Question #3084

A bottled water company sells spring water for water coolers. The company filters the water to keep the mean percent of impurities below a maximum level. To determine if it was meeting this requirement, it selected a random sample of 27 bottles, opened them, and analyzed them for impurities. The sample standard deviation was 0.00161%. The company knew the data were normally distributed, but it did not know the distribution’s population standard deviation. It calculated a confidence interval to be (0.003209, 0.011191). Which of the following is closest to the level of this confidence interval?

Expert's answer

Margin error = (0.011191 - 0.003209)/2 = 0.003991

t-critical value = ME/s*sqrt(27) = 0.003991/0.00161 = 2.47888

tdist(2.47888; df = 27-1 = 26)& = 0.01 -> alpha=0.02

So it is 98% confidence interval.

t-critical value = ME/s*sqrt(27) = 0.003991/0.00161 = 2.47888

tdist(2.47888; df = 27-1 = 26)& = 0.01 -> alpha=0.02

So it is 98% confidence interval.

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