Question

use a normal distribution or a t-distribution to construct a 95% confidence interval for the population 

mean. Justify your decision. If neither distribution can be used, explain why.

Interpret the results. If convenient, use technology to construct the confidence interval.

Body Mass Index.

In a random sample of 50 people, the mean body mass index (BMI) was 27.7 and the standard deviation was 6.12. Assume the body mass indexes are normally distributed.

Answer 1

When the population standard deviation is unknown, the mean has a Student's t-distribution.

The critical value for $\alpha = 0.05$  and $df = n-1 = 49$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.009575.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{X}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{X}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(27.7-2.009575\times\dfrac{6.12}{\sqrt{50}},$$

$$27.7+2.009575\times\dfrac{6.12}{\sqrt{50}})$$

$$=(25.9607, 29.4393)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $25.9607<\mu<29.4393,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(25.9607, 29.4393).$