A random sample of 20 starting salaries in a certain field found an average of $23 per hour, and a standard deviation of $3. Construct a 95% confidence interval for the true starting salary in this field.
Solution :
degrees of freedom = n - 1 = 20 - 1 = 19
t/2,df = t_{0.025 , 19 = 2.093}
Margin of error = E = t_{/2,df} * (s /n)
= 2.093 * ( 3 / 20)
Margin of error = E = 1.40
The 95% confidence interval estimate of the population mean is,
± E
23 ± 1.40
(21.6 , 24.4)
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