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Credit Cards According to creditcard.com, \(29 \%\) of adults do not own a credit card. (a) Suppose a random sample of 500 adults is asked, "Do you own a credit card?" Describe the sampling distribution of \(\hat{p}\), the proportion of adults who do not own a credit card. (b) What is the probability that in a random sample of 500 adults more than \(30 \%\) do not own a credit card? (c) What is the probability that in a random sample of 500 adults between \(25 \%\) and \(30 \%\) do not own a credit card? (d) Would it be unusual for a random sample of 500 adults to result in 125 or fewer who do not own a credit card? Why?

Short Answer

Expert verified
a) Mean = 0.29, SD ≈ 0.02081. b) P(>30%) ≈ 0.3156. c) P(25-30%) ≈ 0.657. d) Unusual, P(≤125) ≈ 0.0274.

Step by step solution

01

Understand the population proportion

Given the proportion of adults who do not own a credit card is 29%. Therefore, \( p = 0.29 \).
02

Define the sample size

The sample size in the problem is 500 adults, so \( n = 500 \).
03

Sampling distribution of \( \hat{p} \)

The sampling distribution of \( \hat{p} \) can be described by its mean and standard deviation. The mean of the sampling distribution is \( \mu_{\hat{p}} = p = 0.29 \). The standard deviation of the sampling distribution is given by \[ \sigma_{\hat{p}} = \sqrt{ \frac{ p(1 - p) }{ n } } = \sqrt{ \frac{ 0.29(1 - 0.29) }{ 500 } } \]
04

Calculate the standard deviation

The standard deviation of \( \hat{p} \) is \[ \sigma_{\hat{p}} \approx 0.02081 \]
05

Probability that more than 30% do not own a credit card

We need to find the probability that \( \hat{p} \) is greater than 0.30.Standardize the value using the Z-score formula: \[ Z = \frac{ \hat{p} - \mu_{\hat{p}} }{ \sigma_{\hat{p}} } = \frac{ 0.30 - 0.29 }{ 0.02081 } \approx 0.48 \]Using Z-tables, the probability \( P(Z > 0.48) \approx 0.3156 \).
06

Probability that between 25% and 30% do not own a credit card

We need to find the probability that \( 0.25 \leq \hat{p} \leq 0.30 \).Standardize the values using the Z-score formula:\( Z_1 = \frac{ 0.25 - 0.29 }{ 0.02081 } = -1.92 \)\( Z_2 = \frac{ 0.30 - 0.29 }{ 0.02081 } = 0.48 \)Using Z-tables, find the probabilities: \( P(Z < -1.92) \approx 0.0274 \)\( P(Z < 0.48) \approx 0.6844 \)Therefore, the probability \( P(-1.92 < Z < 0.48) = 0.6844 - 0.0274 = 0.657 \).
07

Probability of 125 or fewer who do not own a credit card

125 adults out of 500 leads to \( \hat{p} = \frac{ 125 }{ 500 } = 0.25 \).Standardize the value using the Z-score formula:\( Z = \frac{ 0.25 - 0.29 }{ 0.02081 } \approx -1.92 \)Using Z-tables, find the probability \( P(Z < -1.92) \approx 0.0274 \).Since this probability is less than 0.05, it would be unusual to find 125 or fewer non-credit card holders in a sample of 500.

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Key Concepts

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

population proportion
Population proportion refers to the fraction of the entire group (population) that holds a particular characteristic. In the given problem, the population proportion, denoted as \(p\), is 29%, or 0.29. This means that among all adults, 29% do not own a credit card. Understanding population proportion helps us estimate how a sample from this population might behave. It is a starting point for predicting outcomes and calculating probabilities in statistics.
When you take a random sample from a population, the sample will have its own proportion which usually approximates the population proportion. For example, if we randomly survey 500 adults out of all the adults, the fraction of those who do not own a credit card (sample proportion) will likely be close to 0.29.
sample size
Sample size, represented as \(n\), is the number of observations or data points in the sample. In this problem, the sample size is 500 adults. Choosing a large sample size is crucial because it tends to give more reliable and representative results of the population.
A larger sample size reduces the variability of the sample proportion. This means the standard deviation (spread) of the sampling distribution will be smaller, leading to more precise estimates of the population proportion. For instance, with a sample size of 500, we have more confidence in our estimates compared to a smaller sample size. The variability in smaller samples may lead to larger differences between the sample proportion and the population proportion, making predictions less reliable.
The formula used to calculate the standard deviation of the sampling distribution houses the sample size in its denominator, indicating the direct influence of sample size on the result.
z-score
A Z-score measures how many standard deviations an element is from the mean. In terms of sample proportions, it helps standardize the value to understand its position relative to the expected proportion. The Z-score is calculated using the formula:
\[ Z = \frac{ \hat{p} - \mu_{\hat{p}} }{ \sigma_{\hat{p}} } \]
Here, \( \hat{p} \) is the sample proportion, \( \mu_{\hat{p}} \) is the mean of the sampling distribution (which is the same as the population proportion \( p \)), and \( \sigma_{\hat{p}} \) is the standard deviation of the sampling distribution.
In the example provided, if we want to determine how unusual it is to find a sample proportion of 30% adults without a credit card, we can calculate the Z-score. This translates the observed proportion into a standard scale, making it easier to find the associated probability from standard Z-tables. For instance, a Z-score of 0.48 means the observed sample proportion is 0.48 standard deviations above the mean. Using Z-tables, we can find that the probability of observing such a value (or more extreme) in a sample is around 31.56%.

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Most popular questions from this chapter

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