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16 September, 16:22

A fast food restaurant executive wishes to know how many fast food meals adults eat each week. They want to construct a 95% confidence interval with an error of no more than 0.07 meals. A consultant has informed them that a previous study found the mean to be 8.1 fast food meals per week and found the standard deviation to be 2.

What is the minimum sample size required to create the specified confidence interval?

Round your answer up to the next integer

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  1. 16 September, 17:53
    0
    Given Information:

    Confidence level = 95%

    Error = 0.07

    Mean = μ = 8.1

    Standard deviation = σ = 2

    Required Information:

    Sample size = n = ?

    Answer:

    Sample size = n ≤ 2209

    Step-by-step explanation:

    The sample size required to create the specified confidence interval is given by

    error ≤ z * (σ/√n)

    Where z is the corresponding z-score of 95 percentile, n is the sample size and σ is the standard deviation and error is no more than 0.07

    √n ≤ z*σ/error

    n ≤ (z*σ/error) ²

    The z-score for 95% confidence level is 1.645

    n ≤ (1.645*2/0.07) ²

    n ≤ 2209

    Therefore, a sample size of n ≤ 2209 is required to create a 95% confidence interval with an error of no more than 0.07.
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