Question

1)Your master's dissertation concerns recent immigrants' experiences of severe financial strain. A large sample of 960 residents of a medium-sized city indicated that 80% of new immigrants felt that they were extremely stressed financially (which was higher than what is average for the general population). If you were to construct a 99% confidence interval for your finding, what would be the LOWER LIMIT of the interval? (Write as a proportion to 3 decimal places, rather than a percentage)

2) This question is similar to EOC 7.34. Your master's dissertation concerns recent immigrants' experiences of severe financial strain. A large sample of 960 residents of a medium-sized city indicated that 80% of new immigrants felt that they were extremely stressed financially (which was higher than what is average for the general population.) If you were to construct a 99% confidence interval for your finding, what would be the UPPER LIMIT of the interval (written as a proportion to 3 decimal places, rather than a percentage)?

Answer 1

Solution :

Given that,

n = 960

Point estimate = sample proportion = =0.80

1 - = 1 - 0.80=0.20

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

= 0.01

Z = Z_0.01= 2.326

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.326 (((0.80*0.20) /960 )

= 0.030

A 99% confidence interval for population proportion p is ,

- E

0.80 - 0.030

0.770

answer =0.770

(B)

Solution :

Given that,

n = 960

Point estimate = sample proportion = =0.80

1 - = 1 - 0.80=0.20

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

= 0.01

Z = Z_0.01= 2.326

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.326 (((0.80*0.20) /960 )

= 0.030

A 99% confidence interval for population proportion p is ,

+E

0.80 + 0.030

0.830

answer =0.830