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In Exercises 7.1 to \(7.4,\) find the expected counts in each category using the given sample size and null hypothesis. $$ \begin{aligned} &\mathbf{7 . 3} \quad H_{0}: p_{A}=0.50, p_{B}=0.25, p_{C}=0.25 ;\\\ &n=200 \end{aligned} $$

Short Answer

Expert verified
The expected counts for categories A, B, and C are 100, 50, and 50, respectively.

Step by step solution

01

Find Expected Count for Category A

To find out the expected count for category A, multiply its proportion given by the null hypothesis (\(p_A=0.50\)) by the total size of the sample (\(n=200\)). Therefore, the expected count for category A is \(n*p_A = 200*0.50 = 100\)
02

Find Expected Count for Category B

To get the expected count for category B, multiply its proportion given by the null hypothesis (\(p_B=0.25\)) by the total size of the sample (\(n=200\)). Therefore, the expected count for category B is \(n*p_B = 200*0.25 = 50\)
03

Find Expected Count for Category C

To obtain the expected count for category C, multiply its proportion given by the null hypothesis (\(p_C=0.25\)) by the total size of the sample (\(n=200\)). Therefore, the expected count for category C is \(n*p_C = 200*0.25 = 50\)

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Key Concepts

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

Null Hypothesis
The null hypothesis is a key concept in statistics that posits a default assumption about a condition or scenario. When you conduct a statistical test, you start with the null hypothesis, denoted as - **Hâ‚€** This hypothesis assumes that there is no effect or no difference, and any observed effect is due to chance. For example, in your exercise, the null hypothesis states that the proportions of categories A, B, and C are 0.50, 0.25, and 0.25 respectively. This means, under the null hypothesis, you're assuming those percentages accurately describe the population distribution. - **Why is it important?** - It provides a baseline against which to compare your data - It is essential in determining statistical significance Understanding the null hypothesis helps you make informed decisions on whether the data provides enough evidence to reject this initial assumption. Rejection suggests that the observed data is statistically significant, pointing to an effect or difference that is likely not due to random chance.
Sample Size
Sample size refers to the number of observations or measurements taken from your population of interest. It's a crucial factor in statistical analysis as it impacts the power and accuracy of your hypothesis tests. - **Key Insights:** - A larger sample size can lead to more reliable and stable results - It affects the margin of error and confidence intervals In your exercise, a sample size of 200 was used. This means that data was collected from 200 instances or items from the population. Choosing an appropriate sample size depends on several factors including the expected effect size, as well as the level of variability among the population. - **Why does it matter?** - It influences the credibility of the findings - Larger sizes typically reduce sampling errors A suitable sample size ensures your experiment results are a good representation of the actual population, which is crucial for accurate predictions and conclusions.
Proportion
Proportion describes the fraction or percentage of a part compared to the whole. In statistics, it's commonly used to express the likelihood of an event occurring. - **Example from exercise:** - For category A, the proportion is 0.50 - This implies that half of the sample data is expected to come from category A The proportions for categories B and C are both 0.25, each predicting that a quarter of the data will fall into these categories. Understanding and calculating proportions help in determining expected counts, as seen in your exercise. - **Why focus on proportions?** - They offer a way to quantify and analyze categorical data - They help identify the expected distribution of data across different categories Recognizing correct proportions creates the foundational knowledge necessary for calculating expected frequencies in a population. This calculation helps in validating whether the observed data matches expected outcomes based on the assumptions of the null hypothesis.

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Most popular questions from this chapter

{ 7.29 Random } & \text { Digits in } & \text { Students' } & \text { Random }\end{array}\( Numbers? How well can people generate random numbers? A sample of students were asked to write down a "random" four-digit number. Responses from 150 students are stored in the file Digits. The data file has separate variables (RND1, RND2, \)R N D 3,\( and \)R N D 4\( ) containing the digits in each of the four positions. (a) If the numbers are randomly generated, we would expect the last digit to have an equal chance of being any of the 10 digits. Test \)H_{0}\( : \)p_{0}=p_{1}=p_{2}=\cdots=p_{9}=0.10\( using technology and the data in \)R N D 4\(. (b) Since students were asked to produce four-digit numbers, there are only nine possibilities for the first digit (zero is excluded). Use technology to test whether there is evidence in the values of \)R N D 1$ that the first digits are not being chosen by students at random.

Give a two-way table and specify a particular cell for that table. In each case find the expected count for that cell and the contribution to the chi- square statistic for that cell. \((\mathrm{B}, \mathrm{E})\) cell $$ \begin{array}{l|rrrr|r} \hline & \text { D } & \text { E } & \text { F } & \text { G } & \text { Total } \\ \hline \text { A } & 39 & 34 & 43 & 34 & 150 \\ \text { B } & 78 & 89 & 70 & 93 & 330 \\ \text { C } & 23 & 37 & 27 & 33 & 120 \\ \hline \text { Total } & 140 & 160 & 140 & 160 & 600 \\ \hline \end{array} $$

Examining Genetic Alleles in Fast-Twitch Muscles Exercise 7.24 discusses a study investigating the \(A C T N 3\) genotypes \(R R, R X,\) and \(X X .\) The same study also examines the \(A C T N 3\) genetic alleles \(R\) and \(X,\) also associated with fast-twitch muscles. Of the 436 people in this sample, 244 were classified \(R\) and 192 were classified \(X .\) Does the sample provide evidence that the two options are not equally likely? (a) Conduct the test using a chi-square goodnessof-fit test. Include all details of the test. (b) Conduct the test using a test for a proportion, using \(H_{0}: p=0.5\) where \(p\) represents the proportion of the population classified \(R .\) Include all details of the test. (c) Compare the p-values and conclusions of the two methods.

Exercises 7.9 to 7.12 give a null hypothesis for a goodness-of-fit test and a frequency table from a sample. For each table, find: (a) The expected count for the category labeled B. (b) The contribution to the sum of the chi-square statistic for the category labeled \(\mathrm{B}\). (c) The degrees of freedom for the chi-square distribution for that table. $$ \begin{aligned} &H_{0}: p_{a}=p_{b}=p_{c}=p_{d}=0.25\\\ &: \text { Some }\\\ &H_{a}:\\\ &p_{i} \neq 0.25\\\ &\begin{array}{lccc|c} \hline \mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{D} & \text { Total } \\ 120 & 148 & 105 & 127 & 500 \\ \hline \end{array} \end{aligned} $$

Gender and ACTN3 Genotype We see in the previous two exercises that sprinters are more likely to have allele \(R\) and genotype \(R R\) versions of the ACTN3 gene, which makes these versions associated with fast-twitch muscles. Is there an association between genotype and gender? Computer output is shown for this chi-square test, using the control group in the study. In each cell, the top number is the observed count, the middle number is the expected count, and the bottom number is the contribution to the chi-square statistic. What is the p-value? What is the conclusion of the test? Is gender associated with the likelihood of having a "sprinting gene"? \(\begin{array}{lrrrr} & \text { RR } & \text { RX } & \text { XX } & \text { Total } \\ \text { Male } & 40 & 73 & 21 & 134 \\ & 40.26 & 69.20 & 24.54 & \\\ & 0.002 & 0.208 & 0.509 & \\ \text { Female } & 88 & 147 & 57 & 292 \\ & 87.74 & 150.80 & 53.46 & \\ & 0.001 & 0.096 & 0.234 & \\ \text { Total } & 128 & 220 & 78 & 426\end{array}\) \(\mathrm{Chi}-\mathrm{Sq}=1.050, \mathrm{DF}=2, \mathrm{P}\) -Value \(=0.592\)

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