Certain data includes 36 body temperatures for which the sample mean is equal to 98.20 °F and standard deviation of 0.62 °F. Using this sample statistics, construct a 95% confidence interval of the mean body temperature. What does the sample suggest about the use of 98.6°F as the mean body temperature?
95% = mean ± 1.96 SEm
SEm = SD/√n
I'll let you calculate and come to your own conclusions.
To construct a confidence interval for the mean body temperature, we can use the following formula:
Confidence Interval = Sample Mean ± (Critical Value * Standard Error)
Step 1: Calculate the Critical Value
Since we want a 95% confidence level, we need to find the z-score that corresponds to a 95% confidence level. The critical value can be found using a z-table or a statistical calculator. For a 95% confidence level, the critical value is approximately 1.96.
Step 2: Calculate the Standard Error
The standard error measures the variability of the sample mean. It can be calculated using the formula: Standard Error = Standard Deviation / √(Sample Size)
Given:
Sample Mean (x̄) = 98.20 °F
Standard Deviation (σ) = 0.62 °F
Sample Size (n) = 36
So, the standard error is:
Standard Error = 0.62 / √(36) = 0.1033
Step 3: Calculate the Confidence Interval
Using the formula mentioned earlier:
Confidence Interval = Sample Mean ± (Critical Value * Standard Error)
Confidence Interval = 98.20 ± (1.96 * 0.1033) = (97.9989 °F, 98.4011 °F)
Therefore, the 95% confidence interval for the mean body temperature is (97.9989 °F, 98.4011 °F).
Now, let's consider the suggested mean body temperature of 98.6 °F. If this value falls within the confidence interval, it suggests that the true mean body temperature is likely to be 98.6 °F. However, if 98.6 °F falls outside the confidence interval, it suggests that the true mean body temperature is unlikely to be 98.6 °F.
In this case, since 98.6 °F falls outside the 95% confidence interval, the sample data suggests that the mean body temperature is unlikely to be 98.6 °F.