A recent survey of 64 executives who were laid off from thei

A recent survey of 64 executives who were laid off from their previous position revealed that it took them 25 weeks on average to find another position. The standard deviation of the sample was 6.2 weeks. I can assert, with 99% confidence, the an average time for a laid-off executive to find a new job is at most A weeks. What would be the smallest value of A that I can claim?

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Solution

CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=25
Standard deviation( sd )=6.2
Sample Size(n)=64
Confidence Interval = [ 25 ± Z a/2 ( 6.2/ Sqrt ( 64) ) ]
= [ 25 - 2.58 * (0.775) , 25 + 2.58 * (0.775) ]
= [ 23.001,27 ]
the smallest value of A would be 23.001