Question

2.In a sample of 15 random people from a small town, the average salary is $25,778 with a
standard deviation of $11,540. Use a 5% significance level to test the claim that mean
salary in this town is less than $30,000

Answer 1

Ho :   µ =   30000  
Ha :   µ <   30000   (Left tail test)
          
Level of Significance ,    α =    0.050  
sample std dev ,    s =    11540.0000  
Sample Size ,   n =    15  
Sample Mean,    x̅ =   25778.0000  
          
degree of freedom=   DF=n-1=   14  
          
Standard Error , SE = s/√n =   11540/√15=   2979.6152  
t-test statistic= (x̅ - µ )/SE =    (25778-30000)/2979.6152=   -1.4170  
          
critical t value, t* =        -1.7613   [Excel formula =t.inv(α/no. of tails,df) ]
          
p-Value   =   0.0892   [Excel formula =t.dist(t-stat,df) ]
Decision:   p-value>α, Do not reject null hypothesis       
Conclusion: There is not enough evidence to support the claim that mean
salary in this town is less than $30,000