Question

In a random sample of 15 senior-level chemical engineers, the mean annual earnings was 122050 and the standard deviation was 34680. Assume the annual earnings are normally distributed and construct a 95% confidence interval for the population mean annual earnings for senior-level chemical engineers.

1. The critical value: 2. The standard error of the sample mean: 3. The margin of error: 4. The lower limit of the interval: 5. The upper limit of the interval:

Answer 1

solution:-

given that mean = 122050 , standard deviation = 34680

and n = 15

1. critical value

degree of freedom df = n-1 = 15 - 1 = 14

the value of 95% confidence with df from t table

critical value = 2.145

2. standard error

formula

=> standard deviation/sqrt(n)

=> 34680/sqrt(15)

=> 8954.3375

3. margin of error

formula

=> t * standard deviation/sqrt(n)

=> 2.145 * 34680/sqrt(15)

=> 19207.0539

4. lower limit

=> mean - margin of error

=> 122050 - 19207.0539

=> 102842.9461

5. upper limit

mean + margin of error

=> 122050 + 19207.0539

=> 141257.0539

note:- i have rounded all values to four decimal places