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In statistics, the sampling distribution is a critical concept. It refers to the probability distribution of a given statistic based on a random sample. When we take multiple samples from a population, each sample might produce different values. The sampling distribution allows us to understand these potential variations.
For example, consider the proportion of people who have sleepwalked. If we took many samples of 1480 people and recorded the proportion who sleepwalk in each sample, those proportions would form a distribution. This distribution is the sampling distribution of our sample proportion.
In our exercise, we know the population proportion, denoted as \( p \), is 29.2% or 0.292.
The mean of the sampling distribution, \( \mu \), is calculated by \( n \times p \), where \( n \) is the sample size. For our sample size of 1480, the mean is:
\[ \mu = 1480 \times 0.292 = 432.16 \]
The standard deviation of the sampling distribution, \( \sigma \), captures how much those sample proportions vary. It is calculated using:
\[ \sigma = \sqrt{n \times p \times (1 - p)} = \sqrt{1480 \times 0.292 \times 0.708} \approx 17.69 \]
This gives us a measure of how spread out our sample proportions might be.

The z-score is a way of standardizing results from a sample. It tells us how many standard deviations our sample statistic is away from the mean of the sampling distribution.
To calculate the z-score, we use the formula:
\[ z = \frac{x - \mu}{\sigma} \]
Where:
* \( x \) is the observed value from our sample (in this case, 455).
* \( \mu \) is the mean of the sampling distribution (calculated as 432.16).
* \( \sigma \) is the standard deviation of the sampling distribution (calculated as 17.69).
Plugging in our values, we get:
\[ z = \frac{455 - 432.16}{17.69} \approx 1.29 \]
This means our observed value of 455 sleepwalkers is 1.29 standard deviations above the mean of the sampling distribution.

The significance level helps us determine whether our observed result is unusual under the assumed population parameter. Typically, we use a significance level of 0.05. This corresponds to a 5% chance that an observed result is due to random chance.
In our exercise, we found the probability of getting 455 or more sleepwalkers given our population proportion is 0.0985. This probability is calculated by finding the area under the standard normal distribution curve to the right of our z-score, 1.29. Using the cumulative distribution function (CDF) for normal distribution, we got:
\[ P(Z > 1.29) = 1 - P(Z < 1.29) = 1 - 0.9015 = 0.0985 \]
Because 0.0985 is greater than 0.05, it's not considered statistically significant.
This suggests there's no strong evidence to reject the population proportion of 29.2%. Our sample result of 455 sleepwalkers is not significantly high—it falls within a range that could be expected by chance if the true population proportion is indeed 29.2%.