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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?

Short Answer

Expert verified
a) Normal distribution with mean 0.30 and SD 0.0458. b) Probability is 0.0630; not unusual. c) Probability is 0.0044; unusual.

Step by step solution

01

Describe the Sampling Distribution

The sampling distribution of \hat{p} represents the distribution of the sample proportion. Since n = 100, p = 0.30, and q = 1 - p = 0.70, the mean (\mu \_{\hat{p}}) and standard deviation (\sigma \_{\hat{p}}) of the sampling distribution can be described as:\[ \mu_{\hat{p}} = p = 0.30 \] \[ \sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.30 \cdot 0.70}{100}} = \sqrt{0.0021} \approx 0.0458 \] Thus, \hat{p} follows a normal distribution with mean 0.30 and standard deviation 0.0458.
02

Calculate the Probability That More Than 37 Households Plant a Garden

First, convert 37 households to the sample proportion:\[ \hat{p} = \frac{37}{100} = 0.37 \] Then, find the z-score for \hat{p} = 0.37:\[ z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.37 - 0.30}{0.0458} \approx 1.53 \] Using standard normal distribution tables or a calculator, the probability of z > 1.53 is about 0.0630. Since 0.0630 > 0.05, this result is not considered unusual.
03

Calculate the Probability That 18 or Fewer Households Plant a Garden

First, convert 18 households to the sample proportion:\[ \hat{p} = \frac{18}{100} = 0.18 \] Then, find the z-score for \hat{p} = 0.18:\[ z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.18 - 0.30}{0.0458} \approx -2.62 \] Using standard normal distribution tables or a calculator, the probability of z < -2.62 is about 0.0044. Since 0.0044 < 0.05, this result is considered unusual.

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Key Concepts

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

simple random sample
A simple random sample (SRS) is a fundamental sampling method where every member of the population has an equal chance of being selected. This technique ensures each subset of the population has the same opportunity to be included.
In our exercise, we took a simple random sample of 100 households from a town with 5000 households. By doing this, we aim to get a representative group to study the gardening behavior of households.
sample proportion
The sample proportion, denoted as \(\hat{p}\), measures the ratio of the sample displaying a particular characteristic to the sample size. It's a way of estimating a population proportion.
In the given problem, the population proportion (\(p\)) is 0.30, meaning 30% of the households plant a garden. We use \(\hat{p}\) to estimate this proportion in our sample.
For example, if 37 out of 100 households in our sample plant a garden, the sample proportion \(\hat{p}\) is calculated as:
\[ \hat{p} = \frac{37}{100} = 0.37 \]
normal distribution
A normal distribution is a continuous probability distribution that is symmetrical around its mean, forming a bell-shaped curve. It's essential in statistics because many variables are approximately normally distributed.
The mean (\(\mu\)) is the center of the distribution, and the standard deviation (\(\sigma\)) measures how spread out the values are.
In this exercise, when we look at the sampling distribution of \(\hat{p}\), it can be approximated by a normal distribution with:
\[ \mu_{\hat{p}} = p = 0.30\] \[ \sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.30 \cdot 0.70}{100}} = 0.0458 \] This means the distribution of \(\hat{p}\) will be centered around 0.30 with a standard deviation of 0.0458.
z-score calculations
A z-score measures how many standard deviations an element is from the mean of a distribution. It helps determine the probability of observing a value in a normal distribution.
To calculate a z-score:
\[ z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} \] For instance, to find the z-score for \(\hat{p} = 0.37\):
\[ z = \frac{0.37 - 0.30}{0.0458} \approx 1.53 \]
Using the z-score, we can find the probability of \(\hat{p}\) being greater than 0.37.
probability
Probability quantifies the likelihood of an event occurring. It ranges from 0 to 1, with 0 meaning the event will not occur and 1 meaning it certainly will.
In our context, we are interested in the probability of the sample proportion \(\hat{p}\) taking specific values.
For example, the probability that more than 37 households plant a garden (which corresponds to \(\hat{p} = 0.37\)) can be found using the z-score:
\[ P(z > 1.53) \approx 0.0630 \] The probability that 18 or fewer households plant a garden (corresponding to \(\hat{p} = 0.18\)) can be similarly determined:
\[ P(z < -2.62) \approx 0.0044 \]

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Most popular questions from this chapter

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