Question

In Mississippi, a random sample of 110 first graders revealed that 48 of them have been to a dentist at least once.

(a) Find a 95% confidence interval for the population proportion of first-graders in Mississippi who have been to a dentist at least once.
Enter the smaller number in the first box.

A poll is taken in which 367367 out of 600600 randomly selected voters indicated their preference for a certain candidate. Find a 8080% confidence interval for pp.
_____________to ___________________

Answer 1

Solution :

a )Given that

n = 110

x = 48

= x / n = 48 / 110=0.436

1 - = 1 - 0.436 = 0.564

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.960

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

=1.960 * (((0.436 * 0.564) / 110)

= 0.093

A 95 % confidence interval for population proportion p is ,

- E < P < + E

0.436 - 0.093 < p < 0.436 + 0.093

0.343 < p < 0.529

b ) Given that

n = 600

x = 367

= x / n = 367 / 600=0.612

1 - = 1 - 0.612 = 0.388

At 80% confidence level the z is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

Z/2 = Z 0.10 = 1.280

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

=1.280 * (((0.612 * 0.388) / 600)

= 0.025

A 80 % confidence interval for population proportion p is ,

- E < P < + E

0.612 - 0.025 < p < 0.612 + 0.025

0.587 < p < 0.637