A regional hardware chain is interested in estimating the proportion of their customers who own their...

A regional hardware chain is interested in estimating the proportion of their customers who own their own homes. There is some evidence to suggest that the proportion might be around 0.70. Given this, what sample size is required if they wish a 90 percent confidence level with a margin of error of ± .025?

About 355

Almost 1,300

Approximately 910

100

Solutions

Expert Solution

Solution :

Given that,

= 0.70

1 - = 1 - 0.70 = 0.30

margin of error = E = 0.025

At 90% confidence level

= 1 - 90%

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.645 / 0.025)2 * 0.70 * 0.30

=910

Sample size = 910

orchestra answered 1 year ago

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