/*! 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 1 For Exercises 1 to 4, assume tha... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For Exercises 1 to 4, assume that birthdays are equally likely to occur on all possible days in any given year, so there are no seasonal variations or day of the week variations. Suppose you wanted to simulate the birthdays (month and day, not year) of the five children in one family. For each child, you tell the computer to choose a number from 1 to 366 . This covers all possibilities including February 29. a. Would you tell the computer to make all 366 choices equally likely? Explain. b. How would you assign the results to birthdays? For example, what number would correspond with a birthday of December \(31 ?\) c. Would it make sense to tell the computer to allow the same number to be chosen twice or not to allow that? Explain.

Short Answer

Expert verified
Simulate with equal likelihood for each day and allow repeats for shared birthdays.

Step by step solution

01

Understanding the Distribution

Yes, it would be appropriate to tell the computer to make all 366 choices equally likely. Since we do not want to consider the variations in the distribution of birthdays, each day of the year, including February 29, has an equal probability of being selected. Assuming equal probabilities simplifies the simulation to a uniform distribution of days across the year.
02

Assigning Numbers to Dates

To assign numbers to birthdays, we need to map the sequential numbers correctly to the days within each month. - For example, January 1 would correspond to 1, January 2 to 2, ..., and so forth. - December 31 would be day 365 in a non-leap year and day 366 in a leap year. Since we are considering 366 days, December 31 corresponds to 366.
03

Allowing Duplicates in Selection

Yes, it makes sense to allow the same number to be chosen more than once by the computer. Allowing duplicates reflects the real-world scenario where multiple children in a family might share the same birthday.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Uniform Distribution
When simulating the birthdays of five children in a family, it is essential to use a uniform distribution. This concept means that every day of the year, from January 1 to December 31, including the occasional February 29 of a leap year, has an equal chance of being selected.
Applying a uniform distribution in this context ensures that we are not influenced by any actual biases in birthday occurrences, such as more births happening on weekends or certain months.
By assuming all 366 days in a year are equally likely, it allows our model to provide an evenly spread probability across all potential birthdays. This consistency is critical for an accurate simulation.
Random Number Assignment
In probability simulations like the one we are describing, each possible outcome (in this case, each birthday) is assigned a unique number. The range of these numbers must match the range of outcomes—so, numbers from 1 to 366, representing all days in a year.
For example:
  • January 1 can be assigned the number 1.
  • February 1 would be number 32.
  • March 1 as 60.
  • December 31, in our 366-day model, would be number 366.
By linking each day of the year to a unique number, we create a straightforward way for the computer to randomly select a day as if it were rolling a perfectly balanced 366-sided die.
This approach makes the assignment process intuitive and critically reinforces the principle of equal likelihood.
Birthday Paradox
The birthday paradox is a fascinating statistical phenomenon that arises even in cases of a small number of individuals. It demonstrates that the probability of two people sharing a birthday in a group is surprisingly high.
For instance, with just 23 people, the odds of at least two sharing a birthday exceed 50%.
In our exercise, when considering a small family with five children, allowing for duplicate birthdays (meaning the same day can be randomly assigned more than once) accurately reflects this possibility.
This scenario not only enhances the realism of our simulation but also teaches us about probability theory's counterintuitive nature, as the repetition of birthdays can occur more frequently than one might expect in a limited sample size.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose that the chi-square statistic for a chisquare test on a table with 2 rows and 2 columns was computed to be \(5.3 .\) A simulation was run with 10,000 simulated samples, and 217 of them resulted in chi-square statistics of 5.3 or larger. What is the estimated \(p\) -value for the test?

Suppose a teacher observed a correlation of 0.38 between age and number of words children could define on a vocabulary test, for a group of nine children aged 7 to \(15 .\) The teacher wanted to confirm that there was a correlation between age and vocabulary test scores for the population of children in this age range. She performed a simulation with 1000 randomized orders, scrambling the test scores across the ages, similar to the simulation used in Example \(15.3 .\) She was surprised to find that \(29 \%\) of the simulated samples resulted in correlations of 0.38 or larger, even though there should not have been any correlation between ages and the scrambled test scores. a. What null and alternative hypotheses was the teacher testing? b. What is the estimated \(p\) -value for her test? c. Do these results confirm that there is no correlation between age and vocabulary scores for the population of children aged 7 to \(15 ?\) Explain. (Hint: Remember the legitimate conclusions for testing hypotheses.) d. Would a simulation with 10,000 randomized orders be much more likely, much less likely, or about equally likely to enable the teacher to reject the null hypothesis, compared to the simulation performed with 1000 randomized orders? e. The teacher plans to repeat the experiment. What could she do differently to improve the chance of rejecting the null hypothesis?

As a promotion, a cereal brand is offering a prize in each box, and there are four possible prizes. You would like to collect all four prizes, but you only plan to buy six boxes of the cereal before the promotion ends. Assume you have a random number generator that weights all numbers equally in a range you provide. a. Explain how you could simulate the prizes found in one set of six boxes of cereal. b. Explain how you could use simulation to estimate the probability of obtaining all four prizes in six boxes of cereal. Assume that all four prizes are equally likely for any given box, and the choice of prize is independent from one box to the next.

Would probabilities estimated using simulation be considered to be relative- frequency probabilities or personal probabilities? Explain.

Suppose that a randomization distribution resulting from the simulation of a chi-square test had \(7 \%\) of the values at or above the chi-square statistic observed for the real sample. a. What is the estimated \(p\) -value for the test? b. What decision would you make for the test, using a level of \(0.05 ?\) c. What decision would you make for the test, using a level of \(0.10 ?\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.