In: Statistics and Probability
according to the u.s. bureau of labor statistics, the average weekly earnings of a production worker in july 2011 were 657.49. suppose a labor researcher wants to test to determine whether this figure is still accurate today. the researcher randomly selects 55 production workers from across the united states and obtains a representative earnings statement for one week each. the resulting sample average is 671.13. assuming a population standard deviation of 63.90 and a 10% level significance, determine whether the mean weekly earnings of a production worker have changed.
The value of the test statistics is
according to u.s. bureau of labor statistics, the average weekly earnings of a production worker in July 2011 were = 657.49.
and the researcher randomly selects n = 55 production workers from across the united states and obtains a representative earnings statement for one week each. the resulting sample average is = 671.13. assuming a population standard deviation of = 63.90 and a 10% level significance
So, based on the Labor researcher the hypotheses we create are:
Based on the hypothesis it will be a two-tailed test.
Rejection region:
Based on the significance level and type of hypothesis test the critical value is calculated using the excel formula for normal distribution which is =NORM.S.INV(0.95) this results in Z =+/-1.645
So, reject Ho if -Z <-1.645 or Z> 1.645
Test Statistic:
P-value:
P-value is calculated using excel formula for normal distribution at calculated Z score, the formula is =2*(1-NORM.S.DIST(1.583, TRUE))
this results in P-value = 0.1134
Conclusion:
Since the P-value is larger than 0.10 and Z > Zc hence at 0.10 level of significance we failed to reject the null hypothesis and conclude that there is insufficient evidence to warrant the claim that figure is still accurate today