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One-Tailed Tests for Population Standard Deviation of Mean Income of Teachers

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Past records suggest that the mean annual income, u1, of teachers in state of Utah is greater than or equal to the mean annual income, u2, of teachers in Oregon. In a current study, a random sample of 10 teachers from Utah and an independent random sample of 10 teachers from Oregon have been asked to report their mean annual income. The data obtained are as follows.

Utah ----Oregon
27060---41176
32575---43541
32664---31579
38068---35106
32384---27719
34132---37008
32925---30419
40826---41692
29920---37791
44864---30548

The population standard deviation for mean annual income of teachers in Utah and in Oregon are estimated as 6400 and 6500, respectively. It is also known that both populations are approximately normally distributed. At the 0.05 level of significance, is there sufficient evidence to reject the claim that the mean annual income of teachers in state of Utah is greater than or equal to the mean annual income of teachers in Oregon? Perform a one-tailed test. Then fill in the table below.

Carry your intermediate computations to at least three decimal places and round your answers as specified in the table

Answer the following:
Null hypothesis:
Alternative hypothesis:
Type of test statistic:
Value of the test statistic rounded to at least 3 decimal places:
the p value rounded to at least 3 decimal places:
Can we reject the claim that the mean annual income of teachers from Utah is greater than or equal to the mean annual income of teachers from Oregon?
yes or no

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Solution Summary

The solution attaches a .doc file giving the null/alternative hypotheses, value of test statistic, p value and conclusion for this question on mean annual income of teachers in Utah.

Solution provided by:
Changping Wang, MA
Education
  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
Recent Feedback
  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
  • "excellent work"
  • "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
  • "Thank you"
  • "Thank you very much for your valuable time and assistance!"
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