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Ignaz Semmelweiss (1818-1865) was the doctor who first encouraged other doctors to wash their hands with disinfectant before touching patients. Before the new procedure was established, the rate of infection at Dr. Semmelweiss's hospital was about \(10 \%\). Afterward the rate dropped to about \(1 \%\). Assuming the population proportion of infections was \(10 \%\), find the probability that the sample proportion will be \(1 \%\) or less, assuming a sample size of \(200 .\) Start by checking the conditions required for the Central Limit Theorem to apply.

Short Answer

Expert verified
Once you have the z-score, just use it to look up the probability in the standard normal table or use a calculator with a normal distribution function. This will give you the required probability.

Step by step solution

01

Determine conditions for Central Limit Theorem

Check whether the conditions for the CLT are met or not. The two necessary conditions are: 1) the sample drawn from the population must be random and 2) the sample size should be less than 10% of the population when sampling is done without replacement. It can be assumed that these two conditions are met since it is a theoretical question.
02

Verify sample size conditions

The success and failures should be greater than or equal to 10. In this context, 'success' refers to the event that an individual got infected. The expected number of 'successes' is \(n*p = 200*0.10 = 20\), and the expected number of 'failures' (not getting infected) is \(n*(1-p)=200*0.90=180\). Since both of these numbers are above 10, the sample size condition is met.
03

Construct a sampling distribution using Central Limit Theorem

When all these conditions are met, according to the Central Limit Theorem the sampling distribution of the sample proportion will be approximately normally distributed. The mean of this distribution will be the true proportion \(p = 0.10\) and the standard deviation is given by \(\sqrt{ p*(1-p) / n } = \sqrt{ (0.10 * 0.90) / 200 }\). Calculate the standard deviation.
04

Find the z-score

Assuming that conditions for the Central Limit Theorem have been met, a z-score will be calculated for a sample proportion of 0.01. The z-score will represent how many standard deviations below the expected mean the observed \(1 \% \) infection rate is. The formula for calculating the z-score is \((\hat{p}-p) / \sqrt{ p*(1-p) / n }\). Substitute the given values and find the z-score.
05

Find the probability

Use the standard normal (z) table or calculator to find the probability associated with the z-score obtained in the previous step. This will be the probability that the sample proportion will be \(1 \%\) or less.

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Key Concepts

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

Sampling Distribution
In the world of statistics, comprehending the behavior of a sample is crucial for making inferences about an entire population. That's where the concept of a sampling distribution comes into play. Imagine we're selecting a series of smaller groups, or samples, from a larger group, known as the population. The sampling distribution is a probability distribution of all possible sample means (or proportions) from these groups.

It serves as a map, guiding us on what values we can expect if we repeat our sampling multiple times. Significantly, the Central Limit Theorem (CLT) tells us that for a sufficiently large sample size, the sampling distribution of the sample mean will be approximately normally distributed—regardless of the population's original distribution shape. This assumption of normality becomes the foundation for further statistical analysis, such as hypothesis testing or creating confidence intervals.
Probability
Imagine rolling a die: Probability is the likelihood of it landing on a six; or in a deck of cards, the chance of drawing an Ace. It's a way of quantifying the uncertainty inherent in various outcomes. Applied to our medical scenario, it's the measure of how likely it is that a sample would show an infection rate of 1% if the true population rate is 10%.

Calculated as a number between 0 and 1 (or 0% and 100%), this value is the backbone of statistical conclusions. It tells us about expectations in the long run: for example, if we look at many samples from the same population, probability gives us an anticipation of how commonly a specific outcome will occur.
Standard Deviation
Dive into any dataset, and you'll find values spread out in a particular pattern. The standard deviation is your tour guide to this spread, indicating how much the values deviate from the mean on average. In the context of our hospital infection rates, the standard deviation gives us a measure of variability in the sampling distribution. The lower the standard deviation, the more closely packed our sample proportions are around the population proportion (10% in our case).

Understanding the spread helps us determine if a sample's infection rate of 1% is just a regular 'blip' or a significant deviation from what we would typically expect. Essentially, it is the ruler we use to measure variation within our samples.
Z-score
If the standard deviation is our ruler, then the z-score is the mark we make on the ruler to measure our specific situation. A z-score is a statistical tool that describes the position of a raw score in terms of its distance from the mean, measured in standard deviations. It's like saying, 'In this list of heights, Joe is 2 standard deviations taller than the average person.'

