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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

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 class=

Sagot :

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

[tex]Confidence\text{ }interval=p\pm z\cdot\sqrt{\frac{p(p-1)}{n}}[/tex]

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:

[tex]\begin{gathered} \frac{879}{288}=\frac{100\%}{x} \\ . \\ x=100\%\cdot\frac{288}{879} \\ . \\ x=32.765\% \end{gathered}[/tex]

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

[tex]p=\frac{32.765}{100}=0.32765[/tex]

Now, we can use the formula:

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

[tex]\begin{gathered} Lower\text{ }endpoint=0.32765-0.031028=0.296616 \\ Upper\text{ }endpoint=0.32765+0.031028=0.35867 \end{gathered}[/tex]

Thus, the answer is:

Lower endpoint: 0.297

Upper endpoint: 0.359

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