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Chapter 7: Problem 4

Let \(p=0.25\) be the proportion of iPhone owners who have a given app. For a particular iPhone owner, let \(x=1\) if they have the app and \(x=0\) otherwise. For a random sample of 50 owners: a. State the population distribution (that is, the probability distribution of \(X\) for each observation). b. State the data distribution if 30 of the 50 owners sampled have the app. (That is, give the sample proportions of observed 0 s and 1 s in the sample.) c. Find the mean of the sampling distribution of the sample proportion who have the app among the 50 people. d. Find the standard deviation of the sampling distribution of the sample proportion who have the app among the 50 people. e. Explain what the standard deviation in part d describes.

Short Answer

Expert verified
a. Bernoulli(0.25). b. 0.6 (1's), 0.4 (0's). c. 0.25. d. 0.0612. e. It describes variability of sample proportion.

Step by step solution

01

Population Distribution

The population distribution describes the probability distribution of the random variable for each observation. Here, the random variable \(X\) follows a Bernoulli distribution since each iPhone owner either has the app or doesn't. Given \( p = 0.25 \), each observation can be represented as a Bernoulli random variable with \( X \sim \text{Bernoulli}(0.25) \). This means \( P(X=1) = 0.25 \) and \( P(X=0) = 0.75 \).
02

Data Distribution with Sample Proportion

In the sample of 50 owners, 30 have the app, so the sample proportion of owners with the app is \( \hat{p} = \frac{30}{50} = 0.6 \). The proportion without the app is \( 1 - \hat{p} = 0.4 \). The data distribution is represented by these sample proportions where 60% have the app (represented by 1s) and 40% don't have the app (represented by 0s).
03

Mean of the Sampling Distribution

The mean of the sampling distribution of the sample proportion \( \hat{p} \) is equal to the population proportion \( p \). Thus, the mean is \( \mu_{\hat{p}} = p = 0.25 \). This value represents the expected average proportion of iPhone owners in the sample who have the app.
04

Standard Deviation of the Sampling Distribution

The standard deviation of the sampling distribution of the sample proportion is calculated using the formula \( \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \), where \( n = 50 \) and \( p = 0.25 \). Substituting these values, we get \( \sigma_{\hat{p}} = \sqrt{\frac{0.25 \times 0.75}{50}} \approx 0.0612 \).
05

Interpretation of the Standard Deviation

The standard deviation found in the previous step describes the variability or spread of the sample proportion \( \hat{p} \) around the population proportion \( p \). It provides an idea of how much the sample proportions are expected to vary from sample to sample, assuming random sampling from a population where the true proportion is 0.25.

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Key Concepts

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

Population Distribution
In statistical terms, the **population distribution** refers to the probability distribution of a random variable. This describes all possible values and probabilities of a variable within a population. When examining iPhone owners and their likelihood of having a certain app, this distribution is crucial for understanding the framework.
In this case, each iPhone owner either possesses the app or does not, which makes the situation model a Bernoulli distribution. This is because there are only two possible outcomes for each individual:
  • **Having the app (Success)**: The probability is given by the symbol \( p \), which equals 0.25 here. This indicates a 25% chance an owner has the app.
  • **Not having the app (Failure)**: The complementary probability is \( 1 - p = 0.75 \), a 75% chance for owners without the app.
Expressed mathematically, the random variable \( X \) is represented as \( X \sim \text{Bernoulli}(0.25) \). This distribution forms the basis for sampling and analysis that follows.
Sample Proportion
The concept of **sample proportion** introduces the idea of observing a subset of the population to analyze its characteristics. For 50 iPhone owners sampled, we look into the number holding the app, which gives us factual insights about their preferences.
In this scenario, if 30 out of these owners have the app, the sample proportion is calculated as:
  • \( \hat{p} = \frac{30}{50} = 0.6 \)
This means 60% of the sample has the app. The remaining 40% of the sample, calculated as \( 1 - \hat{p} = 0.4 \), represents the proportion without the app.
The sample proportion represents just a snapshot and is crucial because it gives an approximation of the entire population's behavior. However, it鈥檚 important to note that sample proportions can vary between different samples from the same population.
Standard Deviation
**Standard deviation** is a measure of the amount of variation or dispersion within a set of values. When discussing the sample proportion, standard deviation helps in understanding how much the sample proportions deviate from the true population proportion across different samples.
To calculate the standard deviation of the sample proportion, we use:
  • \( \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \)
  • Given \( p = 0.25 \) and \( n = 50 \), substituting in gives:
  • \( \sigma_{\hat{p}} = \sqrt{\frac{0.25 \times 0.75}{50}} \approx 0.0612 \)
This standard deviation signifies how sample proportions (the percentage of individuals with the app in different samples of 50 iPhone owners) are spread around the population proportion (25%). It highlights expected variability, helping educators and statisticians gauge consistency in their samples compared to general population trends.

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Most popular questions from this chapter

In \(2005,\) a study was conducted in West Texas A\&M University. It showed that the average student debt in the United States was \(\$ 18,367\) with a standard deviation of \$4709. (Source: http://swer.wtamu. edu/sites/default/files/Data/15-26-49-178-1-PB.pdf.) a. Suppose 100 students had been randomly sampled instead of collecting data for all of them. Describe the mean, standard deviation, and shape of the sampling distribution of the sample mean. b. Using this sampling distribution, find the \(z\) -score for a sample mean of \(\$ 20,000\) c. Using parts a and b, find the probability that the sample mean would fall within approximately \(\$ 1000\) of the population mean.

The sampling distribution of a sample mean for a random sample size of 100 describes a. How sample means tend to vary from random sample to random sample of size 100 . b. How observations tend to vary from person to person in a random sample of size 100 . c. How the data distribution looks like the population distribution when the sample size is larger than 30 . d. How the standard deviation varies among samples of size 100 .

The scores on the Psychomotor Development Index (PDI), a scale of infant development, have a normal population distribution with mean 100 and standard deviation 15. An infant is selected at random. a. Find the \(z\) -score for a PDI value of 90 . b. A study uses a random sample of 225 infants. Using the sampling distribution of the sample mean PDI, find the \(z\) -score corresponding to a sample mean of 90 . c. Explain why a PDI value of 90 is not surprising, but a sample mean PDI score of 90 for 225 infants would be surprising.

Explain how the standard deviation of the sampling distribution of a sample proportion gives you useful information to help gauge how close a sample proportion falls to the unknown population proportion.

An exam consists of 50 multiplechoice questions. Based on how much you studied, for any given question you think you have a probability of \(p=0.70\) of getting the correct answer. Consider the sampling distribution of the sample proportion of the 50 questions on which you get the correct answer. a. Find the mean and standard deviation of the sampling distribution of this proportion. b. What do you expect for the shape of the sampling distribution? Why? c. If truly \(p=0.70\), would it be very surprising if you got correct answers on only \(60 \%\) of the questions? Justify your answer by using the normal distribution to approximate the probability of a sample proportion of 0.60 or less.
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