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Indicate in which of the following cases the central limit theorem will apply to describe the sampling distribution of the sample proportion. a. \(n=400\) and \(p=.28\) b. \(n=80\) and \(p=.05\) c. \(n=60\) and \(p=.12\) d. \(n=100\) and \(p=.035\)

Short Answer

Expert verified
The Central Limit Theorem applies in cases a and c.

Step by step solution

01

Checking case a

In this case, given \(n=400\) and \(p=.28\). So, we calculate np and n(1-p) for these values: np = \(400 * .28 = 112\) and n(1-p) = \(400 * (1-.28) = 288\). Both values are larger than 5, so the CLT applies in this case.
02

Checking case b

In this case, given \(n=80\) and \(p=.05\). We calculate np and n(1-p) for these values: np = \(80 * .05 = 4\) and n(1-p) = \(80 * (1-.05) = 76\). Although n(1-p) is larger than 5, np is not. So, the CLT does not apply in this case.
03

Checking case c

In this case, given \(n=60\) and \(p=.12\). We calculate np and n(1-p) for these values: np = \(60 * .12 = 7.2\) and n(1-p) = \(60 * (1-.12) = 52.8\). Both values are larger than 5, so the CLT applies in this case.
04

Checking case d

In this case, given \(n=100\) and \(p=.035\). We calculate np and n(1-p) for these values: np = \(100 * .035 = 3.5\) and n(1-p) = \(100 * (1-.035) = 96.5\). Although n(1-p) is larger than 5, np is not. So, the CLT does not apply in this case.

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Key Concepts

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

Sampling Distribution
Sampling distribution refers to the probability distribution of a given statistic based on a random sample. It reflects how the statistic varies from one sample to another. Understanding sampling distribution is crucial when examining how the sample represents the entire population.

In the context of this exercise, we are dealing with the sampling distribution of the sample proportion. This tells us how the proportion of a trait in our sample (like the percentage of people who prefer coffee) will vary if we were to take many samples from the same population.

The central limit theorem (CLT) helps us here because it states that the sampling distribution of the sample proportion becomes approximately normal when the sample size is large enough, regardless of the original distribution of the population.
Sample Proportion
The sample proportion is a key concept in statistics that represents the fraction of items in a sample that have a certain characteristic. It is denoted by \( \hat{p} \) and is calculated by dividing the number of successes (items with the characteristic) by the total number of items in the sample.

For example, if we surveyed 100 people and found that 25 prefer tea over coffee, our sample proportion \( \hat{p} \) would be \( 0.25 \). This value is a point estimate of the true population proportion \( p \), and it varies from sample to sample.

In the current exercise, the aim is to determine whether the central limit theorem applies to the sample proportion, which would imply it has a normal distribution with a mean \( p \) and a standard deviation \( \sqrt{\frac{p(1-p)}{n}} \).
Conditions for CLT
The central limit theorem is a powerful statistical tool that gives us insights into the behavior of sample proportions. However, for it to be applicable, certain conditions must be met.

Firstly, the samples should be independent. This generally means that the sample size is less than 10% of the population; otherwise, sampling it could influence the population's makeup too much.

Secondly, the sample size should be sufficiently large. When dealing with proportions, a common rule of thumb is that both the expected number of successes \( np \) and the number of failures \( n(1-p) \) in the sample must be at least 5. This ensures that the sampling distribution is roughly normal.
Considering these conditions helps us determine cases where the CLT can be safely applied to describe the sampling distribution.
np and n(1-p) Criteria
The criteria \( np \) and \( n(1-p) \) are essential conditions that need to be met for the CLT to apply to sample proportions. They ensure that there are enough successes and failures to approximate a normal distribution.

\( np \) is the expected number of successful outcomes (or the cases that meet our criterion of interest) in the sample. Similarly, \( n(1-p) \) is the expected number of failures (or the cases that do not meet our criterion).

In this exercise, we calculated these for different scenarios to check if both \( np \) and \( n(1-p) \) are greater than or equal to 5. If they are, the sampling distribution of the sample proportion can be assumed to be approximately normal, according to the CLT. If not, the sample size may not be large enough to rely on the normal distribution approximation.

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