In a simple random sample of 144 households in a city in Kentucky, the average number of children in these households was 1.12 children. The standard deviation from this sample was 2.40 children. A 90 percent confidence interval for the mean number of children in all households in this city is:
A) 2.40 ± 0.2.
B) 1.12 ± 0.2.
C) 1.12 ± 2.40.
D) 1.12 ± 0.328.
E) It is impossible to tell without a census.

Question

Quizplus · Accepted Answer

Using the formula for a confidence interval for a population mean with a known standard deviation:

CI = x̄ ± z* (σ/√n)

where:
x̄ = 1.12 (sample mean)
z* = 1.645 (z-value for 90% confidence interval)
σ = 2.40 (sample standard deviation)
n = 144 (sample size)

CI = 1.12 ± 1.645*(2.40/√144)
CI = 1.12 ± 0.328

Therefore, the 90% confidence interval for the mean number of children in all households in this city is 1.12 ± 0.328 children, which is equivalent to (0.792, 1.448) children. Therefore, the best choice is D.

In a Simple Random Sample of 144 Households in a City