Respuesta :
Answer:
The critical t-value for such a test given an alpha level of 0.05 is 2.26
Step-by-step explanation:
Null hypothesis : [tex]H_0:\mu = 7.5[/tex]
Alternate hypothesis :[tex]H_a:\mu \neq 7.5[/tex]
Population mean = [tex]\mu = 7.5[/tex]
Data : 10, 9, 6, 11, 13, 14, 9, 9, 9, and 10
Mean = [tex]\bar{x}=\frac{\text{Sum of all observations}}{\text{no. of observations}}[/tex]
Mean =[tex]\frac{10+9+6+11+13+14+9+9+ 9+ 10}{10}[/tex]
Mean =10
Standard deviation : [tex]\sqrt{\frac{\sum(x-\bar{x})^2}{n}}[/tex]
Standard deviation :[tex]\sqrt{\frac{(10-10)^2+(9-10)^2+(6-10)^2+(11-10)^2+(13-10)^2+(14-10)^2+(9-10)^2+(9-10)^2+(9-10)^2+(10-10)^2}{10}}[/tex]
Standard deviation s :2.144
[tex]t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}t = \frac{10-7.5}{\frac{2.144}{\sqrt{10}}}t=3.687[/tex]
Df = n-1 = 10-1 = 9
T critical =[tex]t_{(df,\alpha)}=t_(9,0.05)=2.26[/tex]
T calculated > T critical
So, We failed to accept null hypothesis
Hence the critical t-value for such a test given an alpha level of 0.05 is 2.26
