Q2-11BSC - Elementary Statistics

Question

Mean Body Temperature Data Set 3 “Body Temperatures” in Appendix B includes a sample of 106 body temperatures having a mean of 98.20°F and a standard deviation of 0.62°F. Construct a 95% confidence interval estimate of the mean body temperature for the entire population. What does the result suggest about the common belief that 98.6°F is the mean body temperature?

Step-by-Step Solution

Answer

The 95% confidence interval of the mean body temperature for the entire population is \(98.08^\circ F < {\rm{ }}\mu {\rm{ }} < 98.32^\circ F\).

 

The result suggests that the common belief that “the mean body temperature is 98.6°F” is not true.

Step 1: Given information

Mean of 106 samples of body temperatures is 98.20°F\(\left( {\bar x} \right)\)and thestandard deviation is 0.62°F (s).

 

The level of confidence is 95%.

 

Step 2: Check the requirements

The sample size of mean body temperatures is 106 which is greater than 30.Therefore, the distribution can be approximated as normal distribution. 

 

Assume that the sample is randomly selected and the population standard deviation is unknown.

 

Thus, t-distribution would be used to find the confidence interval.

Step 3: State the formula for confidence interval

Confidence interval for mean is expressed as\(\bar x - E < \mu  < \bar x + E\).

Here,\(\bar x\) is the sample mean and E is the margin of error.

\(E = {t_{\frac{\alpha }{2}}} \times \frac{s}{{\sqrt n }}\)

 

Where, \({t_{\frac{\alpha }{2}}}\)is the critical value with \(\alpha \)level of significance for normal distribution.

 

Here,\(\bar x\) represents the sample mean of the Mean Body Temperature data and \(\mu \) represents the population mean of the Mean Body Temperature data.

Step 4: Calculate margin of error

The confidence level of 95% implies significance level of 0.05.

 

The critical value \({t_{\frac{\alpha }{2}}}\) is obtained from standard normal table as 1.98.

 

Margin of error is given by,

\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}} \times \frac{s}{{\sqrt n }}\\ = 1.98 \times \frac{{0.62}}{{\sqrt {106} }}\\ = 0.1194\end{array}\)

Step 5: Construct the confidence interval.

The 95% confidence interval for mean body temperature is computed as,

 

\(\begin{array}{c}\bar x - E < \mu  < \bar x + E\\98.20 - 0.1194 < \mu  < 98.20 + 0.1194\\98.08 < \mu  < 98.32\end{array}\)

 

Therefore, 95% confidence interval is\(\left( {98.08{\rm{^\circ F}},98.32{\rm{^\circ F}}} \right)\).

Step 6: Interpret the result

The 95% confidence interval of the mean body temperature for the entire population is \({98.08^0}F < {\rm{ }}\mu {\rm{ }} < {98.32^0}F\).

 

The confidence interval does not contain \({\rm{98}}{\rm{.}}{{\rm{6}}^{\rm{0}}}{\rm{F}}\)in 95% confidence interval.

 

Hence, the result suggests that the common belief that “the mean body temperature is 98.6°F” is not true.