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An automotive manufacturer wants to know the proportion of new car buyers who prefer foreign cars over domestic. Step 2 of 2 : Suppose a sample of 1418 new car buyers is drawn. Of those sampled, 354 preferred foreign over domestic cars. Using the data, construct the 99% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars. Round your answers to three decimal places.

Respuesta :

Answer: 99% confidence interval would be (0.13,0.26).

Step-by-step explanation:

Since we have given that

Sample size n = 1418

x = 354

So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{354}{1418}=0.249[/tex]

At 99% level of significance, z = 2.58

So, interval would be

[tex]\hat{p}\pm z\sqrt{\dfrac{p(1-p)}{n}}}\\=0.249\pm 2.58\times \sqrt\dfrac{0.249\times (1-0.249)}{1418}}}\\\\=0.249\pm 0.01148\\\\=(0.249-0.01148,0.249+0.01148)\\\\=(0.13,0.26)[/tex]

Hence, 99% confidence interval would be (0.13,0.26).

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