Complete Question
According to the Bureau of Labor Statistics, citizens remain unemployed for an average of 15.9 weeks before finding their next job (June, 2008). Suppose you want to show that Louisiana has been effective in getting their unemployed back to work sooner. You take a random sample of 50 citizens who were unemployed six months earlier and ask them to report the duration. You find that the average time spent unemployed was 13.4 weeks with a sample standard deviation of the time unemployed is 6.7 weeks.
1 Which of the following statements is the correct alternative hypothesis?
2 The test statistic for testing the hypothesis is
a. -2.64
b. -2.32
c. -2.11
d. -1.28
e. none of these are correct
Answer:
1
The alternative hypothesis [tex]H_a : \mu < 15.9[/tex]
2
The test statistics [tex]z = -2.64 [/tex]
Step-by-step explanation:
From the question we are told that
The population mean value for time citizens remain unemployed is [tex]\mu = 15.9 \ weeks[/tex]
The sample size is n = 50
The sample standard deviation is 6.7 weeks.
The sample mean value for time citizens remain unemployed is [tex]\mu = 15.9 \ weeks[/tex]
The null hypothesis is [tex]H_o : \mu \ge 15.9[/tex]
The alternative hypothesis [tex]H_a : \mu < 15.9[/tex]
Generally test statistics is mathematically represented as
[tex]z = \frac{ \= x - \mu }{ \frac{s}{\sqrt{n} } }[/tex]
=> [tex] z = \frac{13.4 - 15.9}{\frac{6.7}{\sqrt{50}}}[/tex]
=> [tex]z = -2.64 [/tex]
