Question

The 99% confidence interval for mean is (12, 16).
Then, find the values of sample mean
and margin of error.​

Answer 1

Given :- The 99% confidence interval for mean is (12, 16). Then, find the values of sample mean and margin of error. ?

Answer :-

→ The 99% confidence interval for mean = (12 , 16) .

so,

• lower limit = 12
• upper limit = 16

then,

→ sample mean = ( lower limit + upper limit) / 2 = (12 + 16)/2 = 28/2 = 14 .

therefore,

→ margin of error = (upper limit - lower limit) / 2 = (16 - 12)/2 = 4/2 = 2 .

Hence, value of sample mean is 14 and margin of error is 2 .

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