Respuesta :
Answer:
(25.732,30.868)
Step-by-step explanation:
Given that in a random sample of 42 people, the mean body mass index (BMI) was 28.3 and the standard deviation was 6.09.
Since only sample std deviation is known we can use only t distribution
Std error = [tex]\frac{s}{\sqrt{n} } =\frac{6.09}{\sqrt{42} } \\=0.9397[/tex]
[tex]df = 42-1 =41[/tex]
t critical for 99% two tailed [tex]= 2.733[/tex]
Margin of error[tex]= 2.733*0.9397=2.568[/tex]
Confidence interval lower bound = [tex]28.3-2.568=25.732[/tex]
Upper bound = [tex]28.3+2.568=30.868[/tex]
