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. = ~ N(0,1)
where, = sample proportion of car buyers who preferred foreign over domestic cars = = 0.27
n = sample of new car buyers = 390
p = population proportion
<em>Here for constructing 80% confidence interval we have used One-sample z proportion statistics.</em>
<em />
<u>So, 80% confidence interval for the population proportion, p is ;</u>
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 < < 1.2816) = 0.80
P( < < ) = 0.80
P( < p < ) = 0.80
<u>80% confidence interval for p</u> = [ , ]
= [ , ]
= [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].