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What the sample suggests is that; the mean body temperature couldvery possibly bevery possibly be 98.6 °F.
<h3>What is the Confidence Interval?</h3>
For a 99% Confidence Interval and DF = 103 - 1 = 102, the value of z = 2.33.
The formula for confidence interval is;
CI = x⁻ ± z(σ/√n))
We are given;
Sample mean; x⁻ = 98.9
standard deviation; σ = 0.67
sample size; n = 103
Thus the confidence Interval is;
CI = [98.9 - 2.33(0.67/√103)], [98.9 + 2.33(0.67/√103)]
CI = (98.9 - 0.154, 98.5 + 0.154)
CI = (98.36, 98.64)
Thus, 98.6 is contained within the 99% CI,and so we can conclude that the the mean body temperature could very possibly bevery possibly be 98.6 °F.
Read more about Confidence Interval at; brainly.com/question/17097944
