In a random sample of 20 people, the mean commute time to work was 31.7 minutes and the 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. what is the margin of error of mu? interpret the results.

Solution: We are given:

x bar=31.7

s = 7.3

n=20

The 95% confidence interval for the population mean is given below:

xbar +-

31.7 +- (2.093 x )

31.7 +- 3.42

[31.7-3.42,31.7+3.42]

[28.28 ,35.12 ]

Therefore the 95% confidence interval for the population mean is:

28.28≤μ≤35.12

The margin of error is 3.42

There is 95% chance that the confidence interval 28.28≤μ≤35.12 contains the true population mean.

18/157

Step-by-step explanation

The key to solving this problem is to find the greatest common divisor of 72 and 628.

To find the greatest common divisor of 72 and 628, we list all of the factors of both numbers so we can compare the lists.

When we compared the factors of 72 (1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72) and 628 (1, 2, 4, 157, 314, 628) we found that 4 is the largest factor that they have in common.

Since 4 is the largest common factor of 72 and 628, 4 is also the greatest common divisor of 72 and 628.

Now, we divide both the numerator and denominator in 72/628 by the greatest common divisor as follows:

Numerator: 72/4 = 18

Denominator: 628/4 = 157

Then, we put the numerator and denominator that we are left with back together to get our answer. 72/628 to the simplest form is: 18/157