Answer:
The 95% CI is
[tex]27.5\leq \mu \leq 33.3[/tex]
The margin of error is 2.9 minutes.
Step-by-step explanation:
The question is incomplete:
The sample standard deviation was 7.3 minutes.
Assume the population is normally distributed and use a t-distribution to construct a 95% confidence interval for the population mean mu.
The sample mean is 30.4
The sample standard deviation is 7.3
The sample size is 27.
The degrees of freedom are:
[tex]df=n-1=27-1=26[/tex]
For a 95% CI and df=26, the critical value for t is t=2.056.
The margin of error can be calculated as:
[tex]E=t\cdot s/\sqrt{n}=2.056*7.3/\sqrt{27}=15/5.2=2.9[/tex]
The lower and upper bounds of the 95% CI are:
[tex]LL=\bar x-t\cdot s/\sqrt{n}=30.4-2.9=27.5\\\\UL=\bar x+t\cdot s/\sqrt{n}=30.4+2.9=33.3[/tex]
The 95% CI is then
[tex]27.5\leq \mu \leq 33.3[/tex]
