91Ó°ÊÓ

Foreign Language According to a study done by Wakefield Research, the proportion of Americans who can order a meal in a foreign language is \(0.47 .\) (a) Suppose a random sample of 200 Americans is asked to disclose whether they can order a meal in a foreign language. Is the response to this question qualitative or quantitative? Explain. (b) Explain why the sample proportion, \(\hat{p},\) is a random variable. What is the source of the variability? (c) Describe the sampling distribution of \(\hat{p},\) the proportion of Americans who can order a meal in a foreign language. Be sure to verify the model requirements. (d) In the sample obtained in part (a), what is the probability the proportion of Americans who can order a meal in a foreign language is greater than \(0.5 ?\) (e) Would it be unusual that, in a survey of 200 Americans, 80 or fewer Americans can order a meal in a foreign language? Why?

Step by step solution

01

Title - Determine if the Response is Qualitative or Quantitative

The response to whether an American can order a meal in a foreign language is a qualitative variable. This is because the response is categorical (yes or no), rather than numerical.

02

Title - Explain why the Sample Proportion is a Random Variable

The sample proportion \(\hat{p}\) is a random variable because it varies from sample to sample. The variability comes from the process of selecting different random samples from the population each time.

03

Title - Describe the Sampling Distribution and Verify Model Requirements

The sampling distribution of \(\hat{p}\) follows a normal distribution for large sample sizes due to the Central Limit Theory. We verify the model requirements using the np and n(1-p) conditions. Here, \(n = 200\) and \(p = 0.47\). Calculate \(np = 200 \times 0.47 = 94\) and \(n(1-p) = 200 - 94 = 106\). Both values are greater than 10, so the normal model is appropriate.

04

Title - Calculate the Probability that the Proportion is Greater than 0.5

First, find the mean \(\hat{p}\) and the standard error SE for the sample proportion. The mean is \(p = 0.47\). The SE is calculated as \[ SE = \sqrt{ \frac{p(1-p)}{n} } = \sqrt{ \frac{0.47 \times 0.53}{200} } = 0.0353 \]. Convert the proportion \(0.5\) to a z-score: \[ull z = \frac{0.5 - 0.47}{0.0353} = 0.85\ull\text{}ull\text{}ull \]. Using the z-table, the probability that the proportion is greater than 0.5: \[P(\hat{p} > 0.5) = 1 - P(z < 0.85) = 1 - 0.8023 = 0.1977\ull\text{}ull\text{}ull\text{}ull\text{}ull\text{}ull\text{}ull\text{} \].

05

Title - Determine if it Would be Unusual for 80 or Fewer Americans to Order a Meal in a Foreign Language

Calculate \(\hat{p}\) for 80 out of 200: \(\hat{p} = \frac{80}{200} = 0.4\ull\text{}ull\text{}ull\text{}ull\text{}ull\text{}ull\text{}ull\text{}ull\text{}ull\text{}ull\text{}ull\text{}ull\text{}ull\text{}\). Convert to a z-score: \[ z = \frac{0.4 - 0.47}{0.0353} = -1.98\ull\text{}ull\text{} \]. Using the z-table, find the probability: \[P(z < -1.98) = 0.024\ull\text{}ull\text{}ull\text{} \]. Since 0.024 is lower than 0.05, it would be unusual for 80 or fewer Americans to be able to order a meal in a foreign language.

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Key Concepts

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

Qualitative vs Quantitative Data

Understanding whether data is qualitative or quantitative is essential for selecting the appropriate statistical methods. In the context of our exercise, the question is whether an American can order a meal in a foreign language. This data is qualitative because the responses are categorical (yes or no). Qualitative data describes attributes or characteristics and is often non-numerical. In contrast, quantitative data is numerical and measures quantities or amounts. Examples include height, weight, and age. Knowing the difference helps in choosing how to analyze and interpret the data correctly.

Random Variables in Sampling

A random variable is a variable whose values are determined by the outcome of a random phenomenon. When sampling a population, the sample proportion (denoted as \(\hat{p}\) ) is a random variable. This means that if different samples are taken, \(\hat{p}\) will vary from sample to sample. For example, if we repeatedly take samples of 200 Americans, the proportion of those who can order a meal in a foreign language will not be exactly the same each time. This variability is due to the randomness inherent in the process of selecting the samples. Understanding this helps in estimating population parameters and in making predictions.

Central Limit Theorem

The Central Limit Theorem (CLT) is a fundamental theorem in statistics. It states that the sampling distribution of the sample mean (or proportion) will be approximately normally distributed if the sample size is large enough, regardless of the original distribution of the population. This is crucial because it means we can apply normal probability theory to make inferences about sample means or proportions. In our exercise, we use CLT to describe the sampling distribution of \(\hat{p}\). For a sample size of 200, the CLT is applicable, thus allowing us to use the normal model to estimate probabilities. Verifying the conditions for CLT, we check that both \(\text{np}\) and \(\text{n}(1-\text{p})\) are greater than 10, ensuring our sample size is sufficiently large.

Model Requirements Verification

When using statistical models, it is important to verify that the model requirements are met. For the normal approximation to the binomial distribution (applied in our exercise to \(\hat{p}\)), we verify two conditions: \(\text{np}\) and \(\text{n}(1-\text{p})\) need to be greater than 10. In our exercise, where \(\text{n} = 200\) and \(\text{p} = 0.47\), we calculate \(\text{np} = 200 \times 0.47 = 94\) and \(\text{n}(1-\text{p}) = 200 \times 0.53 = 106\). Both conditions are satisfied, verifying that using a normal approximation is appropriate. This step is crucial to ensure the accuracy and reliability of our predictions and conclusions based on the model.