In a random sample of 1919 people, the mean commute time to work was 34.834.8 minutes and the standard deviation was 7.17.1 minutes. assume the population is normally distributed and use a t-distribution to construct a 9999% confidence interval for the population mean muμ. what is the margin of error of muμ? interpret the results

Solution: We are given:

x bar=34.8

s = 7.1

n=19

The 99% confidence interval for the population mean is given below:

xbar +-

34.8 +- 2.878 ()

34.8 +- 4.69

[34.8-4.69,34.8+4.69]

[30.11, 39.49]

Therefore the 99% confidence interval for the population mean is:

30.11≤μ≤39.49

The margin of error is 4.69

There is 99% chance that the confidence interval 30.11≤μ≤39.49 contains the true population mean.

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2