Answer:
CI = (98.11 , 98.49)
The value of 98.6°F suggests that this is significantly higher
Step-by-step explanation:
Data provided in the question:
sample size, n = 103
Mean temperature, μ = 98.3
°
Standard deviation, σ = 0.73
Degrees of freedom, df = n - 1 = 102
Now,
For Confidence level of 99%, and df = 102, the t-value = 2.62 [from the standard t table]
Therefore,
CI = [tex](Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})[/tex]
Thus,
Lower limit of CI = [tex](Mean - \frac{t\times\sigma}{\sqrt{n}})[/tex]
or
Lower limit of CI = [tex](98.3 - \frac{2.62\times0.73}{\sqrt{103}})[/tex]
or
Lower limit of CI = 98.11
and,
Upper limit of CI = [tex](Mean + \frac{t\times\sigma}{\sqrt{n}})[/tex]
or
Upper limit of CI = [tex](98.3 + \frac{2.62\times0.73}{\sqrt{103}})[/tex]
or
Upper limit of CI = 98.49
Hence,
CI = (98.11 , 98.49)
The value of 98.6°F suggests that this is significantly higher and the mean temperature could very possibly be 98.6°F
