A large company wishes to estimate the average number of sic

A large company wishes to estimate the average number of sick days that

an employee uses each month. Suppose the company knows the standard deviation of the

number of sick days is 0.7

(a) If the company uses a random sample of 49 employees, with what confidence can they

assert their estimate will be off by no more than 0.252?

(b) Find a 95% confidence interval for the mean number of sick days used in a month if a

random sample of size 36 produced a sample mean of 3.

Solution

a)
ME = z*s/sqrt(n)
Then the z score for the confidence is
z = E*sqrt(n)/s = 0.252*sqrt(49)/0.7
z = 2.52
Thus, getting the right tailed area of this, and multiplying by 2,
alpha = 0.011735483
Thus, the confidence level is the complement of this,
confidence level = 0.988 or 98.826%

b)
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=3
Standard deviation( sd )=0.7
Sample Size(n)=36
Confidence Interval = [ 3 ± t a/2 ( 0.7/ Sqrt ( 36) ) ]
= [ 3 - 2.0301 * (0.117) , 3 + 2.0301 * (0.117) ]
= [ 2.763,3.237 ]