Suppose a sample of 510 new car buyers is drawn. Of those sampled, 142 preferred foreign over domestic cars. Using the data, construct the 85% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars. Round your answers to three decimal places.

(0.224, 0.332)

Step-by-step explanation: from the question, the total population is 510 and 142 out of this population prefers foreign cars to domestic ones.

Hence the sample size (n) is 142.

Sample proportion (p) = 142/ 510 = 0.278

q = 1 - 0.278 = 0.722.

We are to construct a 85% confidence interval for sample proportion and this is given by the formulae below.

P = p + Zα/2×(√(pq/n) For upper limit

P = p - Zα/2×(√(pq/n) For lower limit

We are using a z test to get our critical value because sample size is greater than 30 ( n = 142).

The value of Zα/2 from the standard normal distribution table is 1.44 ( this is done at a 15% level of significance).

By substituting the parameters, we have that

For upper limit

P = 0.278 + 1.44 × (√(0.278×0.722/142)

P = 0.278 + 1.44(0.0375)

P = 0.278 + 0.054

P = 0.332

For lower limit

P = 0.278 - 1.44 × (√(0.278×0.722/142)

P = 0.278 - 1.44(0.0375)

P = 0.278 - 0.054

P = 0.224.

Hence the 85% confidence interval for population proportion is (0.224, 0.332)

no

step-by-step explanation:

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