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Musya8 [376]
2 years ago
14

I need help with these

Mathematics
1 answer:
rewona [7]2 years ago
8 0
Also 8 i believe not sure though
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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
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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}

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