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17 August, 16:01

A local company wants to evaluate their quality of service by surveying their customers. Their budget limits the number of surveys to 100. What is their maximum error of the estimated mean quality for a 93% level of confidence and an estimated standard deviation of 7? a. 1.8130 b. 1.1550 c. 1.2824 d. 1.3160

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Answers (2)
  1. 17 August, 17:03
    0
    (C) 1.2824

    Step-by-step explanation:

    Maximum error = t * sd/√n

    sd = 7

    n = 1

    degree of freedom = n - 1 = 100 - 1 = 99

    confidence level = 93%

    t-value corresponding to 99 degrees of freedom and 93% confidence level is 1.855

    Maximum error = 1.855*7/√100 = 12.985/10 = 1.2985

    The closest option is (C)
  2. 17 August, 18:00
    0
    Answer:1.2684

    Step-by-step explanation:

    Formula to find the maximum error of the mean is given by : -

    E=z*/dfrac{/sigma}{/sqrt{n}}

    , where n = sample size.

    z * = Critical value.

    = Population standard deviation

    As it is given, we have

    n = 100

    /sigma = 7

    93%: Confidence interval

    Critical value for 93% confidence as gotten from the z-table is = 1.81 [from z-table ]

    Then, estimated mean quality has the maximum error as

    E = (1.81) / dfrac{7}{/sqrt{100}}

    E = (1.81) / dfrac{7}{10}

    E = (1.81) / dfrac07=1.2684

    Therefore, maximum error required will be = 1.2684
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