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    Confidence Interval and Tests of Hypotheses

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    1. The CDC estimated that influenza vaccination reduced the risk of flu illness in 600 of 1000 randomly sampled individuals during seasons when most circulating flu viruses are well-matched to the flu vaccine. Construct the 97% confidence interval for the population proportion. State your conditions, interpret your interval, and show your work in detail.

    2. Yolanda & Garmin conducted a study with a random sample of 45 asthmatic children. The parents of each asthmatic child completed a questionnaire about their home environment. The researchers discovered that 18 of the asthmatic children lived in homes with dust mites or cockroaches present during the past 12 months. Is this sufficient evidence to conclude that greater than 35 percent of the population of asthmatic children live in homes with dust mites or cockroaches present? Let α = 0.05. Use both the critical value and p-value methods, and show all 5 steps.

    3. A research team would like to know what proportion of surgeons at hospitals wash their hands prior to procedures. Previous literature estimates the population proportion of surgeons that wash their hands prior to procedures to be about 0.57. How large of a sample must the research team use to estimate the true population proportion of surgeons that wash their hands, knowing that the research team requires a width of 0.2 and 98% confidence? Show your work in detail.

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    Solution Preview

    1. The CDC estimated that influenza vaccination reduced the risk of flu illness in 600 of 1000 randomly sampled individuals during seasons when most circulating flu viruses are well-matched to the flu vaccine. Construct the 97% confidence interval for the population proportion. State your conditions, interpret your interval, and show your work in detail.

    n = 1000
    p̂ = 600 / 1000 = 0.6
    α = 1 - 0.97 = 0.03

    Check that all Conditions are met:
    Condition 1) The sample is random as stated in the problem.
    Condition 2a) n * p̂ = 1000 * 0.6 = 600 ≥ 5.
    Condition 2b) n * (1 - p̂) = 1000 * 0.4 = 400 ≥ 5.

    Standard Error = sqrt[ p̂(1 - p̂) / n]
    = sqrt[0.6 * 0.4) / 1000]
    = 0.0155

    To find Critical z, first compute 1 - 0.03/2 = 0.985
    From the standard normal table, Critical z = 2.17

    Margin of Error M.E. = 2.17 * 0.0155
    = 0.0336

    p̂ - M.E. = 0.6 - 0.0336 = 0.57
    p̂ + M.E. = 0.6 + 0.0336 = 0.63

    We are 97% confident that the population proportion is between .57 and .64 .

    A TI 83 was used to ...

    Solution Summary

    Step-by-steps solutions are provided. Computations and formulas are shown. A TI 83 was used to check the work for Problems 1 and 2.

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