Answer:
The width is [tex]w = \$ 729.7[/tex]
Step-by-step explanation:
From the question we are told that
The population standard deviation is [tex]\sigma = \% 1,000[/tex]
The sample size is [tex]n = 50[/tex]
The sample mean is [tex]\= x = \$ 15,000[/tex]
Given that the confidence level is 99% then the level of significance is mathematically represented as
[tex]\alpha = 100 - 99[/tex]
=> [tex]\alpha = 1\%[/tex]
=> [tex]\alpha = 0.01[/tex]
Next we obtain the critical value of [tex]\frac{\alpha }{2}[/tex] from the normal distribution table, the value is
[tex]Z_{\frac{\alpha }{2} } = Z_{\frac{0.01 }{2} } = 2.58[/tex]
Generally margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} * \frac{\sigma }{\sqrt{n} }[/tex]
substituting values
[tex]E = 2.58 * \frac{1000 }{\sqrt{50} }[/tex]
[tex]E = 2.58 * \frac{1000 }{\sqrt{50} }[/tex]
[tex]E = 364.9[/tex]
The width of the 99% confidence interval is mathematically evaluated as
[tex]w = 2 * E[/tex]
substituting values
[tex]w = 2 * 364.9[/tex]
[tex]w = \$ 729.7[/tex]
