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Chapter 8: Problem 13

A simple random sample of size \(n=1000\) is obtained from a population whose size is \(N=1,000,000\) and whose population proportion with a specified characteristic is \(p=0.35\). (a) Describe the sampling distribution of \(\hat{p}\). (b) What is the probability of obtaining \(x=390\) or more individuals with the characteristic? (c) What is the probability of obtaining \(x=320\) or fewer individuals with the characteristic?

Short Answer

Expert verified
Part (a): \(\hat{p} \sim N(0.35, 0.0147)\). Part (b): Probability is 0.0033. Part (c): Probability is 0.0207.

Step by step solution

01

- Describe the sampling distribution of \(\hat{p}\)

The sampling distribution of the sample proportion \(\hat{p}\) is approximately normally distributed with mean \(\mu_{\hat{p}} = p\) and standard deviation \(\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\). Given \(n=1000\) and \(p=0.35\): \[\begin{equation} \mu_{\hat{p}} = 0.35 \sigma_{\hat{p}} = \sqrt{\frac{0.35 \cdot (1 - 0.35)}{1000}} = \sqrt{\frac{0.2275}{1000}} \approx 0.0147 \end{equation}\]
02

- Convert x to \(\hat{p}\)

For part (b), convert \(x=390\) to the sample proportion \(\hat{p}\). \[\begin{equation} \hat{p} = \frac{x}{n} = \frac{390}{1000} = 0.39 \end{equation}\] Likewise, for part (c), convert \(x=320\) to the sample proportion \(\hat{p}\). \[\begin{equation} \hat{p} = \frac{320}{1000} = 0.32 \end{equation}\]
03

- Calculate the z-score for \(\hat{p} = 0.39\)

The z-score for \(\hat{p} = 0.39\) is calculated using the formula: \[\begin{equation} z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.39 - 0.35}{0.0147} \approx 2.72 \end{equation}\]
04

- Calculate the probability of \(x=390\) or more

Using the z-score from Step 3, look up the cumulative probability in the standard normal distribution table. The cumulative probability for a z-score of 2.72 is approximately 0.9967. Thus, the probability of obtaining \(x=390\) or more individuals with the characteristic is: \[\begin{equation} P(X \geq 390) = 1 - P(Z \leq 2.72) = 1 - 0.9967 = 0.0033 \end{equation}\]
05

- Calculate the z-score for \(\hat{p} = 0.32\)

The z-score for \(\hat{p} = 0.32\) is calculated using the formula: \[\begin{equation} z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.32 - 0.35}{0.0147} \approx -2.04 \end{equation}\]
06

- Calculate the probability of \(x=320\) or fewer

Using the z-score from Step 5, look up the cumulative probability in the standard normal distribution table. The cumulative probability for a z-score of -2.04 is approximately 0.0207. Thus, the probability of obtaining \(x=320\) or fewer individuals with the characteristic is: \[\begin{equation} P(X \leq 320) = P(Z \leq -2.04) = 0.0207 \end{equation}\]

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Key Concepts

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

Understanding Sample Proportion
The sample proportion, denoted as \(\backslash\text{\hat{p}}\), is a critical concept in statistics. It represents the proportion of a specified characteristic in a sample. To calculate \(\backslash\text{\hat{p}}\), you can use the formula: \(\backslash\text{\hat{p}} = \frac{x}{n}\), where \(x\) is the number of individuals with the characteristic in the sample, and \(n\) is the total number of individuals in the sample.
In our example exercise, for part (b), we calculated \(\backslash\text{\hat{p}}\) for \(x = 390\) and \(n = 1000\) as follows:
\(\backslash\frac{390}{1000} = 0.39\). Similarly, for part (c), \(x = 320\) and \(n = 1000\) resulted in: \(\backslash\frac{320}{1000} = 0.32\).
Understanding sample proportion is crucial for determining probabilities and making inferences about the population.
Z-Score Calculation
A z-score is a measure that describes a value's position relative to the mean of a group of values. It is specified in terms of standard deviations from the mean. The formula to calculate the z-score is:
\(\backslash{z = \backslashfrac{\backslashtext{\backslashhat{p}} - \backslashmu_{\backslashtext{\backslashhat{p}}}}{\backslashsigma_{\backslashtext{\backslashhat{p}}}}}\).
In our exercise, we used this formula to find the z-scores for \(\backslashtext{\backslashhat{p}} = 0.39\) and \(0.32\).
For \(\backslashtext{\backslashhat{p}} = 0.39\), the z-score calculation was:
\(\backslashfrac{0.39 - 0.35}{0.0147} = \backslashapprox 2.72\)
And for \(\backslashtext{\backslashhat{p}} = 0.32\), it was:
\(\backslashfrac{0.32 - 0.35}{0.0147} = \backslashapprox -2.04\).
Calculating z-scores helps us determine how far, and in which direction, a sample proportion is from the population proportion.
Probability Calculation
Calculating the probability involves finding the likelihood of a sample proportion deviating by a certain amount from the population proportion. Once we have the z-scores, we can use the standard normal distribution table to find the cumulative probabilities.
For our exercise, we calculated the probability for \(P(X \backslashgeq 390)\) and \(P(X \backslashleq 320)\) using their respective z-scores.
  • For \(\backslashtext{\backslashhat{p}} = 0.39\) with z = 2.72, the cumulative probability was 0.9967. So,
    \(P(X \backslashgeq 390) = 1 - 0.9967 = 0.0033\).
  • For \(\backslashtext{\backslashhat{p}} = 0.32\) with z = -2.04, the cumulative probability was 0.0207. Thus,
    \(P(X \backslashleq 320) = 0.0207\).
Understanding how to calculate these probabilities is essential for making informed statistical conclusions.

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Most popular questions from this chapter

The quality-control manager of a Long John Silver's restaurant wants to analyze the length of time that a car spends at the drive-through window waiting for an order. It is determined that the mean time spent at the window is 59.3 seconds with a standard deviation of 13.1 seconds. The distribution of time at the window is skewed right (data based on information provided by Danica Williams, student at Joliet Junior College). (a) To obtain probabilities regarding a sample mean using the normal model, what size sample is required? (b) The quality-control manager wishes to use a new delivery system designed to get cars through the drive-through system faster. A random sample of 40 cars results in a sample mean time spent at the window of 56.8 seconds. What is the probability of obtaining a sample mean of 56.8 seconds or less, assuming that the population mean is 59.3 seconds? Do you think that the new system is effective? (c) Treat the next 50 cars that arrive as a simple random sample. There is a \(15 \%\) chance the mean time will be at or below ____ seconds

Suppose you want to study the number of hours of sleep you get each evening. To do so, you look at the calendar and randomly select 10 days out of the next 300 days and record the number of hours you sleep. (a) Explain why number of hours of sleep in a night by you is a random variable. (b) Is the random variable "number of hours of sleep in a night" quantitative or qualitative? (c) After you obtain your ten nights of data, you compute the mean number of hours of sleep. Is this a statistic or a parameter? Why? (d) Is the mean number of hours computed in part (c) a random variable? Why? If it is a random variable, what is the source of variation?

Describe the sampling distribution of \(\hat{p}\). Assume that the size of the population is 25,000 for each problem. $$ n=1010, p=0.84 $$

True or False: The mean of the sampling distribution of \(\hat{p}\) is \(p\).

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?
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