Dave's Web Corner

Introductory Statistics (openstax.org)

Section 8.3 #118

• Suppose that the car insurance companies did do a survey. They randomly selected \(400\) drivers and found that \(320\) claimed they 'always buckled up'. We are interested in the population proportion of drivers who claim they 'always buckle up'.
  1. For this survey, state
     1. \(x\)
     2. \(n\)
     3. \(p'\)
  2. Define the random variables \(X\) and \(P'\), in words.
  3. Which distribution should you use for this problem? Explain your choice.
  4. Construct a \(95\%\) confidence interval for the population proportion who claim they 'always buckle up'.
     1. State the confidence interval.
     2. Sketch the graph.
     3. Calculate the error bound.
  5. If this survey were done by telephone, list three difficulties the companies might have in obtaining random results.

Solution

1. In this survey:
   1. \(x=320\)
   2. \(n=400\)
   3. \(p'=320/400=0.8000\)
2. The random variable \(X\) is "the number of drivers, in a simple random sample of \(400\) drivers, who say that they 'always buckle up'." The random variable \(P'\) is "the proportion of drivers, in a simple random sample of \(400\) drivers, who say that they 'always buckle up'."
3. Since we are interested in the population proportion of drivers who say that they 'always buckle up', we are going to use the distribution of sample proportions \(P'\).
4. We are going to approximate \(P'\) with the normal distribution \(N(p',\sqrt{p'q'/n})\), where \(p'=0.8\), \(q'=0.2\), and \(n=400\). The confidence level is \(0.95\), so \(\alpha=0.05\), \(\alpha/2=0.025\), and \(z_{\alpha/2}=\mbox{invNorm}(0.975,0.8,\sqrt{(0.8)(0.2)/400})\approx 0.8392\) \begin{equation} \begin{split} EBP & = z_{\alpha/2}\cdot\sqrt{\frac{p'q'}{n}}\\ & \approx (0.8392)\cdot\sqrt{\frac{(0.8)(0.2)}{400}} \\ & \approx 0.0168 \end{split} \end{equation}
   1. The confidence interval \((p'-EBP,p'+EBP)\) is \((0.7832,0.8168)\).
   2. The graph (constructed at desmos.com)
      
   3. The Error Bound for the Porportion, \(EBP\approx 0.0168\), is calculated above
5. A significant problem is that it such a survey would be a convenience sample. The survey might also require many phone calls to reach \(400\) responses. The survey is also recording what people say they do. It might be more valid to observe what people do instead of allowing them to self-report.