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An automotive manufacturer wants to know the proportion of new car buyers who prefer foreign cars over domestic. Suppose a sample of 390 new car buyers is drawn. Of those sampled, 105 preferred foreign over domestic cars. Using the data, construct the 80% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars. Round your answers to three decimal places.

User Malyngo
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Answer:

80% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is [0.240 , 0.298].

Explanation:

We are given that an automotive manufacturer wants to know the proportion of new car buyers who prefer foreign cars over domestic.

Suppose a sample of 390 new car buyers is drawn. Of those sampled, 105 preferred foreign over domestic cars.

Firstly, the pivotal quantity for 80% confidence interval for the population proportion is given by;

P.Q. =
\frac{\hat p -p}{\sqrt{(\aht p(1-\hat p))/(n) } } ~ N(0,1)

where,
\hat p = sample proportion of car buyers who preferred foreign over domestic cars =
(105)/(390) = 0.27

n = sample of new car buyers = 390

p = population proportion

Here for constructing 80% confidence interval we have used One-sample z proportion statistics.

So, 80% confidence interval for the population proportion, p is ;

P(-1.2816 < N(0,1) < 1.2816) = 0.80 {As the critical value of z at 10% level

of significance are -1.2816 & 1.2816}

P(-1.2816 <
\frac{\hat p -p}{\sqrt{(\aht p(1-\hat p))/(n) } } < 1.2816) = 0.80

P(
-1.2816 * {\sqrt{(\aht p(1-\hat p))/(n) } } <
{\hat p -p} <
1.2816 * {\sqrt{(\aht p(1-\hat p))/(n) } } ) = 0.80

P(
\hat p-1.2816 * {\sqrt{(\aht p(1-\hat p))/(n) } } < p <
\hat p+1.2816 * {\sqrt{(\aht p(1-\hat p))/(n) } } ) = 0.80

80% confidence interval for p = [
\hat p-1.2816 * {\sqrt{(\hat p(1-\hat p))/(n) } } ,
\hat p+1.2816 * {\sqrt{(\hat p(1-\hat p))/(n) } }]

= [
(105)/(390) -1.2816 * {\sqrt{((105)/(390)(1-(105)/(390)))/(390) } } ,
(105)/(390) +1.2816 * {\sqrt{((105)/(390)(1-(105)/(390)))/(390) } } ]

= [0.240 , 0.298]

Therefore, 80% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is [0.240 , 0.298].

User Drublic
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