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American Psychological Association conducted a survey for a random sample of psychologists to estimate mean incomes...

American Psychological Association conducted a survey for a random sample of psychologists to estimate mean incomes for psychologists of various types. Of the 10clinical psychologists with 5–9 years of 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.

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Answer #1

Solution :

sample size = n = 10

Degrees of freedom = df = n - 1 = 9

t /2,df = 2.262

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

= 2.262 * ( 33287/ 10)

Margin of error = E = 23810

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

- E < < + E

78500 - 23810 < < 78500 + 23810

54690 < < 102310‬

(54,690, 102,310‬)

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