<!--[if !supportLists]-->1. <!--[endif]-->A teacher recorded the number of minutes it takes her to correct a student’s periodical exam paper. The following constitute her data:

3.1, 2.8, 2.5, 3.6, 2.7, 2, 3.9,  4, 2.5, 2.6, 3.3, 3.8

Find the 99% confidence interval of the population mean.

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2 Answers

Mean=sum of data divided by the pop size=36.8/12=3.067.

Standard deviation=0.622. The given population statistics are 3.067±0.622 minutes, that is, between about 2.4 and 3.7 minutes. (Only 4 data values out of 12 fall outside this range.)

99% means 99.5% on the right side of the mean (Z is positive, higher than the mean) and 99.5% on the left side of the mean (Z is negative, lower than the mean), so the interval between (the CI) is 99% of the distribution. The Z-score for 99.5% is about 2.575.

Z=(X-mean)/standard deviation=(X-3.067)/0.622 (Z measures the number of standard deviations from the mean). X is the extreme data value producing the extreme Z-score. So we have the range 3.067±0.622×2.575, which is 3.067±1.602 minutes, which gives us a CI=[1.465,4.669]. This means that there is a span of about 3.2 minutes (the width of the CI) between the minimum and maximum time it would probably take to correct the student's exam paper on 99% of the occasions the teacher undertakes the exercise. Note that all data in the population fall inside the CI.

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