91Ó°ÊÓ

Ron flips a coin \(n_{1}\) times and Freda flips a coin \(n_{2}\) times. We can assume all coin flips are fair: The coin has an equal chance of landing heads or tails. In each of the following cases, state whether inference for a difference in proportions is appropriate using the methods of this section. If so, give the mean and standard error for the distribution of the difference in proportions \(\left(\hat{p}_{1}-\hat{p}_{2}\right)\) and state whether the normal approximation is appropriate. (a) Let \(\hat{p}_{1}\) be the proportion of Ron's flips that land heads and \(\hat{p}_{2}\) be the proportion of Freda's flips that land heads; \(n_{1}=100\) and \(n_{2}=50\). (b) Let \(\hat{p}_{1}\) be the proportion of Ron's flips that land heads and \(\hat{p}_{2}\) be the proportion of Ron's flips that land tails; \(n_{1}=100\). (c) Let \(\hat{p}_{1}\) be the proportion of Ron's flips that land heads and \(\hat{p}_{2}\) be the proportion of Freda's flips that land tails; \(n_{1}=200\) and \(n_{2}=200\). (d) Let \(\hat{p}_{1}\) be the proportion of Ron's flips that land tails and \(\hat{p}_{2}\) be the proportion of Freda's flips that land tails; \(n_{1}=5\) and \(n_{2}=10\).

Step by step solution

01

- Analyzing case (a)

For this case, since the coin is fair, we have \(p_{1} = p_{2} = 0.5\), \(n_{1} = 100\) and \(n_{2} = 50\). First, we check if we may use Normal approximation by ensuring \(n_{1}p_{1}, n_{1}(1 - p_{1}), n_{2}p_{2}\), and \(n_{2}(1 - p_{2})\) are greater than 5, which is true since they each yield 50 and 25 respectively. This translates to having a normal approximation. The mean \(\mu_{\hat{p}_{1} - \hat{p}_{2}} = p_{1} - p_{2} = 0\) and the standard error \(\sigma_{\hat{p}_{1} - \hat{p}_{2}} = \sqrt{\frac{{p_{1}(1 - p_{1})}}{{n_{1}} } + \frac{{p_{2}(1 - p_{2})}}{{n_{2}}}} = \sqrt{\frac{{0.5*0.5}}{100} + \frac{{0.5*0.5}}{50}} = 0.0816\).

02

- Analyzing case (b)

In this case, we have \(p_{1} = 0.5\) and \(p_{2} = 0.5\), \(n_{1} = 100\). The mean \(\mu_{\hat{p}_{1} - \hat{p}_{2}} = p_{1} - p_{2} = 0\). Because this is not a case of two independent proportions but rather complement proportions, we do not calculate the standard error in this case nor check for normal approximation applicability.

03

- Analyzing case (c)

Given that we have \(p_{1} = p_{2} = 0.5\), \(n_{1} = 200\) and \(n_{2} = 200\), the mean \(\mu_{\hat{p}_{1} - \hat{p}_{2}} = p_{1} - p_{2} = 0\). We check if we may use Normal approximation by ensuring \(n_{1}p_{1}, n_{1}(1 - p_{1}), n_{2}p_{2}\), and \(n_{2}(1 - p_{2})\) are greater than 5, which they all are since they each yield 100. As such, a normal approximation is applicable. The standard error \(\sigma_{\hat{p}_{1} - \hat{p}_{2}} = \sqrt{\frac{{p_{1}(1 - p_{1})}}{{n_{1}} } + \frac{{p_{2}(1 - p_{2})}}{{n_{2}}}} = \sqrt{\frac{{0.5*0.5}}{200} + \frac{{0.5*0.5}}{200}} = 0.050\).

04

- Analyzing case (d)

In this case, we have \(p_{1} = p_{2} = 0.5\), \(n_{1} = 5\) and \(n_{2} = 10\). The mean \(\mu_{\hat{p}_{1} - \hat{p}_{2}} = p_{1} - p_{2} = 0\). However, when we check if we may use Normal approximation by ensuring \(n_{1}p_{1}, n_{1}(1 - p_{1}), n_{2}p_{2}\), and \(n_{2}(1 - p_{2})\) are greater than 5, we find that this is false since they each yield 1.25 and 2.5 respectively. Hence, the normal approximation is not applicable in this case. As a result, we should not compute the standard error in this scenario.

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Key Concepts

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

Coin Tossing Experiment

Imagine flipping a fair coin several times. This simple activity, known as a _coin tossing experiment_, is often used to explore probabilities and statistical concepts. For each flip of the coin, there is an equal chance of landing heads or tails. Thus, the probability for both outcomes is 0.5. This makes the coin a fair one, meaning there's no bias or tilt towards heads or tails.

In our exercise, Ron and Freda each flip their own coins multiple times. Ron flips his coin 100 times, while Freda flips hers 50 times in one scenario. When dealing with these independent flipping experiments, we can analyze the probabilities and outcomes separately for each person. Several foundational statistical analyses often start here, using assumptions or actual flip results.

Key aspects to remember about coin tossing experiments include:

• Each flip is independent from the others, meaning the result of one flip does not influence the next.
• The probability of heads or tails is constant, provided the coin is fair.
• Larger sample sizes can help illustrate patterns or trends notoriously observed in probabilistic setups.

Inference for Difference in Proportions

When we talk about the _inference for the difference in proportions_, we're aiming to understand how two proportions compare statistically. In the context of our exercise, this involves comparing the proportion of heads that Ron gets with the proportion of heads or tails that Freda gets.

To calculate this, we define the proportion of heads for Ron as the sample proportion described as the number of heads over the total flips tools that help us to make statistical inferences. We often look at the difference between two proportions by computing the mean of the difference and its standard error. To do this, we use the following formula: The mean is \(\mu_{\hat{p}_{1} - \hat{p}_{2}} = p_{1} - p_{2}\) and the standard erroris given by \(\sigma_{\hat{p}_{1} - \hat{p}_{2}} = \sqrt{\frac{p_{1}(1-p_{1})}{n_{1}} + \frac{p_{2}(1-p_{2})}{n_{2}}}\).

This inference becomes essential when trying to determine whether the differences observed are just due to chance or reflect a real divergence in outcomes. Remember these points when conducting inference for differences in proportions:

• Identify whether the proportions you're comparing stem from independent or dependent samples.
• Ensure sample sizes are sufficient.
• Determine the relevance and validity of using statistical tools for these inferences.

Normal Approximation Validity

The idea of _normal approximation_ is pivotal in statistical analyses, especially when working with proportions. When conditions permit, we approximate the binomial distribution of coin flipping using a normal distribution. This allows us to employ the normal approximation for making inferences if sample sizes are sufficiently large.

In the context of our exercise, the validity of using a normal approximation depends on whether certain conditions are satisfied. Specifically, for the approximation to be valid, each of the numbers \(n_{1}p_{1}, n_{1}(1-p_{1}), n_{2}p_{2}, \) and \(n_{2}(1-p_{2})\) need to exceed 5. When these conditions are met, we confidently assume the distribution of the difference in proportions resembles a normal distribution.

Which criteria determine the validity for normal approximation?

• Sample size: Larger sample sizes generally improve the accuracy of the normal approximation.
• Expected values: Validate that expected numbers (given by \(n \times p\)) for both proportions and their complements are above the threshold of 5.
• Independence of flips: Each coin flip should be an independent event for calculations to hold.

Understanding these criteria helps in authenticating whether results fall within expected probability distributions, facilitating sound statistical inference.