Answer:
The width is [tex]w = 282.8[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n = 50
The population standard deviation is [tex]\sigma = \$ 1000[/tex]
The sample size is [tex]\= x = \$ 15,000[/tex]
Given that the confidence level is 90% then the level of significance can be mathematically represented as
[tex]\alpha = 100 - 90[/tex]
[tex]\alpha = 10 \%[/tex]
[tex]\alpha = 0.10[/tex]
Next we obtain the critical value of [tex]\frac{\alpha }{2}[/tex] from the normal distribution table, the value is
[tex]Z_{\frac{0.10 }{2} } = 1.645[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{0.10}{2} } * \frac{\sigma }{\sqrt{n} }[/tex]
substituting values
[tex]E = 1.645 * \frac{1000 }{\sqrt{50 }}[/tex]
=> [tex]E = 141.42[/tex]
The width of the 90% confidence level is mathematically represented as
[tex]w = 2 * E[/tex]
substituting values
[tex]w = 2 * 141.42[/tex]
[tex]w = 282.8[/tex]
