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In Exercises 6.11 to 6.14, use the normal distribution to find a confidence interval for a proportion \(p\) given the relevant sample results. Give the best point estimate for \(p,\) the margin of error, and the confidence interval. Assume the results come from a random sample. A \(95 \%\) confidence interval for \(p\) given that \(\hat{p}=\) 0.38 and \(n=500\)

Short Answer

Expert verified
The best point estimate for \(p\) is 0.38, the margin of error is 0.0437 and the 95% confidence interval for \(p\) is [0.3363, 0.4237]

Step by step solution

01

Point Estimate

The best point estimate for the population proportion \(p\) is the sample proportion \(\hat{p}\). Therefore, the best point estimate for \(p\) is 0.38 in this case.
02

Calculate Standard Error

We calculate the standard error of the proportion (SEP) using the formula SEP = \(\sqrt{ \hat{p}(1-\hat{p}) / n}\) where \(n\) is the sample size, and \(\hat{p}\) is the sample proportion. Substituting the given values, SEP = \(\sqrt{0.38(1-0.38)/500} = 0.0223\)
03

Margin of Error

Next, we calculate the margin of error using the formula: Margin of Error = Z * SEP. For a 95% confidence interval, the z-score is 1.96 (from z-table). Hence, Margin of Error = 1.96 * 0.0223 = 0.0437
04

Confidence Interval

Finally, we calculate the confidence interval using the point estimate and the margin of error. The confidence interval = \(\hat{p}\pm\) Margin of Error. Hence, Confidence interval = [0.38 - 0.0437, 0.38 + 0.0437] = [0.3363, 0.4237]

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Key Concepts

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

Normal Distribution
When we talk about the confidence interval for a proportion, we are often considering the normal distribution as the basis for our calculations. Normal distribution, also known as the Gaussian distribution, is a symmetric, bell-shaped distribution where most of the observations cluster around the central peak and the probabilities for values further away from the mean taper off equally in both directions.

It’s important to note that the normal distribution is used in the context of confidence intervals when certain conditions are met. Typically, these conditions require a sufficiently large sample size (usually a rule of thumb is n > 30) such that the sampling distribution of the proportion can be approximated as normal based on the Central Limit Theorem, regardless of the distribution of the population from which the sample was drawn.
Margin of Error
The margin of error is crucial to understanding the range within which we expect the population proportion to lie, given a certain level of confidence. It reflects how much we can expect our estimate to vary if we were to take many samples. A smaller margin of error indicates a more precise estimate.

The formula for the margin of error includes the standard error and a multiplier derived from the z-score: Margin of Error = Z * SEP. Essentially, it adjusts the width of the confidence interval. For a 95% confidence level, common in statistical analysis, the z-score is 1.96, representing the standard deviation limits on either side of a normal distribution curve that contains 95% of the data.
Standard Error of the Proportion
The standard error of the proportion (SEP) quantifies how much we would expect the sample proportion to vary from one random sample to another. It is a pivotal component in calculating the margin of error and, subsequently, the confidence interval.

The SEP is calculated using the formula \( SEP = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } \) where \( \hat{p} \) is the sample proportion and \(n\) is the sample size. A key point to remember is that the SEP decreases as the sample size increases, indicating that larger samples give us more precise estimates of the population proportion.
Z-Score
The z-score is a statistical measure that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations. In the context of confidence intervals, z-scores are used to determine the margin of error by indicating how many standard deviations away from the point estimate the interval extends.

For different confidence levels, there are corresponding z-scores: 1.96 for 95%, 2.58 for 99%, and so on. These scores are critical when working with the normal distribution, as they help us understand the spread and probability of the data within a given range in relation to the mean.
Point Estimate
A point estimate provides an actual value as an estimate of a population parameter based on sample data. When we calculate a confidence interval for a proportion, our point estimate is the sample proportion, often denoted as \( \hat{p} \).

The point estimate is the starting center of the confidence interval, and by itself, it does not convey information about the precision or reliability of the estimate; that's why we use it with the margin of error to construct the confidence interval. The interval thus signifies a range within which we believe the true population proportion may fall, with a certain level of confidence.

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Most popular questions from this chapter

In Exercises 6.109 to 6.111 , we examine the effect of different inputs on determining the sample size needed. Find the sample size needed to give, with \(95 \%\) confidence, a margin of error within ±10 . Within ±5 . Within ±1 . Assume that we use \(\tilde{\sigma}=30\) as our estimate of the standard deviation in each case. Comment on the relationship between the sample size and the margin of error.

When we want \(95 \%\) confidence and use the conservative estimate of \(p=0.5,\) we can use the simple formula \(n=1 /(M E)^{2}\) to estimate the sample size needed for a given margin of error ME. In Exercises 6.40 to 6.43, use this formula to determine the sample size needed for the given margin of error. A margin of error of 0.02 .

A data collection method is described to investigate a difference in means. In each case, determine which data analysis method is more appropriate: paired data difference in means or difference in means with two separate groups. A study investigating the effect of exercise on brain activity recruits sets of identical twins in middle age, in which one twin is randomly assigned to engage in regular exercise and the other doesn't exercise.

Using Data 5.1 on page \(375,\) we find a significant difference in the proportion of fruit flies surviving after 13 days between those eating organic potatoes and those eating conventional (not organic) potatoes. ask you to conduct a hypothesis test using additional data from this study. \(^{40}\) In every case, we are testing $$\begin{array}{ll}H_{0}: & p_{o}=p_{c} \\\H_{a}: & p_{o}>p_{c}\end{array}$$ where \(p_{o}\) and \(p_{c}\) represent the proportion of fruit flies alive at the end of the given time frame of those eating organic food and those eating conventional food, respectively. Also, in every case, we have \(n_{1}=n_{2}=500 .\) Show all remaining details in the test, using a \(5 \%\) significance level. Effect of Organic Potatoes after 20 Days After 20 days, 250 of the 500 fruit flies eating organic potatoes are still alive, while 130 of the 500 eating conventional potatoes are still alive.

In Exercises 6.152 and \(6.153,\) find a \(95 \%\) confidence interval for the difference in proportions two ways: using StatKey or other technology and percentiles from a bootstrap distribution, and using the normal distribution and the formula for standard error. Compare the results. Difference in proportion who favor a gun control proposal, using \(\hat{p}_{f}=0.82\) for 379 out of 460 females and \(\hat{p}_{m}=0.61\) for 318 out of 520 for males. (We found a \(90 \%\) confidence interval for this difference in Exercise 6.144.)

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