Answer:
The confidence interval for the true mean is
[tex]10.86<\mu<11.74[/tex]
Step-by-step explanation:
We have a sample with mean of 11.3 and variance 0.49. The size of the sample is 12.
We can estimate the CI as
[tex]\bar x - t\sqrt{s^2/n} <\mu<\bar x + t\sqrt{s^2/n}[/tex]
We can calculate the parameter t for a 12-1=11 degrees of freedom and 95% confidence. We look up the value in the t-table, and find that for this conditions t=2.2010.
Then we can calculate
[tex]t\sqrt{s^2/n}=2.2010*\sqrt{0.49/12}=2.2010* 0.2021=0.4448[/tex]
The confidence interval becomes
[tex]\bar x - t\sqrt{s^2/n} <\mu<\bar x + t\sqrt{s^2/n}\\\\11.3-0.44<\mu<11.3+0.44\\\\10.86<\mu<11.74[/tex]
