Question

Almost all employees working for financial companies in New York City receive large bonuses at the end of the year. A sample of 62 employees selected from financial companies in New York City showed that they received an average bonus of $47,000 last year with a standard deviation of $15,000. Construct a 90% confidence interval for the average bonus that all employees working for financial companies in New York City received last year.

Answer 1

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

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
`1 - (1 - 0.9)/(2) = 0.95`. So we have T = 1.67

The margin of error is:

`M = T(s)/(√(n)) = 1.67(15000)/(√(62)) = 3181`

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