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21 June, 14:23

A sample of 87 glass sheets has a mean thickness of 4.20 mm with a standard deviation of 0.10 mm. (a) Find a 98% confidence interval for the population mean thickness. (b) What is the level of the confidence interval (4.185, 4.215) ? (c) How many glass sheets must be sampled so that a 98% confidence interval will specify the mean to within ±0.015? (d) Find a 90% confidence upper bound for the mean thickness.

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  1. 21 June, 14:38
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    Step-by-step explanation:

    From the information given,

    Mean, μ = 4.2 mm

    Standard deviation, σ = 0.1 mm

    number of sample, n = 87

    1) For a confidence level of 98%, the corresponding z value is 2.33.

    We will apply the formula

    Confidence interval

    = mean ± z * standard deviation/√n

    It becomes

    4.2 ± 2.33 * 0.1/√87

    = 4.2 ± 2.33 * 0.0107

    = 4.2 ± 0.025

    b) The lower end of the confidence interval is 4.2 - 0.025 = 4.18

    The upper end of the confidence interval is 4.2 + 0.025 = 4.23

    c) 0.015 = 2.33 * 0.1/√n

    0.015/2.33 = 0.1/√n

    0.00644 = 0.1/√n

    √n = 0.1/0.00644 = 16

    n = 16² = 256

    d) For a confidence level of 90%, the corresponding z value is 1.645.

    It becomes

    4.2 ± 1.645 * 0.1/√87

    = 4.2 ± 1.645 * 0.0107

    = 4.2 ± 0.0176

    The upper bound for the mean thickness is

    4.2 + 0.0176 = 4.2176mm
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