Question

An economist is interested in studying the spending habits of consumers in a particular region. The population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average expense of $15,000. What is the width of the 99% confidence interval for the mean of expense? a. 364.28 b. 728.55 c. 329.00 d. 657.99

Answer 1

The width is
`w = \$ 729.7`

Explanation:

From the question we are told that

The population standard deviation is
`\sigma = \% 1,000`

The sample size is
`n = 50`

The sample mean is
`\= x = \$ 15,000`

Given that the confidence level is 99% then the level of significance is mathematically represented as

`\alpha = 100 - 99`

=>
`\alpha = 1\%`

=>
`\alpha = 0.01`

Next we obtain the critical value of
`(\alpha )/(2)` from the normal distribution table, the value is

`Z_{(\alpha )/(2) } = Z_{(0.01 )/(2) } = 2.58`

Generally margin of error is mathematically represented as

`E = Z_{(\alpha )/(2) * (\sigma )/(√(n) )`

substituting values

`E = 2.58 * (1000 )/(√(50) )`

`E = 2.58 * (1000 )/(√(50) )`

`E = 364.9`

The width of the 99% confidence interval is mathematically evaluated as

`w = 2 * E`

substituting values

`w = 2 * 364.9`

`w = \$ 729.7`