Question

The z-score associated with 95% is 1.96. If the sample mean is 200 and the standard deviation is 30, find the upper limit of the 95% confidence interval.

Answer 1

Answer:

The upper limit of the 95% confidence interval is:

C.I_u = 200 + (58.8/$\sqrt{n}$)

Step-by-step explanation:

The formula is given as:

C.I = μ ± Z*σ/$\sqrt{n}$

The upper limit => C.I_u = μ + Z*σ/$\sqrt{n}$

The lower limit => C.I_l = μ - Z*σ/$\sqrt{n}$

The sample size (n) is not stated in the question. Hence, we calculate the upper limit with respect to n.

The upper limit => C.I_u = 200 + 1.96*(30/$\sqrt{n}$)

                                    = 200 + (1.96*30)/$\sqrt{n}$

                                    = 200 + 58.8/$\sqrt{n}$