Question

A sample of 25 different payroll departments found that the employees worked an average of 310.3 days a year with a standard deviation of 23.8 days.

What is the 90% confidence interval for the average days worked by employees in all payroll departments?
A. 301.0 < < 319.6
B. 298.0 < < 322.6
C. 302.2 < < 318.4
D. 314.1 < < 316.8

Answer 1

the average daus worked ny employees is C. 302.2 < < 318.4

Step-by-step explanation:

Hi, since we need the 90% confidence interval, we have to cut 5% in both sides of the normal distribution graph (which accounts for 100%), therefore, the formula to use for both, the upper and lower limit of the interval is as follows.

`UpperLimit=Mean+1.65*(StandardDeviation)/(√(SampleSize) )`

`LowerLimit=Mean-1.65*(StandardDeviation)/(√(SampleSize) )`

Where 1.65 is the Z-score for the probability=0.05 and -1.65 is the Z-score for the probability = 95%

So, things should look like this

`Upper Limit==310.3+1.65*(23.8)/(√(25) ) =318.2`

`LowerLimit==310.3-1.65*(23.8)/(√(25) ) =302.4`

Therefore, the interval is 302.4 << 318.2 and since it is close to answer C. we selected answer C.

Best of luck.