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A USA Today snapshot reports that approximately \(23 \%\) of cell phone owners walked into someone or something while they were talking on their cell phone. \({ }^{12}\) In a random sample of \(n=200\) cell phone owners, what is the probability that the sample proportion of cell phone owners who have walked into someone or something while they were on the phone would be less than \(.15 ?\)

Short Answer

Expert verified
Answer: The probability is approximately 0.29% or 0.0029.

Step by step solution

01

Identify the parameters

We are given that the sample size \(n = 200\) and the proportion of cell phone owners who have walked into someone or something is \(p = 0.23\). Since we are considering a sample, we need to find the sample proportion, \(\widehat{p} = 0.15\).
02

Calculate the mean and standard deviation of the sampling distribution

We can calculate the mean and standard deviation of the sampling distribution by using the formulas: \(\mu_{\widehat{p}} = \frac{p}{n}\) and \(\sigma_{\widehat{p}} = \sqrt{\frac{p(1-p)}{n}}\). Plugging in the given values, we get: Mean: \(\mu_{\widehat{p}} = 0.23\) Standard deviation: \(\sigma_{\widehat{p}} = \sqrt{\frac{0.23 (1-0.23)}{200}} = \sqrt{\frac{0.23 \cdot 0.77}{200}} \approx 0.029\)
03

Calculate the Z-score

To find the probability, we will calculate the Z-score, which is the number of standard deviations away from the mean. The formula for the Z-score is \(Z = \frac{\widehat{p} - \mu_{\widehat{p}}}{\sigma_{\widehat{p}}}\). Plugging in the values, we get: \(Z = \frac{0.15 - 0.23}{0.029} \approx -2.76\)
04

Find the probability

Now that we have the Z-score, we can find the probability using the Z-table (Or standard normal distribution table). We need to find the probability that \(Z \leq -2.76\). From the table, we can find that the area to the left of this Z-score is approximately \(0.0029\). This means the probability that the sample proportion of cell phone owners who have walked into someone or something while they were on the phone is less than \(0.15\) is about \(0.29\%\) or \(0.0029\).

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Key Concepts

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

Sampling Distribution
Understanding the sampling distribution is crucial for interpreting what our data tells us about the larger population. It's a probability distribution of a statistic obtained from a larger number of samples drawn from a specific population. For instance, if we were to repeatedly take samples of a given size from a population and calculate the sample mean for each, the distribution of these means would be the sampling distribution of the mean.

To make this concept clearer, let's consider rolling a six-sided die. If we were to roll it multiple times and record the average (mean) of the outcomes, that series of averages would approximate the expected value, or mean of roll outcomes, as the number of rolls increases. This same principle applies to more complex statistics, such as the sample proportion in our exercise, where each sample's proportion of cell phone owners bumping into someone can vary around the true proportion of the population.

The sampling distribution tells us how the sample proportion will behave as we draw more samples from the population. It's described by a mean, known as the expected value, and a standard deviation. The standard deviation of the sampling distribution is also referred to as the standard error, and it measures how much the sample statistic varies from sample to sample.
Sample Proportion
The sample proportion, often denoted as \(\hat{p}\), is a statistic that estimates the proportion of a certain attribute in the population, based on the sample data. For instance, in our exercise, the sample proportion is the percentage of cell phone owners in the sample that have walked into something while using their phone.

To calculate the sample proportion, we would take the number of successes (in this case, cell phone owners who have had a bump-into-something incident) and divide it by the total number of trials or observations in our sample (i.e., the total number of surveyed cell phone owners). Here, it's important to highlight that sample proportions can differ from sample to sample due to the randomness inherent in which individuals end up in which sample.

However, as the size of the sample increases, the variation of the sample proportion shrinks, and it becomes a better estimator of the true population proportion. This insight is essential because it means that with larger samples, we can be more confident that our sample proportion is close to the actual proportion in the entire population.
Z-score
The Z-score is a powerful tool in statistics for understanding where a particular value lies in relation to the mean of a group of values. It indicates how many standard deviations a given value is from the mean. A Z-score can be positive or negative, denoting whether the value is above or below the mean, respectively.

In probability and statistics, we often use the Z-score to assess the unusualness of a single data point. It plays a pivotal role when working with normal distributions, helping us transform the data into a standardized form so that we can easily calculate probability and make comparisons.

To calculate a Z-score, you subtract the mean of the sampling distribution from the observed value, then divide by the standard deviation of the sampling distribution. In our original problem, we used the Z-score to determine the probability of observing a sample proportion below 15%. Such calculations are vital when applying statistical tests to make inferences about the population, which often hinges on the rarity of the observed result under the assumption that the null hypothesis is true.

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Most popular questions from this chapter

A random sample of \(n\) observations is selected from a population with standard deviation \(\sigma=1 .\) Calculate the standard error of the mean \((S E)\) for the values of \(n\). $$ n=100 $$

Random samples of size \(n\) were selected from a nonnormal population with the means and variances. What can be said about the sampling distribution of the sample mean? Find the mean and standard error for this distribution. $$ n=8, \mu=120, \sigma^{2}=1 $$

Is it appropriate to use the normal distribution to approximate the sampling distribution of \(\hat{p}\) for the situations described in Exercises \(4-8 ?\) $$n=75, p=.4$$

In many states, lists of possible jurors are assembled from voter registration lists and Department of Motor Vehicles records of licensed drivers and car owners. In what ways might this list not cover certain sectors of the population adequately?

A paper manufacturer requires a minimum strength of 20 pounds per square inch. To check on the quality of the paper, a random sample of 10 pieces of paper is selected each hour from the previous hour's production and a strength measurement is recorded for each. Assume that the strength measurements are normally distributed with a standard deviation \(\sigma=2\) pounds per square inch. a. What is the approximate sampling distribution of the sample mean of \(n=10\) test pieces of paper? b. If the mean of the population of strength measurements is 21 pounds per square inch, what is the approximate probability that, for a random sample of \(n=10\) test pieces of paper, \(\bar{x}<20 ?\) c. What value would you select for the mean paper strength \(\mu\) in order that \(P(\bar{x}<20)\) be equal to \(.001 ?\)

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