Question

3. A political candidate wants to estimate the mean income in her district. If the standard deviation for incomes is known to be $10,000, how large a sample must she take if she wishes to be 99% certain that her estimate is within $2000 of the true mean?

4. A company that manufactures golf clubs wants to estimate the proportion of golfers who are left-handed. How large a sample must they take if they want to be 90% certain that their estimate is within 6% of the true proportion?

Answer 1

3.

Solution :

Given that,

standard deviation =s =   =10000

Margin of error = E = 2000

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.58

sample size = n = [Z/2* / E] 2

n = ( 2.58* 10000/ 2000 )2

n =166.41=167

Sample size = n =167 rounded

4.

Solution :

Given that,

= 0.5

1 - = 0.5

margin of error = E = 6% = 0.06

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 / 0.06)2 * 0.5 * 0.5

=188

Sample size = 188