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Chapter 7: Problem 29

The candy machine Suppose a large candy machine has \(45 \%\) orange candies. Imagine taking an SRS of 25 candies from the machine and observing the sample proportion \(\hat{p}\) of orange candies. (a) What is the mean of the sampling distribution of \(\hat{p}\) ? Why? (b) Find the standard deviation of the sampling distribution of \(\hat{p}\). Check to see if the \(10 \%\) condition is met. (c) Is the sampling distribution of \(\hat{p}\) approximately Normal? Check to see if the Large Counts condition is met. (d) If the sample size were 100 rather than 25 , how would this change the sampling distribution of \(\hat{p} ?\)

Short Answer

Expert verified
(a) Mean is 0.45; (b) Standard deviation is 0.099; (c) Distribution is approximately Normal; (d) Larger sample size decreases variability.

Step by step solution

01

Calculate the Mean of the Sampling Distribution

The mean of the sampling distribution of the sample proportion \( \hat{p} \) is equal to the population proportion \( p \). Given: \( p = 0.45 \).Thus, the mean of the sampling distribution of \( \hat{p} \) is \( 0.45 \). This is because the expected proportion of orange candies in any sample is the same as the proportion in the whole population.
02

Calculate the Standard Deviation

The standard deviation of the sampling distribution of \( \hat{p} \) is given by the formula:\[ \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \]Substituting the given values:\[ \sigma_{\hat{p}} = \sqrt{\frac{0.45 \times 0.55}{25}} \approx 0.099 \]Next, check the 10% condition:The sample size \( n = 25 \) should be less than 10% of the population. Assuming a large candy machine, this condition is typically met.
03

Check for Normality Using the Large Counts Condition

The sampling distribution is approximately Normal if both \( np \geq 10 \) and \( n(1-p) \geq 10 \).Calculate:- \( np = 25 \times 0.45 = 11.25 \)- \( n(1-p) = 25 \times 0.55 = 13.75 \)Both values are greater than 10, so the large counts condition is satisfied. Therefore, the sampling distribution of \( \hat{p} \) is approximately Normal.
04

Impact of Increasing Sample Size to 100

If the sample size was increased to 100, the standard deviation of the sampling distribution would decrease. Calculate the new standard deviation:\[ \sigma_{\hat{p}} = \sqrt{\frac{0.45 \times 0.55}{100}} \approx 0.049 \]With a larger sample size, the distribution would be more concentrated around the mean and remain approximately Normal, but with reduced variability.

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Key Concepts

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

Sample Proportion
When analyzing the characteristics of a group within a population, the sample proportion \( \hat{p} \) becomes important. It's a fraction or percentage that represents the part of a sample exhibiting a particular trait. For instance, in this candy machine scenario, the sample proportion measures the share of orange candies in a batch of 25 candies.
Let's break it down:
  • Population Proportion (\( p \)): The actual proportion within the entire population, which is 45% or 0.45 for orange candies in this case.
  • Sample Proportion (\( \hat{p} \)): The proportion you find in your sample, with the hope it resembles the population proportion. If you pick 25 candies, and 12 are orange, then the sample proportion is 12/25.

Why is this important? Because it helps you understand if your small group is a good representation of the whole. By knowing your sample proportion, you can make educated guesses about the broader population.
Standard Deviation
The concept of standard deviation is crucial when discussing the variation in sample proportions. It's a measure of how much individual sample proportions \( \hat{p} \) differ from the mean of the sampling distribution. For the sampling distribution of a sample proportion, the formula is: \[ \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \]
Where:
  • \( p \): Population proportion
  • \( n \): Sample size

In our case, substituting 0.45 for \( p \,\) and 25 for \( n \,\), gives a standard deviation of 0.099. This illustrates the spread or variability of orange candy proportions you might find in different samples of 25 candies.

The smaller this number, the closer your various sample proportions will be to the mean of the population proportion. It provides insight into the reliability of your sample's representation of the total population. Additionally, the 10% condition ensures that the sample size is adequately small relative to the population, keeping the standard deviation calculation valid.
Normal Approximation
To determine if the sample proportion's distribution is approximately normal, we use the Normal Approximation. One essential test is the Large Counts condition, often used with large sample sizes.
This condition states that both \( np \) and \( n(1-p) \) should be at least 10 for normal approximation validity.
  • \( np = 25 \times 0.45 = 11.25 \,\) which is greater than 10
  • \( n(1-p) = 25 \times 0.55 = 13.75 \,\) also greater than 10

These results satisfy the large counts condition, letting us approximate the sampling distribution of \( \hat{p} \) by a normal distribution. Normal approximation aids in predicting outcomes from sample data efficiently. It helps make inferences about the whole population, particularly useful when dealing with larger data sets or making various statistical calculations.

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Most popular questions from this chapter

Making auto parts A grinding machine in an auto parts plant prepares axles with a target diameter \(\mu=40.125\) millimeters \((\mathrm{mm})\). The machine has some variability, so the standard deviation of the diameters is \(\sigma=0.002 \mathrm{~mm} .\) The machine operator inspects a random sample of 4 axles each hour for quality control purposes and records the sample mean diameter \(\bar{x}\). Assuming that the process is working properly, what are the mean and standard deviation of the sampling distribution of \(\bar{x} ?\) Explain.

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