Question

Joe is measuring the widths of doors he bought to install in an apartment complex. He measured 72 doors and found a mean width of 36.1 inches with a standard deviation of 0.3 inches. To test if the doors differ significantly from the standard industry width of 36 inches, he computes a z-statistic. What is the value of Joe's z-test statistic? -1.81 2.83 1.81 -2.83

Answer 1

GIVEN:

Sample size $(n)=72$

Mean width of sample $(x_{bar})=36.1$

Population standard deviation $(\sigma )=0.3$

HYPOTHESIS:

$H_{0}:\mu =36$ (That is, the mean width of doors is not significantly different from standard industry width of 36 inches.)

$H_{0}:\mu \neq 36$ (That is, the mean width of doors is significantly different from standard industry width of 36 inches.)

TEST STATISTIC:

$z=(x_{bar}-\mu _{0})/(\sigma /(n)^{1/2})$

which follows standard normal distribution.

where $\mu _{0}$ is the hypothesized value $36$.

CALCULATION:

$z=(x_{bar}-\mu _{0})/(\sigma /(n)^{1/2})$

  $=(36.1-36)/(0.3 /(72)^{1/2})$

$=0.1/0.0354$

$=2.83$

The value of z-test statistic is $2.83$. Thus option (b) is the correct answer.