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A die is tossed 180 times with the following results: \begin{tabular}{c|cccccc} \(x\) & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \(\boldsymbol{f}\) & 28 & 36 & 36 & 30 & 27 & 23 \end{tabular} Is this a halanced die? Use a 0.01 level of significance.

Short Answer

Expert verified
Yes, given the 0.01 level of significance, it can be argued that the die is balanced as the calculated chi-square value (5.83) is less than the critical chi-square value (15.086).

Step by step solution

01

Observe Frequency Distribution

The observed frequency distribution based on 180 tosses is given: Rolling a 1, 28 times. Rolling a 2, 36 times. Rolling a 3, 36 times. Rolling a 4, 30 times. Rolling a 5, 27 times. Rolling a 6, 23 times.
02

Calculate Expected Frequency

Assuming the die is balanced, the expected frequency for each outcome, 1 through 6, is \((1/6) * 180 = 30\).
03

Calculate the Chi-Square Value

The Chi-square value, denoted by \(\chi^2\), is calculated using the formula: \(\chi^2= \sum ((O-E)^2 /E)\), where \(O\) stands for observed frequency and \(E\) stands for expected frequency. Plugging in our observed and expected frequencies in this formula we get: \(\chi^2 = ((28-30)^2/30)+(36-30)^2/30)+((36-30)^2/30)+((30-30)^2/30)+((27-30)^2/30)+((23-30)^2/30) = 1.5+1.2+1.2+0+0.3+1.63 = 5.83\).
04

Determine the Critical Chi-Square Value

To find the critical chi-square value, we need the degrees of freedom, which is \(df=n-1\), where \(n\) is the number of outcomes. For a die roll, the degrees of freedom is \(6-1=5\). Using a chi-square table or calculator for a 0.01 level of significance and 5 degrees of freedom, the critical \(\chi^2\) value is 15.086.
05

Compare Calculated and Critical Chi-Square Values

Compare the calculated chi-square value with the critical chi-square value to decide whether to accept or reject the null hypothesis. Here, the calculated \(\chi^2\) (5.83) is less than the critical \(\chi^2\) (15.086), therefore null hypothesis is accepted and the die can be assumed to be balanced.

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Key Concepts

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

Understanding Hypothesis Testing
Hypothesis testing is a method used to make decisions based on data analysis. Imagine you have a balanced die, which means each number between 1 and 6 should appear equally often when rolled many times. In hypothesis testing, we start with a statement called the null hypothesis. For this die, the null hypothesis is that it is balanced.

The next step is to collect data by rolling the die multiple times, then analyze whether the results support this hypothesis. If the results seem unlikely compared to what we expect from a balanced die, we might reject the null hypothesis. However, if the results align with our expectations, we accept the null hypothesis. It’s a structured way to decide if what we observe matches what we expect based on chances.
Examining Frequency Distribution
Frequency distribution shows how often each outcome occurs in a data set. In our die rolling example, frequency distribution is how often each side of the die appears in 180 rolls.

Let's break it down:
  • Rolling a 1: 28 times
  • Rolling a 2: 36 times
  • Rolling a 3: 36 times
  • Rolling a 4: 30 times
  • Rolling a 5: 27 times
  • Rolling a 6: 23 times
To determine if the die is balanced, we compare these observed frequencies with the expected frequency, which is 30 times for each number if the die is fair. The difference between observed and expected values helps us understand any discrepancies.
Role of Degrees of Freedom
Degrees of freedom (df) refer to the number of values that are free to vary when calculating a statistic. In the context of the Chi-square test, it is calculated as the number of outcomes minus one.

For the die, there are six outcomes (1 through 6). Therefore, the degrees of freedom is calculated as:
  • df = Number of outcomes - 1 = 6 - 1 = 5
Degrees of freedom impact the critical value of our test, which determines if the observed difference is statistically significant. It helps us know how much wiggle room we have in our data analysis, contributing to making sound statistical judgments.
Significance of the Level of Significance
The level of significance is a threshold that determines how extreme the data must be to reject the null hypothesis. It is often denoted as alpha (α).

For our die roll test, we use a significance level of 0.01, meaning there's only a 1% risk of rejecting the true null hypothesis. It sets a rigorous standard for deciding if our observed outcomes are unusual enough to declare the die unbalanced.

In practical terms, after calculating the Chi-square value, we compare it to the critical value from the Chi-square distribution table based on our degrees of freedom and significance level. If the calculated value is greater, we reject the null hypothesis. Otherwise, we accept it, concluding the die is balanced at the 1% significance level.

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Most popular questions from this chapter

According to a dietary study, a high sodium intake may be related to ulcers, stomach cancer, and migraine headaches. The human requirement for salt is only 220 milligrams per day, which is surpassed in most single servings of ready- to-eat cereals. If a random sample of 20 similar servings of of certain cereal has a mean sodium content of 244 milligrams and a standard deviation of 24.5 milligrams, does this suggest at the 0.05 level of significance that the average sodium content for a single serving of such cereal is greater than 220 milligrams? Assume the distribution of sodium contents to be normal.

A coin is tossed 20 times, resulting in 5 heads. Is this sufficient evidence to reject the hypothesis that tile coin is balanced in favor of the alternative that heads occur less than \(50 \%\) of the time? Quote a Pvalue.

A fabric manufacturer believes that the proportion of orders for raw material arriving late is \(p=0.6\) If a random sample of 10 orders shows that 3 or fewer arrived late, the hypothesis that \(p=0.6\) should be rejected in favor of the alternative \(p<0.6\). Use the binomial distribution. (a) Find the probability of committing a type 1 error if the true proportion is \(p=0.6\) (b) Find the probability of committing a type II error for the alternatives \(p=0.3, p-0.4,\) and \(p=0.5\)

In a study to estimate the proportion of residents in a certain city and its suburbs who favor the construction of a nuclear power plant, it is found that 63 of 100 urban residents favor the construction while only 59 of 125 suburban residents are in favor. Is there a significant difference between the proportion of urban and suburban residents who favor construction of the nuclear plant? Make use of a P-value.

In a research report by Richard H. Weindruch of the UCLA Medical School, it is claimed that mice with an average life span of 32 months will live to be about 40 months old when \(40 \%\) of the calories in their food are replaced by vitamins and protein. Is there any reason to believe that \(\mu<40\) if 64 mice that are placed on this diet have an average life of 38 months with a standard deviation of 5.8 months? Use a P-value in your conclusion.

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