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A requirement for using the normal distribution to approximate the \(\hat{p}\) distribution is that both \(n p > 5\) and \(n q > 5\) since we usually do not know \(p,\) we estimate \(p\) by \(\hat{p}\) and \(q\) by \(\hat{q}=1-\hat{p}\) Then we require that \(n \hat{p}>5\) and \(n \hat{q}>5 .\) Show that the conditions \(n \hat{p}>5\) and \(n \hat{q}>5\) are equivalent to the condition that out of \(n\) binomial trials, both the number of successes \(r\) and the number of failures \(n-r\) must exceed 5 Hint: In the inequality \(n \hat{p}>5,\) replace \(\hat{p}\) by \(r / n\) and solve for \(r .\) In the inequality \(n \hat{q}>5,\) replace \(\hat{q}\) by \((n-r) / n\) and solve for \(n-r\).

Short Answer

Expert verified
Both the number of successes and failures in the trials must exceed 5.

Step by step solution

01

Condition for Successes

We start with the inequality given in the problem, namely \( n \hat{p} > 5 \). By substituting \( \hat{p} = \frac{r}{n} \), we rewrite the inequality as follows:\[n \left( \frac{r}{n} \right) > 5\]This simplifies to:\[ r > 5 \]This implies that the number of successes, \( r \), must be greater than 5.
02

Condition for Failures

Next, we examine the inequality \( n \hat{q} > 5 \). Since \( \hat{q} = 1 - \hat{p} = \frac{n-r}{n} \), we can substitute this into the inequality:\[n \left( \frac{n-r}{n} \right) > 5\]Simplifying, we get:\[ n - r > 5 \]This inequality indicates that the number of failures, \( n - r \), must also be greater than 5.
03

Equivalence of Conditions

From both conditions derived:1. The condition \( r > 5 \) translates directly to needing more than 5 successes in \( n \) trials.2. The condition \( n - r > 5 \) translates to needing more than 5 failures in \( n \) trials.Therefore, these conditions are equivalent to ensuring that in the \( n \) binomial trials, both the number of successes \( r \) and the number of failures \( n-r \) are greater than 5.

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Key Concepts

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

Binomial Distribution
The binomial distribution is a probability distribution that describes scenarios where there are two possible outcomes. These outcomes can be thought of as 'success' and 'failure'. This distribution becomes particularly handy when we're dealing with a fixed number of trials, often labeled as "n", and we want to count the number of successes within those trials.

Think of flipping a coin. Each flip can either land as heads (success) or tails (failure). If you were to flip a coin 10 times, you might be interested in figuring out the probability of getting heads exactly 3 times. This is where the binomial distribution comes in! It expands to real-life scenarios beyond coin flips, such as quality control in factories or predicting election outcomes.

The binomial distribution helps us calculate probabilities using the formulae of the binomial coefficient and the probabilities of success and failure. In mathematical terms, it involves finding the probability of achieving exactly "r" successes in "n" trials, with the constant probability of success being "p" and failure "q = 1 - p". It's the building block for more advanced approximations, like using the normal distribution.
Sample Size Requirements
When approximating the binomial distribution with a normal distribution, having a suitable sample size is crucial. Why? Because it ensures the approximation is accurate. However, not all sample sizes are created equal!

Primarily, the sample size "n" should be large enough to satisfy two tireless inequalities:
  • \( n\hat{p} > 5 \)
  • \( n\hat{q} > 5 \)
Meeting these inequalities means that both the predicted number of successes \((n\hat{p})\) and failures \((n\hat{q})\) exceed 5.

Why is this important? Because when the sample or number of trials is just right, it shapes the binomial distribution more closely to the symmetrical, bell-shaped curve of the normal distribution. Otherwise, our approximation could be skewed, misleading, or simply incorrect! By ensuring a large sample size, we amplify the reliability of our statistical predictions. So, before diving deep into data analysis, checking these conditions saves a lot of future headaches!
Success and Failure Conditions
For a successful approximation of the binomial distribution to the normal distribution, both success and failure conditions must be examined. These conditions ensure that within our trials, we have a robust number of both successes and failures.

