#8 **Activity 17: One Sample Inference for Proportions****Objective:** The objective of this activity is to gain experience with hypothesis testing for a proportion. We will do this by studying the classic experiment proposed by French naturalist Buffon in 1733. This experiment is popularly known as "Buffon’s Needle."**Topics covered:**1. One sample hypothesis test for a population proportion2. One sample confidence interval for a population proportion3. Duality between confidence intervals and hypothesis testingFirst, we put Buffon’s original question from 1733 in our context. We would like to know what is the probability that a standard 2.5 inch toothpick will fall on a line when the lines are parallel. In 1777, Buffon showed that the probability is $ p = \frac{2}{\pi} \approx 0.636619 $ when the lines are also 2.5 inches apart.1. Suppose we don’t believe Buffon’s proof. That is, we think that the probability of landing on the line is most definitely not $ p = \frac{2}{\pi} $. State the hypotheses for our research claim. \[ H_0: p = 0.636619 \] \[ H_A: p \neq 0.636619 \]2. Next, we collect data to test our research question. Remove the last page of this activity (Page 75) with parallel lines that are 2.5 inches apart. Drop a standard 2.5 inch toothpick on the page. Record whether it falls on a line or not. Repeat the process ten times and record your answers below using - 0 = “not on line” and 1 = “landed on line”. | Drop Result | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | Total | |-------------|---|---|---|---|---|---|---|---|---|----|-------|3. To get a better estimate, combine your data with the class and record the values below. - Total number of tosses: $ n = 10 \times \text{number of students} = 250 $ - Total number landing on the line: $ x = \text ### Page 74 Transcript for Educational Website---**Exercise Details:****5.** Calculate the p-value of the test.**6.** Interpret the p-value.**7.** State the conclusion of the test in terms of the problem.**8.** Based on the hypothesis test alone, do you expect that the parameter \( p = \frac{2}{\pi} \approx 0.636619 $ would be in a 95% confidence interval for $ p $? Explain without computing the interval.**9.** This experiment is often used to estimate the value of $ \pi $.\[ p = \frac{2}{\pi} \]So, with algebra we get:\[ \pi = \frac{2}{p} \]Plug the class estimate $ \hat{p} $ into the second equation. Do you get a good estimate for $ \pi \approx 3.14159265359 $?Calculation:\[ \pi = \frac{2}{0.596} = 3.3557047 \]**Note:** The estimate is fairly close, but is still off, so this would not be a very good estimate.---This text is designed to guide students through understanding hypothesis testing, p-value calculation, and estimation processes, particularly in relation to π (pi).

Name: #8 **Activity 17: One Sample Inference for Proportions****Objective:**
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Description: #8 **Activity 17: One Sample Inference for Proportions****Objective:** The objective of this activity is to gain experience with hypothesis testing for a proportion. We will do this by studying the classic experiment proposed by French naturalist Buf

Here it is asked that the parameter p = 2/π ≈0.636619 would be in a 95% confidence interval for p. Given information : x = 149. n = 250. Since you've not equal sign in alternative hypotheses it is two tailed test. It seems that you've calculated z score = -1.334. Now for this z score we have to find p-value for two tailed test. We can easily find this by using MS-Excel. Note that ,MS-Excel gives P-value for left tailed test. The function to find P-value in excel is as follows; =NORMSDIST ( Test statistic , z). Replace z with their value to get the P-value for left tailed test. It will look like , =NORMSDIST (-1.334) now hit on enter to get the P-value. Thus , P-value for left tailed test = 0.091102. Next , find P-value for two tailed test = 2*P-value for left tailed test. = 2 *0.091102 = 0.182204. The P-value decision rule : If P-value ≤ significance level , then we reject null and supports alternative . If P-value > significance level , then we fail to reject null that is we support null and reject alternative. In our example P-value = 0.182204 and there is no value given for significance level so we will consider this as 0.05. Here P-value is greater than 0.05 so we fail to reject the null hypotheses and it simply means that we are in favor of accepting null hypotheses. Decision : Fail to reject the null hypotheses. Since the null hypotheses p = 2/π ≈0.636619 is not rejected , we expect that the parameter p = 2/π ≈0.636619 would be in a confidence interval for p.

Answered: 8. Based on the hypothesis test alone,…

Math

Statistics

8. Based on the hypothesis test alone, do you expect that the parameter p = 2 2 0.636619 would be in a 95% confidence interval for p? Explain without computing the interval.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

**Activity 17: One Sample Inference for Proportions**

**Objective:** The objective of this activity is to gain experience with hypothesis testing for a proportion. We will do this by studying the classic experiment proposed by French naturalist Buffon in 1733. This experiment is popularly known as "Buffon’s Needle."

**Topics covered:**

1. One sample hypothesis test for a population proportion
2. One sample confidence interval for a population proportion
3. Duality between confidence intervals and hypothesis testing

First, we put Buffon’s original question from 1733 in our context. We would like to know what is the probability that a standard 2.5 inch toothpick will fall on a line when the lines are parallel. In 1777, Buffon showed that the probability is $ p = \frac{2}{\pi} \approx 0.636619 $ when the lines are also 2.5 inches apart.

1. Suppose we don’t believe Buffon’s proof. That is, we think that the probability of landing on the line is most definitely not $ p = \frac{2}{\pi} $. State the hypotheses for our research claim.

\[
H_0: p = 0.636619
\]
\[
H_A: p \neq 0.636619
\]

2. Next, we collect data to test our research question. Remove the last page of this activity (Page 75) with parallel lines that are 2.5 inches apart. Drop a standard 2.5 inch toothpick on the page. Record whether it falls on a line or not. Repeat the process ten times and record your answers below using
- 0 = “not on line” and 1 = “landed on line”.

| Drop Result | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | Total |
|-------------|---|---|---|---|---|---|---|---|---|----|-------|

3. To get a better estimate, combine your data with the class and record the values below.

- Total number of tosses: $ n = 10 \times \text{number of students} = 250 $
- Total number landing on the line: \( x = \text

### Page 74 Transcript for Educational Website

---

**Exercise Details:**

**5.** Calculate the p-value of the test.

**6.** Interpret the p-value.

**7.** State the conclusion of the test in terms of the problem.

**8.** Based on the hypothesis test alone, do you expect that the parameter $ p = \frac{2}{\pi} \approx 0.636619 $ would be in a 95% confidence interval for $ p $? Explain without computing the interval.

**9.** This experiment is often used to estimate the value of $ \pi $.
\[ p = \frac{2}{\pi} \]
So, with algebra we get:
\[ \pi = \frac{2}{p} \]

Plug the class estimate $ \hat{p} $ into the second equation. Do you get a good estimate for $ \pi \approx 3.14159265359 $?

Calculation:
\[ \pi = \frac{2}{0.596} = 3.3557047 \]

**Note:** The estimate is fairly close, but is still off, so this would not be a very good estimate.

---

This text is designed to guide students through understanding hypothesis testing, p-value calculation, and estimation processes, particularly in relation to π (pi).

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