Respuesta :
Solution :
Given :
[tex]$\mu = 24$[/tex]
n = 15
[tex]$\overline x = 24.89$[/tex]
s = 0.5902
The hypothesis :
[tex]$H_0 : \mu = 25$[/tex]
[tex]$H_a : \mu \neq 25$[/tex]
This is a 2 tailed test.
The significance level is [tex]$95 \% \ ( \alpha =0.05)$[/tex]
The test statistic :
[tex]$Z=\frac{\overline x - \mu}{\frac{\sigma}{\sqrt n}}$[/tex]
[tex]$Z=\frac{24.89-25}{\frac{0.5902}{\sqrt 15}}$[/tex]
= -0.72
The p value : The p value for Z = -0.72 is 0.4716
The critical value : the critical value at α = 0.05 is +1.96 to -1.96
The decision rule :
If [tex]$Z_{observed} > Z_{critical}$[/tex] or if [tex]$Z_{observed} < -Z_{critical}$[/tex], then reject [tex]$H_0$[/tex].
Also if p value is less than α, then reject [tex]$H_0$[/tex].
The decision :
Since the Z falls in between +1.96 and -1.96, we fail to reject the [tex]$H_0$[/tex]. Also since p value is greater than α, we fail to reject [tex]$H_0$[/tex].
The conclusion :
There is not sufficient evidence at the 95% significance level to warrant rejection of the claim that the pills come from a population in which the amount of the atorvastatin is equal to 25 mg.
Now calculating the mean and the standard deviation :
[tex]$\text{Mean} = \frac{\text{sum of observation}}{\text{total observations}}$[/tex]
Standard deviation = [tex]$\sqrt{\text{variance}}$[/tex]
Variance = [tex]$\frac{\text{sum of squares(SS)}}{n-1}$[/tex]
Where, SS = [tex]$\sum (X - \text{mean})^2$[/tex]
X Mean [tex]$(X-\text{mean})^2$[/tex]
24.1 24.89 0.62
24.4 24.89 0.24
24.3 24.89 0.35
24.9 24.89 0
24.1 24.89 0.62
24.2 24.89 1.72
24.1 24.89 0.04
26.2 24.89 0.04
25.1 24.89 0.24
25 24.89 ��0.01
24.7 24.89 0.04
25.1 24.89 0.04
25.3 24.89 0.17
25.5 24.89 0.37
25.5 24.89 0.37
n 15
Sum 373.3
Average 24.89
SS 4.8775
Variance = [tex]$\frac{SS}{n-1}$[/tex] 0.348392857
Standard deviation 0.5902
