In: Statistics and Probability
A labor union claims that the mean income of its members is no
more than $20,000 per year. The Employers' Association believes
that this claim is too low and that the actual mean income is
higher. A random sample of 64 union members yields a mean income of
$20,278.50 with a standard deviation of $793.81. Calculate the
standardized test statistic and find the P-value.
The null and alternative hypothesis is ,
The test is right tailed test.
Since , the population standard deviation is not known.
Therefore , use t-distribution.
Now , df=degrees of freedom=n-1=64-1=63
The p-value is ,
p-value= ; The excel function is , =TDIST(2.8067,63,1)
Decision : Here , p-value =0.0033<0.05 significance level.
Therefore , reject Ho.
Conclusion : There is not sufficient evidence to support the a labor union claims that the mean income of its members is no more than $20,000 per year.