Remember previously, that a sample distribution will have:
If we apply the standard score formula to the sample distribution, we get this:
\(\displaystyle \begin{aligned} \frac{\hat{p} -\mu}{\sigma} &= \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}} \end{aligned} \)
When a sample is normally distributed (meets criteria in "Random Sampling"), this statistic will be approximately standard normal. That is:
\(\displaystyle \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}} \sim N(0, 1^2) \)
If a sample is normally distributed, this statistic will be the standard normal distrubution. If the sample is not normally distributed, and a question asks you to 'calculate the value' of the statistic, simply substitute values for each variable accordingly.
Conducting a census is very costly and exhaustive. The samples proportion value (\(\hat{p}\)), can be used to estimate the truly probability (\(p\)). When it is used this way, it is named the point estimate of the population proportion \(p\).
The notation for the point estimate is: \(\hat{p}_0\)
Although the term 'point estimate' seems confusing - don't be, it is simply means to substitute \(\hat{p}\) for \(p\) if that value is unkown. Not that hard, we may have done it before.
A normal distribution of sample data (distributions of \(\hat{p}\)) will contain somewhere within itself, the true probability \(p\). This is never, ever guaranteed.
A confidence interval (CI) is a range of values within a sample distribution. The concept of examining a strict range within a distribution of sample data is named an interval estimate. Instead of a point-estimate, which estimates the probability equal to the proportion, an interval estimate results in an output range from which the parameter in question may or may not reside within.
In this course, the parameter of interest is typically the mean.
The width of a CI is twice the error margin, which itself is the distance between the middle of the CI (the mean) and the edge.
Therefore, the boundaries of a CI is always equally distant from the mean, for normal distribution (dont worry about CIs for other distributions, they are not in this course).
The level of confidence is reffered to as the, understandingly, the 'confidence level', and is measured as a percentage. A 99% confidence interval is wider then a 95% confidence interval, and as a result is more likely to capture the true mean, but in turn, the interval is less accurate. A more narrow interval is less likely to capture the true mean, but if so, it will be more accurate.
Now for some tasty formulas!!
\(\displaystyle E(\hat{p} = p)\)
\(\displaystyle \sigma = \sqrt{\frac{p(1-p)}{n}}\)
\(\displaystyle \sigma^2 = \frac{p(1-p)}{n}\)
\(\displaystyle e = z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)
Confidence Interval: \(\hat{p} - e \le p \le \hat{p} + e\)
The z-score required for the confidence interval is related to the level of confidence. To find this value, navigate to eActivity > "Z score for CI", and solve for the z value.
Follow these steps:
This method results in less lines of working, and thus should only be used within a question with 2-3 marks. However this is risky - although it may be correct, you may not get marks for finding the error margin and laying it out like the main method. If you're daring, and for use in samples.
We can also use the average of the confidence interval (\(\frac{L+U}{2}\))
Finally, like all the other topics, a question is not always as simple as "find the confidence interval". Although you would expect basic questions like this, most question are designed to test your understanding of the formulas, and processes. For instance, you may be given some confidence interval data, and are expected to traverse the formulas to find, say, n. From there on you may use n for another application/formula to find the actual answer to the question.
3) In a sample, 37 out of 50 houses were found to have a maple coloured front doot.
a. What is the point estimate for the houses with maple coloured front doors?
\(\begin{aligned} \hat{p} &= \frac{37}{50} \\[5pt] &= 0.74 \end{aligned} \)
b. Calculate a 90% interval for the sample.
\(\begin{aligned} z &= 1.6449 \\[5px] e &= z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \\[5px] &= 1.6449\sqrt{\frac{0.74(1-0.74)}{50}} \\[5px] &= 0.062032 \\[10px] \hat{p} - e \le &p \le \hat{p} + e \\[5px] 0.74 - 0.06203 \le &p \le 0.74 + 0.06203 \\[5px] 0.6780 \le &p \le 0.8020 \end{aligned} \)
c. In another sample of \(n\) houses, 55 houses had maple front doors. A confidence interval for the true proportion of houses was \(0.5748 \le p \le 0.7816\). Find the sample size.
\(\begin{aligned} p &= \frac{0.7816+0.5748}{2} \\[5px] &= 0.6782 \\[5px] \frac{55}{n} &= 0.6782 \\[5px] n &= 81.97 \\[5px] n &= 82 \end{aligned} \)
d. What is the level of confidence from (c), given a true error margin of 0.101?
\(\begin{aligned} e &= 0.101 \\[5px] \text{ClassPad} \ &\gt \ \text{"Sample n for CI"}: z = 1.9458 \\[5px] \text{ClassPad} \ &\gt \ \text{"Z score for CI"} \\[5px] c &=0.9483 \\[5px] &= 0.95 \end{aligned} \)
This has been the Methods course for year 12!!! Thanks very much if you have stuck to this site for use as additional teaching, for tips, or for the tasty questions and hand written solutions - which I am currently yet to add. It's been a pleasure, and I wish you success for the WACE :)
ps. please tell your friends and kids from other year groups about this site if you have found it useful, that is the best way to support this project :)
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