Question

suppose that a random sample of 10 students in grade six was taken and the confidence interval corresponding to their mean height (in inches) was found. the confidence interval was calculated to be (63.1, 65.3) with a known population standard deviation of 2.6 inches. what is the confidence level (as percentage) corresponding to this confidence interval ?

Answer 1

Ans : Given that ,

sample size n = 10

population standard deviation $\sigma$ = 2.6 inches

and we have to given  the confidence interval corresponding to their mean height

the confidence interval is

(63.1, 65.3)  

Now to find what is the confidence level corresponding to this confidence interval

The formula for the confidence interval margin of error :

МОЕ = *t.(C, n-1) n

First we find Margin of error

margin of error = (Upper confidence limit - Lower confidence limit ) /2

Margin of error = (65.3 - 63.1 )/2

margin of error = 1.1

$\Rightarrow$1.1 = (2.6 / $\sqrt{10}$ )*t_c (C,10-1)

1.1 = 0.8222*t_c (C,9)

t_c (C,9) = 1.34

here sample size is 10 which is less than 30 then we use a critical t table too see that C level corresponding to 1.34 and degree of freedom 9

$\alpha$ = 0.2

C = 80 %

the confidence level (as percentage) corresponding to this confidence interval is 80 %