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 in…

Question

asked Oct 24, 2021 186k views

1 Answer

TRosenflanz · Answer 1 · 2021-10-29T22:47:01+0000

Answer:

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 = 364.9

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

w = 2 * E

substituting values

w = 2 * 364.9

$w = \$ 729.7$