Question

if the sample mean is 70 and standard error of the sample mean is 8, what will be the 90% confidence interval assuming normal distribution? the sample size was 30.

Answer 1

The 90% Confidence interval will be (56.84  - 83.16).

The value of the sample is 30 and

the value of sample mean (X) = 70

The standard error(SE) = 8

for 90% CI in normal distribution ,  we have the z-value = 1.645

and the margin of error will be = ± SE x z-value

                                                = ± 8 x 1.645

                                                = ±  13.16

The confidence interval is given by  = mean ±  Margin of error

∴ The lower limit will  be =  mean - margin of error

                                          = 70 - 13.16

                                          = 56.84

The upper limit will be = mean + margin of  error

                                     =  70 + 13.16

                                     =  83.16

∴ 90% CI of normal distribution is ( 56.84  -   83.16 )

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