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Suppose we want a 90% confidence interval for the average amount spent on books by freshmen in their first year at college. The amount spent has a normal distribution with standard deviation $16.

a) How large should the sample be if the margin of error is to be less than $2?

Respuesta :

Answer:

sample should be 172

Explanation:

given data

confidence interval = 90%

standard deviation =  $16

error is less than =  $2

solution

we know that here z value for 90% confidence interval is by the z table is

z value = 1.64

so here Margin of Error is express as

Margin of Error = z × [tex]\frac{SD}{\sqrt{n} }[/tex]    ...........................1

here SD is standard deviation  and n is no of sample

so put here value and we get

2 = 1.64 × [tex]\frac{16}{\sqrt{n} }[/tex]

n = 172.13

so sample should be 172

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