The American Psychological Associated conducted a survey in 2003 of psychologists to estimate the average incomes for psychologists based on their academic degree and years of experience. 36 psychologists responded who had a just received their masters degree in 2003. The average was $48,800 with a standard deviation of $16,800.

The 95% Confidence Interval is ($, $ )

Answer:

at 95%, confidence interval is ( $43312, $54288 )

Step-by-step explanation:

Given the data in the question;

sample size n = 36

mean x' = 48,800

standard deviation σ = 16,800

at 95% confidence level;

∝ = 1 - 95% = 1 - 0.95 = 0.05

∝/2 = 0.05 / 2 = 0.025

z critical value, $Z_{\alpha/2$ = 1.96

now, confidence interval for population mean is given by the formula

x' ± $Z_{\alpha/2$ ( σ/√n )

so we substitute

⇒ 48800 ± 1.96( 16800/√36 )

⇒ 48800 ± 1.96( 2800 )

⇒ 48800 ± 5488

⇒ 48800 - 5488, 48800 + 5488

⇒ ( 43312, 54288 )

Therefore, at 95%, confidence interval is ( $43312, $54288 )

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