Respuesta :
Answer:
Following are the answer to this question:
Step-by-step explanation:
Given:
n = 30 is the sample size.
The mean [tex]\bar X[/tex] = 7.3 days.
The standard deviation = 6.2 days.
df = n-1
[tex]= 30-1 \\ =29[/tex]
The importance level is [tex]\alpha[/tex] = 0.10
The table value is calculated with a function excel 2010:
[tex]= tinv (\ probility, \ freedom \ level) \\= tinv (0.10,29) \\ =1.699127\\ = t_{al(2x-1)}= 1.699127[/tex]
The method for calculating the trust interval of 90 percent for the true population means is:
Formula:
[tex]\bar X - t_{al 2,x-1} \frac{S}{\sqrt{n}} \leq \mu \leq \bar X+ t_{al 2,x-1} \frac{S}{\sqrt{n}}[/tex]
[tex]=\bar X - t_{0.5, 29} \frac{6.2}{\sqrt{30}} \leq \mu \leq \bar X+ t_{0.5, 29} \frac{6.2}{\sqrt{30}}\\\\=7.3 -1.699127 \frac{6.2}{\sqrt{30}}\leq \mu \leq7.3 +1.699127 \frac{6.2}{\sqrt{30}}\\\\=7.3 -1.699127 (1.13196)\leq \mu \leq7.3 +1.699127 (1.13196) \\\\=5.37 \leq \mu \leq 9.22 \\[/tex]
It can rest assured that the true people needs that middle managers are unavailable from 5,37 to 9,23 during the years.