Question

Suppose that the salaries, in thousands of dollars, of the managers for a industry are normally distributed with an unknown mean and standard deviation. The salaries of 26 randomly sampled managers are used to estimate the mean of the population. What t-score should be used to find the 95% confidence interval for the population mean

Answer 1

Answer: t _(α/2) = 2.060

Step-by-step explanation:

Given that;

sample size n = 26

confidence interval = 95%

level of significance α = ( 1 - 95/100 ) = 0.05

but from the question the salaries (26) are use to estimate the mean of the population so, its a two tail test

therefore our level if significance α will be divided by 2 = 0.05/2 = 0.025

now our degree of freedom df = 26 - 1 = 25

FROM standard t-table value

the critical value of t for the level of significance 0.025

and 25 degree of freedom is 2.060

therefore, t _(α/2) = 2.060