Answer:
The correct option is A
Step-by-step explanation:
From the question we are told that
The average number of meetings hours per week is [tex]\mu= 18 \ hours[/tex]
The standard deviation is [tex]\sigma = 5.2 \ hours[/tex]
The sample size is n= 35
The sample average per week is [tex]p = 16.8 \ hours[/tex]
From each solution statement we can deduce that the confidence level is
[tex]t = 95[/tex]%
Thus the significance level is [tex]\alpha = 0.05[/tex]= 5%
The z value for the significance level is gotten as 1.96 from the z-table
The confidence level interval for the sample mean is mathematically evaluated as
[tex]\= x = \mu \pm (1.96 * \frac{\sigma }{\sqrt{n} } )[/tex]
Sustituting values
[tex]\= x = 18 \pm (1.96 * \frac{5.2 }{\sqrt{35} } )[/tex]
[tex]\= x = 18 \pm1.7[/tex]
=> [tex]18 - 1.7 < \= x < 18 +1.7[/tex]
[tex]16.3 < \= x < 19.7[/tex]
