It is claimed that the average income in a particular community is $45,000.

Question

It is claimed that the average income in a particular community is $45,000. The standard deviation of that income is known to be $4,000. A sample of 49 individuals is selected, and a sample mean of $46,000 is derived. If the goal is to test whether the average income for the community is greater than $45,000, what is the value of the calculated test statistic at the 5% level of significance?
a.
z = 0.25
b.
z = 1.75
c.
t = 0.25
d.
t = 1.75

Answer 1

you can compute the test statistic here by taking

(x-bar - u) / (o / sqrt(n))

for this scenario that means taking

(46,000 - 45,000) / (4000 / sqrt(49))

= 1000 / 571.43

= 1.75

then because the population standard deviation (o = 4000) is known on this one and the sample size (n = 49) is larger than 30, that means that the computed value here is a Z stat

Z = 1.75