In: Statistics and Probability
It is stated that the average wage for a particular profession is $60K however a researcher believes that the actual wage of this profession is less. A random sample of N = 30 individuals working in this profession are taken and it is found that the sample mean is $57K and a sample standard deviation of $8K. Test this hypothesis at the 1% significance level.
Reject the null
Fail to Reject
Solution :
Given that,
Population mean = = 60
Sample mean = = 57
Sample standard deviation = s = 8
Sample size = n = 30
Level of significance = = 0.01
This is a left (One) tailed test,
The null and alternative hypothesis is,
Ho: 60
Ha: 60
The test statistics,
t = ( - )/ (s/)
= ( 57 - 60 ) / ( 8 / 30 )
= -2.054
P-value = 0.0245
The p-value is p = 0.0245 > 0.01, it is concluded that the null hypothesis is fail to rejected.
Fail to Reject