An entertainment complex is eager to debut a thrilling new feature. To gauge the potential popularity of this attraction, they seek to understand what fraction of people experiences acrophobia. A team from the park conducted a survey among 500 individuals, finding that 40% expressed a fear of heights. Another crew gathered data from 600 respondents, with 30% sharing this fear. Is it possible to gauge the overall fear of heights in the population from these figures?
This text explores the idea of sample proportions, including their symbols, calculations, significance, and real-world applications.
Explaining Sample Proportions
The figures mentioned previously indicate the percent of the group exhibiting a given trait, such as a fear of heights, which equates to a proportion.
The sample proportion is the percentage of participants in a sample who exhibit a targeted trait.
Representing the Sample Proportion
The symbol \(p\) denotes the proportion of a whole population, while the sample proportion is symbolized by \(\widehat{p}\), calculated by the quantity of successful cases in the sample over the total number of samples \(n\).
\[\widehat{p}=\frac{\text{successful instances in the sample}}{n}.\]
Generally, the sample proportion \(\widehat{p}\) varies from the overall population proportion \(p\).
Grasping Sample Proportions
Imagine a pack of 40 candies, with equal parts sour and sweet. Each candy is assigned a unique identifier.
| Sour candies | \(1, 2, 3, 4, 5, 6, 7, 8, 9, 10,\)\(11, 12, 13, 14, 15, 16, 17, 18, 19, 20\) |
| Sweet candies | \(21, 22, 23, 24, 25, 26, 27, 28, 29, 30,\)\(31, 32, 33, 34, 35, 36, 37, 38, 39, 40\) |
Table 1. an example of data, sample proportions.
Without knowledge of the actual flavor distribution, if you're interested in the quantity of sweet candies and pick a sample size of 4, selecting candies 1, 13, 14, and 29, where a sweet candy counts as a success, yielding a sample proportion of:
\[\widehat{p}=\frac{1}{4}=0.25\]
Additional samplings will yield further insights.
| Sampling instance | Candies chosen | \(\widehat{p}\) |
| \(1\) | \(1, 13, 14, 29\) | \(0.25\) |
| \(2\) | \(11, 12, 13, 14\) | \(0\) |
| \(3\) | \(1, 2, 26, 37\) | \(0.5\) |
| \(4\) | \(2, 14, 26, 38\) | \(0.5\) |
| \(5\) | \(2, 13, 15, 28\) | \(0.25\) |
| \(6\) | \(3, 4, 15, 36\) | \(0.25\) |
| Sampling session | Candies picked | \(\widehat{p}\) |
| \(7\) | \(3, 26, 27, 38\) | \(0.75\) |
| \(8\) | \(4, 26, 38, 39\) | \(0.75\) |
| \(9\) | \(15, 26, 27, 38\) | \(0.75\) |
| \(10\) | \(5, 26, 37, 39\) | \(0.75\) |
| \(11\) | \(26, 27, 28, 29\) | \(1\) |
| \(12\) | \(26, 37, 38, 40\) | \(1\) |
Table 2. an example of data, sample proportions.
Sample proportions clearly vary with different selections.
Visualization 1. A histogram illustrating the variability in sample proportions for sweet candies.
Understanding the variance in sample proportion \(\widehat{p}\) becomes simpler through visualization.
Value of Sample Proportions
Assembling comprehensive data to ascertain the proportion of subjects or items showcasing a certain attribute within a population is often unfeasible or unrealistic.
The rationale behind sample proportions then is to predict the population's proportion. Therefore, understanding the standard of the proportion is crucial.
In reference to the candies scenario, the depicted graph provides an approximation of the sample proportion \(\widehat{p}\)'s distribution. Considering all potential samples of size \(4\) would be necessary to achieve the actual sampling proportion distribution graph!
Necessities for the Proportion's Sampling Distribution
To ensure the sampling distribution for the proportion \(\widehat{p}\) mirrors the actual population proportion \(p\), certain prerequisites must be fulfilled:
1. Random selection imperative: The fundamental criterion for the sampling distribution to be valid is the randomness of the sample selection.
