In estimating the mean score on a fitness exam, we use an original sample of size n = 30 and a bootstrap distribution containing 5000 bootst

Question

In estimating the mean score on a fitness exam, we use an original sample of size n = 30 and a bootstrap distribution containing 5000 bootstrap samples to obtain a 95% confidence interval of 67 to 73. A change in this process is described below. If all else stays the same, which of the following confidence intervals (A, B, or C) is the most likely result after the change:

Using 10,000 bootstrap samples for the distribution:

A. 66 to 74 8.

B. 67 to 73

C. 67.5 to 72.

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Sophia 2 hours 2021-09-10T07:45:53+00:00 1 Answer 0

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    2021-09-10T07:47:12+00:00

    Answer:

    C. 67.5 to 72.

    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 zscore that has a pvalue of 1 - \frac{\alpha}{2}.

    The width of the interval is determined by it’s margin of error, which is given by the following formula:

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

    So, as n increases, the margin of error decreases, and the interval gets smaller.

    Using 10,000 bootstrap samples for the distribution:

    We increase the sample size, which means that the interval gets smaller.

    We had 67 to 73, since it got smaller, it will be from a value higher than 67 to a value lower than 73.

    So the correct answer is:

    C. 67.5 to 72.

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45:7+7-4:2-5:5*4+35:2 =? ( )