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Chapter 8: Problem 1

In a town of 500 households, 220 have a dog. The population proportion of dog owners in this town (expressed as a decimal) is _____.

Short Answer

Expert verified
The population proportion of dog owners is 0.44.

Step by step solution

01

- Identify the Total Number of Households

In the town, the total number of households is given as 500.
02

- Identify the Number of Households with Dogs

The number of households that have a dog is given as 220.
03

- Calculate the Proportion

To find the population proportion of dog owners, divide the number of households with dogs by the total number of households: \[ \text{Proportion} = \frac{220}{500} \]
04

- Simplify the Fraction

Simplify the fraction to find the decimal form: \[ \text{Proportion} = \frac{220}{500} = 0.44 \]

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Key Concepts

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

Fraction to Decimal Conversion
When you need to convert a fraction to a decimal, you essentially perform a division. In our example, we have the fraction \( \frac{220}{500} \). To find the decimal form, divide 220 by 500.

Follow these steps:
  • Perform the division: \( 220 \div 500 = 0.44 \).
  • So, \( \frac{220}{500} \) converts to 0.44 as a decimal.
This conversion helps you express numbers in a more commonly understood format. It鈥檚 easier to compare and interpret decimal values than fractions.
Sample Size
The sample size is crucial in any statistical calculation. It represents how many observations you have, which in our example refers to the 500 households.

Here鈥檚 why sample size matters:
  • A larger sample size generally means more reliable results.
  • It reduces the impact of outliers or anomalies.
In this exercise, the sample size (total households) is 500, which gives us enough data to confidently estimate population proportions.
Statistical Calculation
Statistical calculations often involve determining proportions, means, medians, and more. For our problem, we performed a proportion calculation, which is a common statistical method.

To calculate a proportion:
  • Identify the part of the population you are studying (e.g., 220 households with dogs).
  • Divide this number by the total population (500 households) to get the proportion: \( \frac{220}{500} = 0.44 \).
This proportion tells us that 44% of the households have a dog, which helps in understanding the characteristics of the town鈥檚 population.

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Most popular questions from this chapter

Gardeners Suppose a simple random sample of size \(n=100\) households is obtained from a town with 5000 households. It is known that \(30 \%\) of the households plant a garden in the spring. (a) Describe the sampling distribution of \(\hat{p}\) (b) What is the probability that more than 37 households in the sample plant a garden? Is this result unusual? (c) What is the probability that 18 or fewer households in the sample plant a garden? Is this result unusual?

Suppose a simple random sample of size \(n=36\) is obtained from a population with \(\mu=64\) and \(\sigma=18\) (a) Describe the sampling distribution of \(\bar{x}\). (b) What is \(P(\bar{x}<62.6) ?\) (c) What is \(P(\bar{x} \geq 68.7) ?\) (d) What is \(P(59.8<\bar{x}<65.9) ?\)

Suppose a simple random sample of size \(n=40\) is obtained from a population with \(\mu=50\) and \(\sigma=4 .\) Does the population need to be normally distributed for the sampling distribution of \(\bar{x}\) to be approximately normally distributed? Why? What is the sampling distribution of \(\bar{x} ?\)

The shape of the distribution of the time required to get an oil change at a 10 -minute oil-change facility is unknown. However, records indicate that the mean time for an oil change is 11.4 minutes and the standard deviation for oil-change time is 3.2 minutes. (a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required? (b) What is the probability that a random sample of \(n=40\) oil changes results in a sample mean time less than 10 minutes?

The quality-control manager of a Long John Silver's restaurant wishes to analyze the length of time a car spends at the drive-through window waiting for an order. According to records obtained from the restaurants, 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 drivethrough 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 the population mean is 59.3 seconds? Do you think that the new system is effective?
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