Question

Alma is estimating the proportion of students in her school district who, in the past month, read at least 1 book. From a random sample of 50 students, she found that 32 students read at least 1 book last month. Assuming all conditions for inference are met, which of the following defines a 90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month?

a.
`32 \pm 1.645\sqrt{((32)(18))/(50)}`b.
`32 \pm 1.96\sqrt{((32)(18))/(50)}`c.
`0.64 \pm 1.282\sqrt{((0.64)(0.36))/(50)}`d.
`0.64 \pm 1.645\sqrt{((0.64)(0.36))/(50)}`e.
`0.64 \pm 1.96\sqrt{((0.64)(0.36))/(50)}`

Answer 1

Answer: D

Explanation:

Confidence interval is written as

Sample proportion ± margin of error

Margin of error = z × √pq/n

Where

z represents the z score corresponding to the confidence level

p = sample proportion. It also means probability of success

q = probability of failure

q = 1 - p

p = x/n

Where

n represents the number of samples

x represents the number of success

From the information given,

n = 50

x = 32

p = 32/50 = 0.64

q = 1 - 0.64 = 0.36

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

α = 1 - 0.9 = 0.1

α/2 = 0.01/2 = 0.05

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

1 - 0.05 = 0.95

The z score corresponding to the area on the z table is 2.05. Thus, confidence level of 90% is 1.645

Therefore, the 90% confidence interval is

0.64 ± 1.645√(0.64)(0.36)/50