To establish these requirements for success:
Start with the successes condition: \( n\hat{p} > 5 \). By substituting, it means finding how many successes \(r\) are in reality greater than 5 \(( r > 5 )\).

Then, flip to consider the failures: \( n\hat{q} > 5 \). This can be rearranged to show that the number of failures, \((n-r)\), must also surpass 5 \(( n - r > 5 )\).

These conditions simplify the need for having more than 5 successes and more than 5 failures in our binomial trials. This dual-condition simultaneously ensures that our trials lead to a reliable normal approximation.

Always check these conditions before relying on normal distribution to predict outcomes from your binomial trials. They serve as the backbone for ensuring results are as accurate as they can be, making data insights all the more dependable! Buoyed by these success and failure checks, you're primed to see your statistical approximations truly shine.

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Most popular questions from this chapter

S. C. Jett is a professor of geography at the University of California, Davis. He and a colleague, V. E. Spencer, are experts on modern Navajo culture and geography. The following information is taken from their book Navajo Architecture: Forms, History, Distributions (University of Arizona Press). On the Navajo Reservation, a random sample of 210 permanent dwellings in the Fort Defiance region showed that 65 were traditional Navajo hogans. In the Indian Wells region, a random sample of 152 permanent dwellings showed that 18 were traditional hogans. Let \(p_{1}\) be the population proportion of all traditional hogans in the Fort Defiance region, and let \(p_{2}\) be the population proportion of all traditional hogans in the Indian Wells region. (a) Can a normal distribution be used to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(99 \%\) confidence interval for \(p_{1}-p_{2}\) (c) Examine the confidence interval and comment on its meaning. Does it include numbers that are all positive? all negative? mixed? What if it is hypothesized that Navajo who follow the traditional culture of their people tend to occupy hogans? Comment on the confidence interval for \(p_{1}-p_{2}\) in this context.

For a binomial experiment with \(r\) successes out of \(n\) trials, what value do we use as a point estimate for the probability of success \(p\) on a single trial?

In order to use a normal distribution to compute confidence intervals for \(p,\) what conditions on \(n p\) and \(n q\) need to be satisfied?

Independent random samples of professional football and basketball players gave the following information (References: Sports Encyclopedia of Pro Football and Official NBA Basketball Encyclopedia ). Note: These data are also available for download at the Companion Sites for this text. Assume that the weight distributions are mound-shaped and symmetric. (a) Use a calculator with mean and standard deviation keys to verify that \(\bar{x}_{1}=259.6, s_{1}=12.1, \bar{x}_{2}=205.8,\) and \(s_{2}=12.9\) (b) Let \(\mu_{1}\) be the population mean for \(x_{1}\) and let \(\mu_{2}\) be the population mean for \(x_{2} .\) Find a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) (c) Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the \(99 \%\) level of confidence, do professional football players tend to have a higher population mean weight than professional basketball players? (d) Which distribution (standard normal or Student's \(t\) ) did you use? Why?

Confidence Interval for \(\mu_{1}-\mu_{2}\) Consider two independent normal distributions. A random sample of size \(n_{1}=20\) from the first distribution showed \(\bar{x}_{1}=12\) and a random sample of size \(n_{2}=25\) from the second distribution showed \(\bar{x}_{2}=14\) (a)If \(\sigma_{1}\) and \(\sigma_{2}\) are known, what distribution does \(\bar{x}_{1}-\bar{x}_{2}\) follow? Explain. (b) Given \(\sigma_{1}=3\) and \(\sigma_{2}=4,\) find a \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) (c) Suppose \(\sigma_{1}\) and \(\sigma_{2}\) are both unknown, but from the random samples, you know \(s_{1}=3\) and \(s_{2}=4 .\) What distribution approximates the \(\bar{x}_{1}-\bar{x}_{2}\) distribution? What are the degrees of freedom? Explain. (d) With \(s_{1}=3\) and \(s_{2}=4,\) find a \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) (e) If you have an appropriate calculator or computer software, find a \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using degrees of freedom based on Satterthwaite's approximation. (f) Based on the confidence intervals you computed, can you be \(90 \%\) confident that \(\mu_{1}\) is smaller than \(\mu_{2} ?\) Explain.

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