Respuesta :
Answer:
80% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is [0.240 , 0.298].
Step-by-step 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. = [tex]\frac{\hat p -p}{\sqrt{\frac{\aht p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of car buyers who preferred foreign over domestic cars = [tex]\frac{105}{390}[/tex] = 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 < [tex]\frac{\hat p -p}{\sqrt{\frac{\aht p(1-\hat p)}{n} } }[/tex] < 1.2816) = 0.80
P( [tex]-1.2816 \times {\sqrt{\frac{\aht p(1-\hat p)}{n} } }[/tex] < [tex]{\hat p -p}[/tex] < [tex]1.2816 \times {\sqrt{\frac{\aht p(1-\hat p)}{n} } }[/tex] ) = 0.80
P( [tex]\hat p-1.2816 \times {\sqrt{\frac{\aht p(1-\hat p)}{n} } }[/tex] < p < [tex]\hat p+1.2816 \times {\sqrt{\frac{\aht p(1-\hat p)}{n} } }[/tex] ) = 0.80
80% confidence interval for p = [[tex]\hat p-1.2816 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] , [tex]\hat p+1.2816 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex]]
= [ [tex]\frac{105}{390} -1.2816 \times {\sqrt{\frac{\frac{105}{390}(1-\frac{105}{390})}{390} } }[/tex] , [tex]\frac{105}{390} +1.2816 \times {\sqrt{\frac{\frac{105}{390}(1-\frac{105}{390})}{390} } }[/tex] ]
= [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].
