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In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion, find the standard error of the distribution of sample proportions. Samples of size 30 from a population with proportion 0.27

Short Answer

Expert verified
The standard error of the distribution of sample proportions is approximately 0.089.

Step by step solution

01

Understand and note down given parameters

The provided proportion of the population (\( p \)) is 0.27 and the sample size (\( n \)) is 30.
02

Substitute the values into the standard error equation.

Now we have to place these provided values into the standard error equation, which is \( \sqrt{\frac{p(1 - p)}{n}} \). So now it becomes \( \sqrt{\frac{0.27(1 - 0.27)}{30}} \)
03

Perform the calculation

Upon performing the calculation, the standard error becomes approximately 0.089.

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Key Concepts

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

Sample Proportions
Understanding sample proportions is essential when working with statistical sampling. A sample proportion is a decimal value that represents the fraction of items in a sample that have a certain trait. For instance, let's say we want to know the proportion of left-handed students within a school. If we survey a class of 30 students and find that 9 are left-handed, the sample proportion (\( \text{denoted as} \( \(\hat{p}\) \) \)) would be the number of left-handed students (9) divided by the total number of surveyed students (30), resulting in a sample proportion of 0.30.

In the context of our exercise, we are dealing with a sample of size 30, and we want to estimate the population proportion of a certain characteristic that is present at 0.27 rate. Understanding the variability or precision of this estimate is where the standard error comes into play, which we calculate from the sample proportion.
Population Proportion
The population proportion (\( p \) is an underlying value which represents the ratio of members in a population who have a particular characteristic. Considering our original problem, the population proportion (0.27) suggests that if we could look at every member within the population, 27% of them would have the characteristic we are interested in.

When studying a whole population is not feasible, we use samples to estimate population parameters like proportion. However, we must acknowledge that every sample we take is subject to sampling variability, which leads us to estimate the standard error to quantify the uncertainty of our sample proportion as an estimate of the true population proportion.
Standard Error Equation
The standard error equation is crucial for understanding the variability of an estimate from sample data. It helps us understand how tightly clustered or spread out the sampling distribution of our estimate might be. It is calculated using the formula \(\sqrt{\frac{p(1 - p)}{n}}\), where \(\(p\)\) is the population proportion, and \(\(n\)\) is the sample size.

In the exercise, we're given the population proportion of 0.27 and a sample size of 30. Plugging these values into the standard error formula provides us with an estimate of how much we expect our sample proportion to deviate from the population proportion due to random sampling alone. The lower the standard error, the better our estimate typically is, because it indicates less variability or uncertainty in our sample proportion.
Statistical Sampling
Statistical sampling involves selecting a subset of individuals from within a larger population to estimate characteristics about the whole group. The key is that the sample must be random and representative of the population in order to provide reliable estimates. If it is not, the results could be biased.

In exercises like the one we’ve examined, where we are given sample sizes and population proportions, we are implicitly operating under the assumption of random sampling. The concept of standard error is inherently linked to this practice; it quantifies the expected fluctuation of sample statistics if we were to repeat our sampling process many times. It’s a tool that tells us about the reliability of our sample-based estimates and, by extension, the quality of our statistical conclusions.

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Most popular questions from this chapter

In Exercises 6.150 and \(6.151,\) use StatKey or other technology to generate a bootstrap distribution of sample differences in proportions and find the standard error for that distribution. Compare the result to the value obtained using the formula for the standard error of a difference in proportions from this section. Sample A has a count of 90 successes with \(n=120\) and Sample \(\mathrm{B}\) has a count of 180 successes with \(n=300\).

Use the t-distribution and the given sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distributions are relatively normal. Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}>\mu_{2}\) using the sample results \(\bar{x}_{1}=56, s_{1}=8.2\) with \(n_{1}=30\) and \(\bar{x}_{2}=51, s_{2}=6.9\) with \(n_{2}=40\).

Metal Tags on Penguins and Breeding Success Data 1.3 on page 10 discusses a study designed to test whether applying metal tags is detrimental to penguins. Exercise 6.148 investigates the survival rate of the penguins. The scientists also studied the breeding success of the metal- and electronic-tagged penguins. Metal-tagged penguins successfully produced offspring in \(32 \%\) of the 122 total breeding seasons, while the electronic-tagged penguins succeeded in \(44 \%\) of the 160 total breeding seasons. Construct a \(95 \%\) confidence interval for the difference in proportion successfully producing offspring \(\left(p_{M}-p_{E}\right)\). Interpret the result.

The dataset ICUAdmissions, introduced in Data 2.3 on page \(69,\) includes information on 200 patients admitted to an Intensive Care Unit. One of the variables, Status, indicates whether each patient lived (indicated with a 0 ) or died (indicated with a 1 ). Use technology and the dataset to construct and interpret a \(95 \%\) confidence interval for the proportion of ICU patients who live.

Split the Bill? Exercise 2.153 on page 105 describes a study to compare the cost of restaurant meals when people pay individually versus splitting the bill as a group. In the experiment half of the people were told they would each be responsible for individual meal costs and the other half were told the cost would be split equally among the six people at the table. The data in SplitBill includes the cost of what each person ordered (in Israeli shekels) and the payment method (Individual or Split). Some summary statistics are provided in Table 6.20 and both distributions are reasonably bell-shaped. Use this information to test (at a \(5 \%\) level ) if there is evidence that the mean cost is higher when people split the bill. You may have done this test using randomizations in Exercise 4.118 on page 302 .

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