Question

Suppose a sample of 879 new car buyers is drawn. Of those sampled, 288 preferred foreign over domestic cars. Using the data construct a 95% confidence interval for the population proportion of new car buyers who prefer for foreign cars over domestic cars. Round your answers to three decimal places

Answer 1

To find the confidence interval for a proportion, we use the following formula:

`Confidence\text{ }interval=p\pm z\cdot\sqrt{(p(p-1))/(n)}`

Where:

p is the sample proportion

z the chosen z-value

n sample size

Since we want to make a confidence interval of 95%, we need to use z = 1.96. The sample size is n = 879.

We can use cross multiplication to find p, which is the percentage of the total sample size that preferred foreign cars:

`\begin{gathered} (879)/(288)=(100\%)/(x) \\ . \\ x=100\%\cdot(288)/(879) \\ . \\ x=32.765\% \end{gathered}`

p is the proportion in decimal, we need to divide by 100:

`p=(32.765)/(100)=0.32765`

Now, we can use the formula:

`Confidence\text{ }interval=0.32765\pm1.96\sqrt{(0.32765(1-0.32765))/(879)}=0.32765\pm0.031028`

`\begin{gathered} Lower\text{ }endpoint=0.32765-0.031028=0.296616 \\ Upper\text{ }endpoint=0.32765+0.031028=0.35867 \end{gathered}`

Thus, the answer is:

Lower endpoint: 0.297

Upper endpoint: 0.359