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Is your random number generator working? Use your calculator's RandInt function to generate 200 digits from 0 to 9 and store them in a list. (a) State appropriate hypotheses for a chi-square test for goodness of fit to determine whether your calculator's random number generator gives each digit an equal chance to be generated. (b) Carry out a test at the \(\alpha=0.05\) significance level. For parts (c) and (d), assume that the students' random number generators are all working properly. (c) What is the probability that a student who does this exercise will make a Type I error? (d) Suppose that 25 students in an AP Statistics class independently do this exercise for homework. Find the probability that at least one of them makes a Type I error.

Short Answer

Expert verified
a) Uniform distribution vs alternative; b) Depends on chi-square value; c) 0.05; d) ≈0.722.

Step by step solution

01

State the Hypotheses

The null hypothesis \(H_0\) is that the calculator's random number generator provides each digit with an equal chance of occurring, i.e., the distribution of digits is uniform. The alternative hypothesis \(H_a\) is that the distribution is not equal, i.e., that at least one digit does not have a uniform chance of occurring.
02

Collect the Data

Use the calculator's RandInt function to generate 200 random digits between 0 and 9. Note each digit's frequency to form our observed data set.
03

Calculate Expected Frequencies

If each digit is equally likely, the expected frequency for each digit in 200 draws should be \( \frac{200}{10} = 20 \).
04

Compute Chi-Square Statistic

Use the formula \[ \chi^2 = \sum_{i=1}^{n} \frac{(O_i - E_i)^2}{E_i} \] where \( O_i \) is the observed frequency and \( E_i \) is the expected frequency for each digit. Calculate this for all 10 digits.
05

Determine the Critical Value

Look up the critical value for the chi-square distribution at \( \alpha = 0.05 \) for 9 degrees of freedom (one less than the number of categories, which is 10). The critical value is approximately 16.92.
06

Make a Decision

Compare the calculated chi-square statistic to the critical value. If the statistic exceeds 16.92, reject \( H_0 \). Otherwise, fail to reject \( H_0 \).
07

Determine Probability of Type I Error

The probability of committing a Type I error (rejecting the null hypothesis when it's true) is equal to the significance level, \( \alpha = 0.05 \).
08

Calculate Probability for 25 Students

The probability that at least one of the 25 students makes a Type I error is given by \[ 1 - (1 - \alpha)^{25} = 1 - 0.95^{25} \approx 0.722 \].

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Key Concepts

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

Random number generation
When we talk about random number generation, it simply means creating a sequence of numbers that lack any pattern. True random numbers are generated by physical processes, like radioactive decay or atmospheric noise. However, in most computational applications, we use a technique called pseudo-random number generation.

Computers and calculators generally can't generate truly random numbers, so they use algorithms to produce sequences of numbers that only appear random. The sequences are determined by a starting point called a seed, and a known formula. This makes them predictable to some extent, which is why they're termed 'pseudo-random'.

In the exercise context, you're using your calculator's RandInt function, which is an example of a pseudo-random number generator. This function is designed to give every possible digit an equal chance of appearing, which should result in a fair representation of each digit in any lengthy series of numbers.
Null hypothesis
The null hypothesis, commonly denoted as \( H_0 \), is a critical concept in statistics. It represents the assumption that there is no effect or no difference in the context of your test. It's a statement you aim to either reject or fail to reject based on your data analysis.

In our exercise, the null hypothesis is that the random number generator of your calculator is fair. This means each digit from 0 to 9 should appear with equal probability in the randomly generated numbers. If the data supports this hypothesis, you say you "fail to reject the null hypothesis".

If the test results indicate that certain digits appear more or less frequently than expected by chance alone, then you might reject the null hypothesis in favor of the alternative hypothesis. It's important to note that rejecting the null hypothesis doesn't prove the alternative hypothesis; it simply suggests evidence against \( H_0 \).
Type I error
A Type I error occurs when we mistakenly reject the null hypothesis when it is actually true. It's like a false alarm in statistical testing. Imagine you think there's something significant happening when it's simply due to random chance. This error can lead to incorrect conclusions, which are crucial to avoid especially in serious decisions like medical tests or quality control.

The probability of making a Type I error is denoted by the significance level \( \alpha \). In our case, it's set at 0.05, meaning there's a 5% risk of rejecting the null hypothesis when it is actually true.

Understanding Type I errors is vital in research and hypothesis testing, as it emphasizes caution in interpreting results. Reducing the likelihood of such errors often involves increasing the significance of findings or using larger sample sizes.
Significance level
The significance level, represented by \( \alpha \), is a threshold used to decide whether the evidence against the null hypothesis is strong enough. It's like setting a strictness level for deciding when results are statistically significant.

In statistical hypothesis testing, \( \alpha = 0.05 \) is commonly used, implying a 5% risk of making a Type I error. If the p-value (the probability of observing a test statistic as extreme as the one computed, assuming the null hypothesis is true) is less than 0.05, you reject the null hypothesis. If it's greater or equal, you fail to reject it.

This threshold balances the risk of Type I errors against the power of the test. Researchers choose significance levels based on the context and consequences of making such errors. A lower \( \alpha \) can reduce the chance of Type I errors but might increase Type II errors, where a true effect is missed.

