Question

The mayor of a town saw an article that claimed the national unemployment rate is 8%. They wondered if this held true in their town, so they took a sample of 200 residents to test H_0 : p = 0.08 versus H_a : p != 0.08, where p is the proportion of residents in the town that are unemployed. The sample included 22 residents who were unemployed.Identify the correct test statistic for this significance test to one decimal place.

Answer 1

Answer:

Z = $\frac{0.11-0.08}{\sqrt{0.08*0.92/200} }$   = 1.6  

we reject the Null hypothesis

Step-by-step explanation:

The correct test statistic

Assume :

Null Hypothesis :  H_0 : p = 0.08

Alternative hypothesis : H_a : p ≠ 0.08

p = proportion of residents that are unemployed

x = unemployed residents = 22

n = sample size = 200

To get the sample of the population that is unemployed we use this equation

$p = \frac{X}{n}$  =  22 / 200   = 0.11

now we apply the test of of significance of single proportion

= Z = $\frac{p - P}{\sqrt{pq/n} }$

therefore Z = $\frac{0.11-0.08}{\sqrt{0.08*0.92/200} }$   = 1.6

Therefore we reject the Null hypothesis