Question

In 2018 a random sample of 1,200 American adults was asked the question, “What do you think is the ideal number of children for a family to have?” Results showed that 40% of respondents said three or more children is ideal. I. Calculate a 95% confidence interval for the proportion of Americans who think that three or more children is the ideal family size. The multiplier for a 95% confidence interval is 1.960. Give your answer to 4 decimal places. II. A 90% confidence interval produced from the same survey results would be (i) narrower (ii) wider (iii) the same width as the interval computed in part I.

Answer 1

Solution :

Given that,

n = 1200

Point estimate = sample proportion = = 0.40

1 - = 1 - 0.40 = 0.60

  At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

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

= 1.96 * (((0.40 * 0.60) / 1200)

= 0.0277

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.40 - 0.0277 < p < 0.40 + 0.0277

0.3723 < p < 0.4277

The 95% confidence interval for the population proportion p is : (0.3723 , 0.4277) .

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z _0.05 = 1.645

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

= 1.645 * (((0.40 * 0.60) / 1200)

= 0.0233

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.40 - 0.0233 < p < 0.40 + 0.0233

0.3767 < p < 0.4233

The 90% confidence interval for the population proportion p is : (0.3767 , 0.4233) .

A 90% confidence interval produced from the same survey results would be narrower .

Because if we decrease the confidence level the confidence interval is narrower .

(i) narrower