An automotive manufacturer wants to know the proportion of new car buyers who prefer foreign cars over domestic. In an earlier study, the population proportion was estimated to be 0.33. How large a sample would be required in order to estimate the fraction of new car buyers who prefer foreign cars at the 90% confidence level with an error of at most 0.02?

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rodolforodriguezr rodolforodriguezr · Answer 1 · 2019-11-16T23:09:21+01:00

Answer:

1,496 new car buyers

Step-by-step explanation:

The sample size n in Simple Random Sampling is given by

[tex] \bf n=\frac{z^2p(1-p)}{e^2}[/tex]

where

z = 1.645 is the critical value for a 90% confidence level (*)

p= 0.33 is the population proportion.

e = 0.02 is the margin of error

so

[tex] \bf n=\frac{(1.645)^20.33*0.67}{0.02^2}=1,495.76\approx 1,496[/tex]

(*)This is a point z such that the area under the Normal curve N(0,1) inside the interval [-z, z] equals 90% = 0.9

It can be obtained in Excel or OpenOffice Calc with

NORMSINV(0.95)

adefioyelaoye adefioyelaoye · Answer 2 · 2022-04-01T15:55:58+02:00

1,496 new car buyers prefer foreign cars with an error of at most 0.02.

This is defined as part of the sampling technique in which each sample has an equal probability of being chosen.

Sample size n = z²p (1-p)

where

z = 1.645 is the critical value for a 90% confidence level.

p= 0.33 is the population proportion.

e = 0.02 is the margin of error.

n = (1.645) ˣ 0.33 ˣ 0.67 / 0.02²

= 1,496.

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