Question

Use the confidence level and sample data to find a confidence interval

for estimating the population p. Round your answer to the same


number of decimal places as the sample mean,


Test scores: n = 104, X = 95.3, o = 6.5; 99% confidence

Answer 1

The confidence interval at at 99% level of confidence is 93.7 ≤ μ ≤ 96.9.

Explanation:

Step 1:

We must first determine the z-value at a confidence level of 99%.

Therefore,

99% = 100%(1 - 0.01)

Thus,

α = 0.01

Therefore, the z-value will be

z_(α/2) = z_(0.01/2) = z_0.005 = 2.58

(The z-value is read-off from the z table from the standard normal probabilities.)

Step 2:

We can now write the confidence interval:

X - z_(α/2) [s/√(n)] ≤ μ ≤ X + z_(α/2) [s/√(n)]

95.3 - 2.58(6.5/√(104)) ≤ μ ≤ 95.3 + 2.58(6.5/√(104))

93.7 ≤ μ ≤ 96.9

Therefore, confidence interval is 93.7 ≤ μ ≤ 96.9 which means that we are 99% confident that the true mean population lies is at least 93.7 and at most 96.9.