Question

Using TI-83 Calculator Only . In 1999, American Psychological Association conducted a survey for a random sam- ple of psychologists to estimate mean incomes for psychologists of various types. Of the 10 clinical psychologists with 5–9 years experience who were in a medical psychological group practice, the mean income was $78, 500 with a standard deviation of $33, 287. Assuming that the data is normal, construct a 95% confidence interval for the population mean. Write your answer in English and find the margin of error.

Answer 1

Solution :

Given that,

= $78,500

s = $33,287

n = 10

Degrees of freedom = df = n - 1 = 10 - 1 = 9

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,9 = 2.262

Margin of error (E) = t_/2,df * (s /n)

= 2.262 * (33,287 / 10)

= 23810.43

The 95% confidence interval estimate of the population mean is,

- E < < + E

78,500 - 23,810.43 < < 78,500 + 23,810.43

54689.57 < < 102310.43

(54,689.57, 102,310.43)