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a. Is the normal approximation to the sampling distribution of \(\hat{p}\) appropriate when \(n=400\) and \(p=.8 ?\) b. Use the results of part a to find the probability that \(\hat{p}\) is greater than \(.83 .\) c. Use the results of part a to find the probability that \(\hat{p}\) lies between .76 and .84

Short Answer

Expert verified
Answer: Yes, the normal approximation is appropriate when \(n=400\) and \(p=0.8\). The probability that \(\hat{p}\) is greater than \(0.83\) is \(0.0668\), and the probability that \(\hat{p}\) lies between \(0.76\) and \(0.84\) is \(0.9544\).

Step by step solution

01

Check if normal approximation is appropriate

To check if the normal approximation is appropriate, we need to test the following conditions: 1. \(np \geq 10\) 2. \(n(1-p) \geq 10\) Here, \(n=400\) and \(p=0.8\). Let's test the conditions: 1. \(np = 400 \times 0.8 = 320 \geq 10\) 2. \(n(1-p) = 400 \times (1-0.8) = 80 \geq 10\) Both conditions are met, so the normal approximation to the sampling distribution of \(\hat{p}\) is appropriate when \(n=400\) and \(p=.8\).
02

Find the mean and standard deviation

Since we are dealing with the normal approximation to \(\hat{p}\), we can find the mean and standard deviation using the following formulas: Mean, \(\mu = p = 0.8\) Standard deviation, \(\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.8 \times 0.2}{400}}\) Calculate the standard deviation: \(\sigma = \sqrt{\frac{0.16}{400}} = 0.02\)
03

Find the probability that \(\hat{p}\) is greater than \(0.83\)

We will use the standard normal distribution (Z) table to find the probability that \(\hat{p}\) is greater than \(0.83\). First, we need to find the Z-score for \(0.83\) using the formula: \(Z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.83 - 0.8}{0.02}\) Calculate the Z-score: \(Z = \frac{0.03}{0.02} = 1.5\) Now, we can use the Z table to find the probability that \(\hat{p}\) is greater than \(0.83\). This is equivalent to finding the area under the curve to the right of \(Z=1.5\). From the standard normal table, we find that the area to the left of Z is equal to \(0.9332\). To find the area to the right, subtract it from 1: \(P(\hat{p}>0.83) = 1 - 0.9332 = 0.0668\) So, the probability that \(\hat{p}\) is greater than \(0.83\) is \(0.0668\).
04

Find the probability that \(\hat{p}\) lies between \(0.76\) and \(0.84\)

We will use the Z table to find the probability that \(\hat{p}\) lies between \(0.76\) and \(0.84\). First, find the Z-scores for \(0.76\) and \(0.84\): \(Z_1 = \frac{0.76 - 0.8}{0.02}\) \(Z_2 = \frac{0.84 - 0.8}{0.02}\) Calculate the Z-scores: \(Z_1 = \frac{-0.04}{0.02} = -2\) \(Z_2 = \frac{0.04}{0.02} = 2\) Now, we can use the Z table to find the area under the curve between \(Z_1=-2\) and \(Z_2=2\). From the standard normal table, we find that: \(P(Z<-2) = 0.0228\) \(P(Z<2) = 0.9772\) To find the probability that \(\hat{p}\) lies between \(0.76\) and \(0.84\), subtract the two probabilities: \(P(0.76<\hat{p}<0.84) = P(Z<2) - P(Z<-2) = 0.9772 - 0.0228 = 0.9544\) So, the probability that \(\hat{p}\) lies between \(0.76\) and \(0.84\) is \(0.9544\).

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Key Concepts

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

Sampling Distribution
In statistics, a sampling distribution refers to the probability distribution of a given statistic based on a random sample. When we focus on the sampling distribution of sample proportions, such as \(\hat{p}\), it's about understanding how different sample proportions might vary when you draw different samples from the same population.
The Central Limit Theorem states that, given a sufficiently large sample size, the sampling distribution of the sample mean will be normally distributed. This also applies to sample proportions, \(\hat{p}\), when specific conditions are met:
  • \(np \geq 10\)
  • \(n(1-p) \geq 10\)
These criteria ensure that the sample size is large enough for the normal approximation to be valid. If these conditions are satisfied, as seen in our problem with \(n=400\) and \(p=0.8\), the sampling distribution of the sample proportion can be approximated by a normal distribution. This leads us smoothly to the next concept—the Standard Normal Distribution.
Standard Normal Distribution
The standard normal distribution is a special normal distribution with a mean of 0 and a standard deviation of 1. It is often denoted as \(Z\). This distribution is used as a reference to compare and interpret values from other normal distributions.
For any normal distribution, you can transform a random variable into a standard normal variable using the Z-score. The Z-score is a way to measure the number of standard deviations a data point is from the mean. By converting to the standard normal distribution, we utilize the well-established properties of Z-tables to find probabilities easily.
For problems involving the normal approximation to a sampling distribution, once it is confirmed as appropriate, one can use the standard normal distribution to find probabilities related to the sample proportion \(\hat{p}\). The process involves calculating a Z-score, which simplifies our calculations substantially by placing them within an established framework.
Z-score
The Z-score is an essential concept when working with normal distributions. It allows you to find out how far a particular score is from the mean in terms of standard deviations.
To calculate the Z-score for a sample purpose, you use the formula:\[Z = \frac{\hat{p} - \mu}{\sigma}\]where \(\hat{p}\) is the sample proportion, \(\mu\) is the mean (population proportion), and \(\sigma\) is the standard deviation of the sampling distribution.
By transforming your problem into Z-scores, you can utilize the standard normal distribution. For example, finding the probability that \(\hat{p}\) is greater than 0.83 simply becomes identifying the area under the standard normal curve beyond your calculated Z value. Similarly, finding that \(\hat{p}\) lies between two values involves finding the area between two Z-scores. This makes complex probability calculations much more manageable by using pre-calculated Z-tables.

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Most popular questions from this chapter

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