A manufacturer wishes to estimate the proportion of washing machines leaving the factory that is defective. How large a sample should she check in order to be 90% that the true …

Question

asked Jul 9, 2020 104k views

1 Answer

Nathansizemore · Answer 1 · 2020-07-14T01:37:26+0000

Given:

ME = 1.5% = 0.015

P = 0.5 ( maximises P)

Answer and Step-by-step explanation:

Now,

Margin of Error (ME) is given by the formula:

$ME = t* \ sigma$

where,

t = critical value = 1.645

$\sigma$ = standard deviation

So, squaring both sides and then rearranging:

$ME^(2) = t^(2)* {(P(1 - P))/(n)}$

$n = t^(2)* {(P(1 - P))/(ME^(2))}$

n = (1.645)^{2}\times {\frac{0.5(1 - 0.5)}{0.015^{2}}}

n = 3006.69

n = 3007 (approx)