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6 October, 04:50

The makers of a soft drink want to identify the average age of its consumers. A sample of 16 consumers is taken. The average age in the sample was 22.5 years with a standard deviation of 5 years. a) Construct a 95% confidence interval for the true average age of the consumers. b) Construct an 80% confidence interval for the true average age of the consumers. c) Discuss why the 95% and 80% confidence intervals are different.

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  1. 6 October, 05:59
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    Step-by-step explanation:

    Assuming a normal distribution for the age of the consumers. We want to determine a 95% confidence interval for the true average age of the consumers.

    Number of sample, n = 16

    Mean, u = 22.5 years

    Standard deviation, s = 5 years

    For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

    We will apply the formula

    Confidence interval

    = mean ± z * standard deviation/√n

    It becomes

    22.5 ± 1.96 * 5/√16

    = 22.5 + / - 1.96 * 1.25

    = 22.5 + / - 2.45

    The lower end of the confidence interval is 22.5 - 2.45 = 20.05

    The upper end of the confidence interval is 22.5 + 2.45 = 24.95

    For 80% confidence interval,

    the corresponding z value is 1.28. This is determined from the normal distribution table.

    We will apply the formula

    Confidence interval

    = mean ± z * standard deviation/√n

    It becomes

    22.5 ± 1.28 * 5/√16

    = 22.5 + / - 1.28 * 1.25

    = 22.5 + / - 1.6

    The lower end of the confidence interval is 22.5 - 1.6 = 20.9

    The upper end of the confidence interval is 22.5 + 1.6 = 24.1

    95% provides a wider interval than 80%. The wider the possible values of the true mean, the more confident we are. The lesser the possible values of the true mean, the less confident we are
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