Answer: [tex](9.27025,\ 11.12975)[/tex]
Step-by-step explanation:
Given : Sample size : n= 9
Degree of freedom = df =n-1 =8
Sample mean : [tex]\overline{x}=10.2[/tex]
sample standard deviation : [tex]s= 1.5[/tex]
Significance level ; [tex]\alpha= 1-0.90=0.10[/tex]
Since population standard deviation is not given , so we use t- test.
Using t-distribution table , we have
Critical value = [tex]t_{\alpha/2, df}=t_{0.05 , 8}=1.8595[/tex]
Confidence interval for the population mean :
[tex]\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}[/tex]
90% confidence interval for the mean value will be :
[tex]10.2\pm (1.8595)\dfrac{1.5}{\sqrt{9}}[/tex]
[tex]10.2\pm (1.8595)\dfrac{1.5}{3}[/tex]
[tex]10.2\pm (1.8595)(0.5)[/tex]
[tex]10.2\pm (0.92975)[/tex]
[tex](10.2-0.92975,\ 10.2+0.92975)[/tex]
[tex](9.27025,\ 11.12975)[/tex]
Hence, the 90% confidence interval for the mean value= [tex](9.27025,\ 11.12975)[/tex]