In the Semmelweiss example, calculating a z-score for the 1% infection rate tells us how many standard deviations this rate lies below the expected proportion of infections. This score will then convert into a probability that helps evaluate the effectiveness of hand-washing, providing a numerical argument for this healthcare improvement.
Population Proportion
Flipping back to the beginning of our storybook on statistics, the population proportion is a fundamental element. It's the fraction of the larger population that has a particular attribute, like the percentage of people with brown eyes in a country or, in Semmelweiss's scenario, the infection rate in the hospital. The population proportion is a critical starting point for our journey, as it anchors our expectations about what the sample proportions should be if the sampling process is unbiased.

In our medical example, the population proportion is the pre-existing infection rate of 10%. It serves as the reference point against which the new infection rate (post-handwashing) is assessed, offering a quantitative look into how much an intervention impacts a population's health.

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Most popular questions from this chapter

A 2017 survey of U.S. adults found the \(64 \%\) believed that freedom of news organization to criticize political leaders is essential to maintaining a strong democracy. Assume the sample size was 500 . a. How many people in the sample felt this way? b. Is the sample large enough to apply the Central Limit Theorem? Explain. Assume all other conditions are met. c. Find a \(95 \%\) confidence interval for the proportion of U.S. adults who believe that freedom of news organizations to criticize political leaders is essential to maintaining a strong democracy. d. Find the width of the \(95 \%\) confidence interval. Round your answer to the nearest whole percent. e. Now assume the sample size was increased to 4500 and the percentage was still \(64 \%\). Find a \(95 \%\) confidence interval and report the width of the interval. f. What happened to the width of the confidence interval when the sample size was increased. Did it increase or decrease?

A Harris poll asked a sample of U.S. adults if they agreed with the statement "Artificial intelligence will widen the gap between the rich and poor in the U.S." Of those aged 18 to \(35,69 \%\) agreed with the statement. Of those aged 36 to \(50,60 \%\) agreed with the statement. A \(95 \%\) confidence interval for \(p_{1}-p_{2}\) (where \(p_{1}\) is the proportion of those aged \(18-35\) who agreed and \(p_{2}\) is the proportion of those aged \(36-50\) who agreed) is \((0.034,0.146)\). Does the interval contain 0 ? What does this tell us about the proportion of adults in these age groups who agree with the statement?

According to a 2018 Pew Research report, \(40 \%\) of Americans read print books exclusively (rather than reading some digital books). Suppose a random sample of 500 Americans is selected. a. What percentage of the sample would we expect to read print books exclusively? b. Verify that the conditions for the Central Limit Theorem are met. c. What is the standard error for this sample proportion? d. Complete this sentence: We expect ____\(\%\) of Americans to read print books exclusively, give or take ______\(\%\) .

A Harris poll asked Americans in 2016 and 2017 if they were happy. In \(2016,31 \%\) reported being happy and in 2017 , \(33 \%\) reported being happy. Assume the sample size for each poll was 1000 . A \(95 \%\) confidence interval for the difference in these proportions \(p_{1}-p_{2}\) (where proportion 1 is proportion happy in 2016 and proportion 2 is the proportion happy in 2017) is \((-0.06,0.02)\). Interpret this confidence interval. Does the interval contain \(0 ?\) What does this tell us about happiness among American in 2016 and \(2017 ?\)

Has trust in the executive branch of government declined? A Gallup poll asked U.S. adults if they trusted the executive branch of government in 2008 and again in 2017 . The results are shown in the table. $$\begin{array}{|l|r|}\hline & \mathbf{2 0 0 8} & \mathbf{2 0 1 7} \\ \hline \text { Yes } & 623 & 460 \\\\\hline \text { No } & 399 & 562 \\\\\hline \text { Total } & 1022 & 1022 \\ \hline\end{array}$$ a. Find and compare the sample proportion for those who trusted the executive branch in 2008 and in 2017 . b. Find the \(95 \%\) confidence interval for the difference in the population proportions. Assume the conditions for using the confidence interval are met. Based on the interval, has there been a change in the proportion of U.S. adults who trust the executive branch? Explain.

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