Question

6.5 Of 100 customers who received promotional materials for a marketing campaign, 10 responded to the promotion. For the following confidence levels, construct a confidence interval for the population proportion who would respond to the promotion.

1. (a) 90%

2. (b) 95%

3. (c) 99%

4. (d) For each of the confidence intervals above, calculate and interpret the margin

   of error.

5. (e) Refer to part (d) above. Describe the relationship between margin of error and

   confidence level.

Answer 1

6.5)

Solution :

Given that,

n = 100

x = 10

Point estimate = sample proportion = = x / n = 10 / 0.100 = 0.100

1 - = 1 - 0.100 = 0.900

(a)

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 * (((0.100 * 0.900) / 100/10)

= 0.049

Margin of error = E = 0.049

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.100 - 0.049 < p < 0.100 + 0.049

0.051 < p < 0.149

(0.051 , 0.149)

(b)

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 * (((0.100 * 0.900) / 100/10)

= 0.059

Margin of error = E = 0.059

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.100 - 0.059 < p < 0.100 + 0.059

0.041 < p < 0.159

(0.014 , 0.159)

(c)

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * (((0.100 * 0.900) / 100/10)

= 0.077

Margin of error = E = 0.077

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.100 - 0.077 < p < 0.100 + 0.077

0.023 < p < 0.177

(0.023 , 0.177)

d) See margin of error above and e) See the relation of andmargin of error