Consider a random sample of ten children selected from the population of infants

Question

Consider a random sample of ten children selected from the population of infants receiving antacids that contain aluminum. The distribution of plasma aluminum levels is known to be approximately normal; however, its mean µ and the standard deviation are not known. The mean aluminum level for the sample of ten infants is = 37.2 µg/l and the sample standard deviation is s = 7.13 µg/l.

What would the 95% confidence interval be if the population standard deviation was known to be equal to the sample value of7.13 µg/l?

Explanation / Answer

Here mean is 37.2, sd is 7.13 so

Formula:

If (n>=30), CI = x ± Z/2 × (/n) If (n<30), CI = x ± t/2 × (/n) Where, x = Mean = Standard Deviation = 1 - (Confidence Level/100) Z/2 = Z-table value t/2 = t-table value CI = Confidence Interval

So 95% confidence interval is 32.09951<=x<=42.30049

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