Answer:
The 90% confidence interval for the average bonus that all employees working for financial companies in New York City received last year is between $43,819 and $50,181
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
T interval
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 62 - 1 = 61
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 61 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 1.67
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 1.67\frac{15000}{\sqrt{62}} = 3181[/tex]
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 47000 - 3181 = $43,819
The upper end of the interval is the sample mean added to M. So it is 47000 + 3181 = $50,181
The 90% confidence interval for the average bonus that all employees working for financial companies in New York City received last year is between $43,819 and $50,181
