This is not designed to be a comprehensive review. There may be items on the exam that are not covered in this review. Similarly, there may be items in this review that are not tested on this exam. You are strongly encouraged to review the readings, homework exercises, and other activities from Units 1-3 as you prepare for the exam. In particular, you should go over the Review for Exam 1 and the Review for Exam 2.
By the end of this lesson, you should be able to:
The sample proportion \(\hat{p}\) is computed by dividing the number of successes (\(x\)) found in a sample by the sample size (\(n\)), \[ \text{Sample Proportion} \quad \widehat{p} = \frac{x}{n} \]
The sample proportion \(\widehat{p}\) is interpreted as the probability of a success occurring in the sample. It is a point estimate of the true proportion, \(p\).
Two graphical summaries are typically used to show sample proportions.
The sampling distribution of the sample proportion can be considered to be normally distributed when both \(np \ge 10\) and \(n(1-p) \ge 10\). The value of \(np\) gives the expected number of successes for our sample, while the value of \(n(1-p)\) gives the expected number of failures for our sample.
The mean of the sampling distribution of the sample proportion is \(\mu_{\widehat{p}} = p\). In other words, sample proportions are centered around the true proportion.
The standard deviation of the sampling distribution of the sample proportion is given by \(\sigma_{\widehat{p}} = \displaystyle{\sqrt{\frac{p\cdot(1-p)}{n}}}\). This quantifies how far sample proportions spread away from the true proportion.
If \(np \ge 10\) and \(n(1-p) \ge 10\), probability calculations using the Normal Probability Applet can be computed using the equation \[ \displaystyle {z = \frac{\textrm{value} - \textrm{mean}}{\textrm{standard deviation}} = \frac{\widehat p - p}{\sqrt{\frac{p \cdot (1-p)}{n}}}} \]
To convert a z-score into a probability, we enter the z-score into the Normal Probability Applet and shade in the typical way.
By the end of this lesson you should be able to do the following.
The estimator of \(p\) is \(\widehat p\). \(\displaystyle{ \widehat p = \frac {x}{n}}\) and is used for both confidence intervals and hypothesis testing.
You will use the Excel spreadsheet Math 221 Statistics Toolbox to perform hypothesis testing and calculate confidence intervals for problems involving one proportion.
The requirements for a confidence interval are \(n \widehat p \ge 10\) and \(n(1-\widehat p) \ge 10\). The requirements for hypothesis tests involving one proportion are \(np\ge10\) and \(n(1-p)\ge10\).
We can determine the sample size we need to obtain a desired
margin of error using the formula \(\displaystyle{ n=\left(\frac{z^*}{m}\right)^2
p^*(1-p^*)}\) where \(p^*\) is a
prior estimate of \(p\). If no prior estimate is available, the
formula \(\displaystyle{
\left(\frac{z^*}{2m}\right)^2}\) is used.
By the end of this lesson, you should be able to do the following.
When conducting hypothesis tests using two proportions, the null hypothesis is always \(p_1=p_2\), indicating that there is no difference between the two proportions. The alternative hypothesis can be left-tailed (\(<\)), right-tailed(\(>\)), or two-tailed(\(\ne\)).
For a hypothesis test and confidence interval of two proportions, we use the following symbols: \[ \begin{array}{lcl} \text{Sample proportion for group 1:} & \hat p_1 = \displaystyle{\frac{x_1}{n_1}} \\ \text{Sample proportion for group 2:} & \hat p_2 = \displaystyle{\frac{x_2}{n_2}} \end{array} \]
For a hypothesis test only, we use the following symbols:
\[ \begin{array}{lcl} \text{Overall sample proportion:} & \hat p = \displaystyle{\frac{x_1+x_2}{n_1+n_2}} \end{array} \]
Whenever zero is contained in the confidence interval of the difference of the true proportions we conclude that there is no significant difference between the two proportions.
You will use the Excel spreadsheet Math 221 Statistics
Toolbox to perform hypothesis testing and calculate confidence
intervals for problems involving two proportions.
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