Respuesta :
Answer:
The absolute value is [tex]|z| = 1.92[/tex]
Step-by-step explanation:
From the question we are told that
The mean is [tex]\mu = 20.1[/tex]
The standard deviation is [tex]\sigma = 1.4[/tex]
For the null hypothesis [tex]H_0[/tex]
The mean remains [tex]\mu = 20.1[/tex]
For the alternative hypothesis [tex]H_a[/tex]
The mean is [tex]\mu < 20[/tex]
This mean that the claim drops
The test statistic(the calculated value ) (z) is mathematically obtained with the following formula
[tex]z = \frac{\= x -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]
Where [tex]\= x[/tex] is the mean for the the alternative hypothesis
substituting values
[tex]z = \frac{18.9 - 20.1}{\frac{1.4}{\sqrt{5} } }[/tex]
[tex]= -1.92[/tex]
the absolute value of the calculated value is
[tex]|z| = 1.92[/tex]
Answer:
[tex]\mid z \mid = 1.92[/tex]
Step-by-step explanation:
The null hypothesis is that the average number of complaints per week is 20.1
If H₀ = Null hypothesis
H₀: μ = 20.1
After the training, the complaints drop to 18.9 < 20.1. This means that the alternative hypothesis, [tex]H_{a} : \mu < 20.1[/tex]
[tex]\bar{x} = 18.9[/tex]
To know if the alternative hypothesis is true, we need to calculate the absolute z value using the test statistic
number of days, n = 5
Standard deviation, [tex]\sigma = 1.4[/tex]
[tex]z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex]
[tex]z = \frac{18.9 - 20.1}{\frac{1.4}{\sqrt{5} } }[/tex]
z = -1.92
[tex]\mid z \mid = 1.92[/tex]