2. Independence (the \(10\%\) rule): Sampled entities ought to be independent. This is manageable by keeping sample sizes below \(10\%\) of the entire population size.
As demonstrated with the candies, random selections can be made blindly from a pack or by drawing numbers \(1-40\) on slips and picking one at random, ensuring the chosen batch of \(4\) meets the independence criterion as it represents \(10\%\) of the candy total.
Calculating Mean and Standard Deviation for Sample Proportions
Assuming \(p\) signifies the success rate within a population and \(\widehat{p}\) the success rate in a sampled fraction size \(n\), sample proportions' mean and standard deviation are calculated as follows:
\[\mu_{\widehat{p}}=p\] and \[ \sigma_{\widehat{p}}=\sqrt{\frac{p(1-p)}{n}}.\]
When the number \(np\geq 10\) and \(n(1-p)\geq 10\), the sample proportion's spread \(\widehat{p}\) approaches a normal distribution.
Thus, if the normality requirement is satisfied, a sample proportion \(\widehat{p}\) can be standardized to a \(z\)-score (consult the write-up on \(z\)-scores for depth) with the formula
\[ z=\frac{\widehat{p}-\mu_\widehat{p}}{\sigma_\widehat{p}}=\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}.\]
For a population success proportion of \(p=0.35\), identify the mean and variance of the sample proportion \(\widehat{p}\) from random samples of size \(n=70\).
Answer:
Applying the formulas provided, we find
\[\mu_{\widehat{p}}=0.35,\]
and the variance as
\[\sigma_{\widehat{p}}=\sqrt{\frac{(0.35)(0.65)}{70}} \approx 0.057.\]
Descriptive Examples of Sample Proportions
Let's delve into an example illustrating the computation of probabilities concerning a sample proportion's distribution.
Consider a firm claiming a mere \(10\%\) fault rate in its production line. An auditor selects a batch of \(200\) items.
(a) What's the likelihood of discovering less than \(12\%\) defectives?
(b) How probable is finding a defect rate bracketed between \(9\%\) and \(11\%\)?
Answer:
(1) Given the parameters
\[np=200(0.10)=20>10\]
and
\[n(1-p)=200(0.90)=180>10,\]
the fault proportion \(\widehat{p}\) suits a normal distribution approximation. Hence, norms of normal distribution apply for these probability computations.
(2) Calculating the mean and standard deviation for \(\widehat{p}\) through the formulas yields
\[\mu_\widehat{p}=0.10\] and
\[\sigma_\widehat{p}=\sqrt{\frac{(0.10)(0.90)}{200}} \approx 0.021.\]
(3) To determine \(z\)-scores: for (a)
\[ \begin{align}P(\widehat{p}<0.12) &= P\left(z<\frac{0.12-0.10}{0.021}\right) \\ &= P(z<0.95) \\ &=0.8289. \end{align} \]
And for (b)
\[ \begin{align}P(0.09<\widehat{p}<0.11) &= P\left(\frac{0.09-0.10}{0.021}<z<\frac{0.11-0.10}{0.021}\right) \\&= P(-0.48<z<0.48) \\ &= P(z<0.48)-P(z<-0.48) \\ &=0.6844-0.3156 \\ &=0.3688.\end{align} \]
Hence, the chance of a defect rate at or below \(12\%\) is \(0.8289\), and finding a defect rate from \(9\%\) to \(11\%\) stands at \(0.3688\).
Principal Insights on Sample Proportion
- The scrutiny of sample proportions serves to predict the proportions within the entire population.
- The symbol \(\widehat{p}\) denotes the sample proportion.
- The methodologies for evaluating the mean and standard deviation of the sample proportion distribution \(\widehat{p}\) are formulated as\[\mu_{\widehat{p}}=p\,\]and\[\sigma_{\widehat{p}}=\sqrt{\frac{p(1-p)}{n}}.\]
- If \(np\geq 10\) and \(n(1-p)\geq 10\), then the sampling distribution of \(\widehat{p}\) aligns with a normal distribution shape.