Question

Suppose that 590 people were surveyed and a 95% confidence interval for the mean spending amount by U.S. citizens on summer vacations was calculated to be ($1980, $2040).
a. If a 99% confidence interval was calculated instead, would it be wider or narrower? __________________
b. If 1,000 people were sampled instead of 590, would the interval be wider or narrower? ___________________
c. BONUS: what is the margin of error for this confidence interval? ______________

Answer 1

Solution :

n = 1000

x = 590

= x / n = 590 / 1000 = 0.590

1 - = 1 - 0.590 = 0.410

a ) At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z0.05 = 1.645

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

= 1.645 * (((0.590 * 0.410) / 1000)

= 0.030

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.590 - 0.030< p < 0.590 + 0.030

0.560 < p < 0.2620

b ) At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576

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

= 2.576 * (((0.590 * 0.410) / 1000)

= 0.040

A 99% confidence interval for population proportion p is ,

- E < P < + E

0.590 - 0.040 < p < 0.590 + 0.040

0.550 < p < 0.630
c ) 99% confidence interval is wider Confidence level increased  the margin of error is increased