Question

Records show that the average commute time for a person living in Atlanta in 2000 was 42 minutes. An analyst claims that the average commute time in 2007 was longer. The analyst takes a simple random sample of thirteen commuters in 2007 and calculates an average of 46.4 and a standard deviation of 3.8. (It is reasonable to assume that the population of commute times has a normal distribution.)

Which of the following is the appropriate test statistic?

t = (46.4 - 42.0)/(3.8/√13)

z = (42.0 - 46.4)/(3.8/√13)

z = (46.4 - 42.0)/(3.8/√13)

t = (42.0 - 46.4)/(3.8/√13)

None of the above

Answer 1

Answer:

Option a)  t = (46.4 - 42.0)/(3.8/√13)

Step-by-step explanation:

Here we are given that Records show that the average commute time for a person living in Atlanta in 2000 was 42 minutes i.e. $\mu$ = 42

And to test this a sample of thirteen commuters in 2007 has been taken and calculates an average of 46.4 and a standard deviation of 3.8 i.e.;

Sample Mean, X bar = 46.4             Sample standard deviation,s = 3.8

Sample size, n = 13

Since, here we know nothing about population standard deviation so we will use t statistics here ;

  Test statistics =  $\frac{Xbar - \mu }{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

Therefore, appropriate test statistic = $\frac{46.4 - 42.0 }{\frac{3.8}{\sqrt{13} } }$ ~ $t_1_2$ .