Assume the interval 2022 is a 998 confidence interval for me

Assume the interval (20,22) is a 99.8% confidence interval for mean with n=16 (assume the population standard deviation is unkown), find:

(a) The sample mean.

(b) The margin of error.

(d) The 90% confidence interval for the mean.

(a) The sample mean.

(20+22)/2 =21

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(b) The margin of error.

(22-20)/2 =1

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The degree of freedom =n-1=16-1=15

Given a=1-0.998=0.002, t(0.001, df=15) =3.73 (from student t table)

t*s/vn = 1

--> 3.73*s/sqrt(16) =1

So s =4/3.73 =1.072386

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(d) The 90% confidence interval for the mean

Given a=1-0.9=0.1, t(0.05, df=15) =1.75 (from student t table)

So the lower bound is

xbar - t*s/vn=21 - 1.75*1.072386/sqrt(16) =20.53083

So the upper bound is

xbar + t*s/vn =21 + 1.75*1.072386/sqrt(16)=21.46917