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People end up tossing \(12 \%\) of what they buy at the grocery store (Reader 's Digest, March, 2009 ). Assume this is the true population proportion and that you plan to take a sample survey of 540 grocery shoppers to further investigate their behavior. a. Show the sampling distribution of \(\bar{p},\) the proportion of groceries thrown out by your sample respondents. b. What is the probability that your survey will provide a sample proportion within ±.03 of the population proportion? c. What is the probability that your survey will provide a sample proportion within ±.015 of the population proportion?

Short Answer

Expert verified
a. Normal distribution centered around 0.12 with SE ≈ 0.0143. b. Probability ≈ 0.9642. c. Probability ≈ 0.7062.

Step by step solution

01

Define the Parameters

The true population proportion \( p \) is given as 0.12. The sample size \( n \) is 540.
02

Calculate the Standard Error of the Sampling Distribution

The standard error of the sample proportion \( \bar{p} \) is computed using the formula: \[ SE = \sqrt{\frac{p(1-p)}{n}} \] Substitute \( p = 0.12 \) and \( n = 540 \): \[ SE = \sqrt{\frac{0.12 \times 0.88}{540}} \approx 0.0143 \]
03

Determine the Z-Score for ±0.03

We calculate the Z-score to find the probability that the sample proportion is within ±0.03 of 0.12. Using the formula: \[ Z = \frac{\text{margin}}{SE} \] For a margin of 0.03: \[ Z = \frac{0.03}{0.0143} \approx 2.10 \]
04

Calculate the Probability for ±0.03

Using the standard normal distribution table, find the probability for \( Z = 2.10\). The cumulative probability for \( Z = 2.10\) is approximately 0.9821. The probability that the sample proportion is within ±0.03 is: \[ 2 \times (0.9821 - 0.5) = 0.9642 \]
05

Determine the Z-Score for ±0.015

Using the same method as before, for a margin of 0.015: \[ Z = \frac{0.015}{0.0143} \approx 1.05 \]
06

Calculate the Probability for ±0.015

Using the standard normal distribution table for \( Z = 1.05 \), the cumulative probability is approximately 0.8531. Thus, the probability that the sample proportion is within ±0.015 is: \[ 2 \times (0.8531 - 0.5) = 0.7062 \]

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Key Concepts

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

Population Proportion
The concept of population proportion is an important aspect of statistics. Simply put, it refers to the fraction or percentage of a population that exhibits a particular characteristic. In the context of our exercise, the population proportion signifies the percentage of groceries that people typically waste, which is given as 12% or 0.12. This figure represents an assumed truth about the entire population, based on previous study or data collection from sources like Reader's Digest.
In exercises like this one, the population proportion ( \( p \) ) gives us a baseline to compare our sample data against. It helps in understanding how closely our sample represents the population. For instance, in our problem, we explore the proportion of groceries thrown away by shoppers to see if it matches or deviates from the actual population.
  • Population proportion provides a baseline.
  • It represents truth or assumed reality for statistical analysis.
  • Used for comparisons with sample characteristics.
Standard Error
The standard error is a crucial statistic that tells us how much variability we can expect in sample data. It measures how precisely a sample proportion reflects the population proportion. In the formula for standard error, we see:\[ SE = \sqrt{\frac{p(1-p)}{n}} \]where \( p \) is the population proportion, and \( n \) is the sample size.
In our scenario, using the values of \( p = 0.12 \) and \( n = 540 \), the standard error is calculated as approximately 0.0143. This value indicates the likely deviation range between different sample proportions if you took many samples from the same population. Lower SE means more reliable results and better estimation of the population parameter.
  • Standard error evaluates sample statistic reliability.
  • Calculated using population proportion and sample size.
  • A smaller SE suggests less variability in sample proportions.
Z-Score Calculation
Z-Score is a statistical metric that shows the number of standard deviations a data point is from the mean. In sampling distribution, it helps determine how unusual a sample proportion is compared to the population proportion. The formula for the Z-score is:\[ Z = \frac{\text{Margin of Error}}{SE} \]For instance, to figure out how likely it is that the sample proportion differs from the population proportion by ±0.03, we need the Z-score. With our calculated standard error of 0.0143 and a margin of 0.03, the Z-score comes out to be approximately 2.10.
Higher Z-scores indicate a greater deviation from the mean, quantifying the rarity of a sample outcome.
  • Z-score measures deviation from the mean.
  • Essential for evaluating sample proportion accuracy.
  • Calculated using margin of error and standard error.
Normal Distribution
Normal distribution, often symbolized by the bell curve, is foundational in statistics. It's a probability distribution that has a symmetrical shape, where most occurrences take place near the mean, and probabilities taper off as they move away. This characteristic is crucial when dealing with sample proportions because it helps us apply the Z-score and evaluate probabilities.
In the context of our exercise, after finding the Z-scores, we use the properties of the normal distribution to find probabilities that a sample proportion lies within a specific range of the population proportion. For example, with a Z-score of 2.10 derived from a ±0.03 range, the probability that the sample proportion is within this range is around 0.9642. This means there's a high likelihood that the sample accurately reflects the population's waste habits.
Why is this vital?
  • Normal distribution allows precise probability calculations.
  • Aids in understanding how sample results relate to population.
  • Forms the basis for many statistical inference methods.

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