Question

In a recent survey of 200 elementary students, many revealed they preferred math than English. Suppose that 80 of the students surveyed were girls and that 120 of them were boys. In the survey, 60 of the girls, and 80 of the boys said that they preferred math more.

Required:
a. Calculate an 80% confidence interval for the difference in proportions.
b. What is the standard error of the difference in the probability between that girls prefer math more and boys prefer math more?

1. 0.4097
2. 0.0042
3. 0.0833
4. 0.0647734

c. What is the difference in the probability between that girls prefer math more and boys prefer math more?

1. 0.0833
2. 0.5
3. 0.0042
4. 0.4097

Answer 1

Explanation:

a) Confidence interval for the difference in the two proportions is written as

Difference in sample proportions ± margin of error

Sample proportion, p= x/n

Where x = number of success

n = number of samples

For the girls,

x = 60

n1 = 80

p1 = 60/80 = 0.75

For the boys

x = 80

n2 = 120

p2 = 80/120 = 0.67

Margin of error = z√[p1(1 - p1)/n1 + p2(1 - p2)/n2]

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.8 = 0.2

α/2 = 0.2/2 = 0.1

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.1 = 0.9

The z score corresponding to the area on the z table is 1.282. Thus, z score for confidence level of 80% is 1.282

Margin of error = 1.282 × √[0.75(1 - 0.75)/80 + 0.67(1 - 0.67)/120]

= 1.282 × √0.00418625

= 0.081

Confidence interval = 0.75 - 0.67 ± 0.081

= 0.08 ± 0.081

b) The formula for determining the standard error of the distribution of differences in sample proportions is expressed as

Standard error = √{(p1 - p2)/[(p1(1 - p1)/n1) + p2(1 - p2)/n2}

Therefore,

Standard error = √{(0.

75 - 0.67)/[0.75(1 - 0.75)/80 + 0.67(1 - 0.67)/120]

Standard error = √0.08/0.00418625

Standard error = 4.37

c) the difference in the probability between that girls prefer math more and boys prefer math more is

0.75 - 0.67 = 0.08