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The sample proportion is a fundamental concept in statistics. It represents the proportion of individuals in a sample with a particular characteristic, analogous to the population proportion, which represents this in the total population.
The term is denoted by \( \hat{p} \) and is calculated as \( \hat{p} = \frac{x}{n} \), where \( x \) is the number of successes in the sample and \( n \) is the sample size.
In the exercise provided, the sample proportion after introducing a hand wash procedure dropped to \( 1\% \). But before this introduction, the observed hospital infection rate was \( 10\% \), equivalent to the population proportion. So the exercise involves comparing this dramatic reduction to the standard population-level infection rate.

Standard deviation is a measure of the dispersion or spread of a set of values. For our sample proportion, it's crucial to understand how much variation exists in this proportion from sample to sample.
The formula for standard deviation of a sample proportion is \( \sigma = \sqrt{\frac{p(1-p)}{n}} \). Here, \( p \) is the population proportion, and \( n \) is the sample size.
In this context, with \( p = 0.10 \) and \( n = 200 \), the calculated standard deviation is \( 0.0212 \). This small value suggests there is little variation in the proportion under the given conditions.

The z-score quantifies how many standard deviations an element is from the mean. It's instrumental in determining the position of a sample proportion relative to the assumed population proportion.
To compute the z-score, use the formula \( z = \frac{x - \mu}{\sigma} \), where \( x \) is our sample proportion, \( \mu \) is the average (or mean) proportion, and \( \sigma \) is the standard deviation.
In the example, the z-score for a \( 1\% \) sample proportion was calculated as \( -4.25 \), indicating that \( 0.01 \) is \( 4.25 \) standard deviations below the mean proportion (\( 0.10 \)). Such a large deviation implies a very unlikely occurrence under normal conditions.

Probability calculation translates the z-score into a meaningful probability that expresses how likely or unlikely a particular outcome is under the assumed normal distribution of the population.
In our problem, the objective was to find the probability that the sample proportion is \( 1\% \) or less given the original population proportion of \( 10\% \). After calculating the z-score as \( -4.25 \), the corresponding probability from standard z-tables was nearly \( 0 \) or about \( 0.00001 \).
This practically zero probability indicates that, under normal circumstances and without the hand wash procedure, observing such a low infection rate would be virtually impossible, highlighting the effectiveness of this intervention.