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Most popular questions from this chapter

No chi-square A school's principal wants to know if students spend about the same amount of time on homework each night of the week. She asks a random sample of 50 students to keep track of their homework time for a week. The following table displays the average amount of time (in minutes) students reported per night: $$ \begin{array}{lccccccc} \hline \text { Night: } & \text { Sunday } & \text { Monday } & \text { Tuesday } & \text { Wednesday } & \text { Thursday } & \text { Friday } & \text { Saturday } \\ \text { Average } & 130 & 108 & 115 & 104 & 99 & 37 & 62 \\ \text { time: } & & & & & & & \\ \hline \end{array} $$ Explain carefully why it would not be appropriate to perform a chi-square test for goodness of fit using these data.

When analyzing survey results from a two-way table, the main distinction between a test for independence and a test for homogeneity is (a) how the degrees of freedom are calculated. (b) how the expected counts are calculated. (c) the number of samples obtained. (d) the number of rows in the two-way table. (e) the number of columns in the two-way table.

Multiple choice: Select the best answer for Exercises 19 to 22 Exercises 19 to 21 refer to the following setting. The manager of a high school cafeteria is planning to offer several new types of food for student lunches in the following school year. She wants to know if each type of food will be equally popular so she can start ordering supplies and making other plans. To find out, she selects a random sample of 100 students and asks them, "Which type of food do you prefer: Asian food, Mexican food, pizza, or hamburgers?" Here are her data: $$ \begin{array}{lcccc} \hline \text { Type of Food: } & \text { Asian } & \text { Mexican } & \text { Pizza } & \text { Hamburgers } \\ \text { Count: } & 18 & 22 & 39 & 21 \\ \hline \end{array} $$ An appropriate null hypothesis to test whether the food choices are equally popular is (a) \(H_{0}: \mu=25,\) where \(\mu=\) the mean number of students that prefer each type of food. (b) \(H_{0}: p=0.25,\) where \(p=\) the proportion of all students who prefer Asian food. (c) \(H_{0}: n_{A}=n_{M}=n_{P}=n_{H}=25,\) where \(n_{A}\) is the number of students in the school who would choose Asian food, and so on. (d) \(H_{0}: p_{A}=p_{M}=p_{P}=p_{H}=0.25,\) where \(p_{A}\) is the proportion of students in the school who would choose Asian food, and so on. (e) \(\quad H_{0}: \hat{p}_{\mathrm{A}}=\hat{p}_{M}=\hat{p}_{P}=\hat{p}_{H}=0.25,\) where \(\hat{p}_{\mathrm{A}}\) is the pro- portion of students in the sample who chose Asian food, and so on.

Refer to the following setting. The National Longitudinal Study of Adolescent Health interviewed a random sample of 4877 teens (grades 7 to 12 ). One question asked was "What do you think are the chances you will be married in the next ten years?" Here is a two-way table of the responses by gender: \({ }^{28}\) $$ \begin{array}{lcc} \hline & \text { Female } & \text { Male } \\ \text { Almost no chance } & 119 & 103 \\ \text { Some chance, but probably not } & 150 & 171 \\ \text { A 50-50 chance } & 447 & 512 \\ \text { A good chance } & 735 & 710 \\ \text { Almost certain } & 1174 & 756 \\ \hline \end{array} $$ The appropriate null hypothesis for performing a chi-square test is that (a) equal proportions of female and male teenagers are almost certain they will be married in 10 years. (b) there is no difference between the distributions of female and male teenagers' opinions about marriage in this sample. (c) there is no difference between the distributions of female and male teenagers' opinions about marriage in the population. (d) there is no association between gender and opinion about marriage in the sample. (e) there is no association between gender and opinion about marriage in the population.

Benford's law Faked numbers in tax returns, invoices, or expense account claims often display patterns that aren't present in legitimate records. Some patterns are obvious and easily avoided by a clever crook. Others are more subtle. It is a striking fact that the first digits of numbers in legitimate records often follow a model known as Benford's law. \({ }^{3}\) Call the first digit of a randomly chosen record \(X\) for short. Benford's law gives this probability model for \(X\) (note that a first digit can't be 0 ): $$ \begin{array}{lccccccccc} \hline \text { First digit: } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \text { Probability: } & 0.301 & 0.176 & 0.125 & 0.097 & 0.079 & 0.067 & 0.058 & 0.051 & 0.046 \\ \hline \end{array} $$ A forensic accountant who is familiar with Benford's law inspects a random sample of 250 invoices from a company that is accused of committing fraud. The table below displays the sample data. $$ \begin{array}{lcrrrrrrrr} \hline \text { First digit: } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \text { Count: } & 61 & 50 & 43 & 34 & 25 & 16 & 7 & 8 & 6 \\ \hline \end{array} $$ (a) Are these data inconsistent with Benford's law? Carry out an appropriate test at the \(\alpha=0.05\) level to support your answer. If you find a significant result, perform a follow-up analysis. (b) Describe a Type I error and a Type II error in this setting, and give a possible consequence of each. Which do you think is more serious?

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