Question

Suppose a large telephone manufacturer has a problem with excessive customer complaints and consequent returns of the phones for repair or replacement. The manufacturer wants to estimate the magnitude of the problem in order to design a quality control program. How many telephones should be sampled and checked in order to estimate the proportion defective to within 9 percentage points with 89% confidence

Answers

  1. Answer:

    80 telephones should be sampled

    Step-by-step explanation:

    In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

    \pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

    In which

    z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

    The margin of error is of:

    M = z\sqrt{\frac{\pi(1-\pi)}{n}}

    89% confidence level

    So \alpha = 0.11, z is the value of Z that has a p-value of 1 - \frac{0.11}{2} = 0.945, so Z = 1.6.

    How many telephones should be sampled and checked in order to estimate the proportion defective to within 9 percentage points with 89% confidence?

    n telephones should be sampled, an n is found when M = 0.09. We have no estimate for the proportion, thus we use \pi = 0.5

    M = z\sqrt{\frac{\pi(1-\pi)}{n}}

    0.09 = 1.6\sqrt{\frac{0.5*0.5}{n}}

    0.09\sqrt{n} = 1.6*0.5

    \sqrt{n} = \frac{1.6*0.5}{0.09}

    (\sqrt{n})^2 = (\frac{1.6*0.5}{0.09})^2

    n = 79.01

    Rounding up(as 79 gives a margin of error slightly above the desired value).

    80 telephones should be sampled

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