Answer:
The 90% confidence interval is [tex]\$ \ 30313.9< \mu < \$ \ 33686.13[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n = 64
The sample mean is [tex]\= x = \$ 32, 000[/tex]
The standard deviation is [tex]\sigma= \$ 8, 200[/tex]
Given that the confidence interval is 90% then the level of significance is mathematically evaluated 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{\alpha }{2} } = 1.645[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{ \sigma }{ \sqrt{n} }[/tex]
=> [tex]E = 1.645 * \frac{ 8200 }{ \sqrt{64} }[/tex]
=> [tex]E = 1686.13[/tex]
The 90% confidence interval is mathematically represented as
[tex]\= x - E < \mu < \= x + E[/tex]
=> [tex]32000 - 1689.13 < \mu < 32000 + 1689.13[/tex]
=> [tex]\$ \ 30313.9< \mu < \$ \ 33686.13[/tex]
