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7 November, 16:42

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

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  1. 7 November, 17:21
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    The confidence interval at at 99% level of confidence is 93.7 ≤ μ ≤ 96.9.

    Step-by-step 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.
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