Question

US consumers are increasingly using debit cards as a substitute for cash and checks. From a sample of 100 consumers, the average amount annually spent on debit cards is $7,790. Assume that this average was based on a sample of 100 consumers and that the population standard deviation is $500.
A. At 99% confidence, what is the margin of error?
B. Construct the 99% confidence interval for the population mean amount spent annually on a debit card.

Answer 1

Answer:

A. Margin of error = 128.79

B. The 99% confidence interval for the population mean is (7661.21, 7918.79).

Step-by-step explanation:

We have to calculate a 99% confidence interval for the mean.

The population standard deviation is know and is σ=500.

The sample mean is M=7790.

The sample size is N=100.

As σ is known, the standard error of the mean (σM) is calculated as:

$\sigma_M=\dfrac{\sigma}{\sqrt{N}}=\dfrac{500}{\sqrt{100}}=\dfrac{500}{10}=50$

The z-value for a 99% confidence interval is z=2.576.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_M=2.576 \cdot 50=128.79$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 7790-128.79=7661.21\\\\UL=M+t \cdot s_M = 7790+128.79=7918.79$

The 99% confidence interval for the population mean is (7661.21, 7918.79).