 37.3 \\ 32.9 & 41.3 & 10.8 \\ 27.8 & 29.9 & 32.5 \\ 38.3 & 39.5 & 15.8 \\ 36.1 & & 42.6 \\ 20.2 & & \end{array} $$ Shall we reject or not reject the claim that there is no difference in population mean annual profits per employee in each of the three types of companies? Use a \(1 \%\) level of significance.

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