The following table lists aluminum content (mu g) per liter of water, measured i
ID: 3226013 • Letter: T
Question
The following table lists aluminum content (mu g) per liter of water, measured in three different PA counties. The data has been collected over five different, alternating years. a.) Perform an ANOVA analysis for the main factor "county", constructing a one-way ANOVA table summarizing your calculations. (Use the format outlined in the slides). State the hypothesis and conclusion based on your analysis. b). Second, perform a hypothesis test to determine whether the year in which the data was acquired mattered (made a significant difference), assuming alpha = 0.05. State the hypothesis and conclusion based on your analysis. Are the results of your conclusion for the "county" main factor altered? c). Using the tools of this class to-date, apply another statistical test to discern whether there is a difference in aluminum concentrations, between counties across years. And, compare your result and conclusion with result and conclusion from part (a) above.Explanation / Answer
The primary reason of applying ANOVA (Analysis of Variance) is when we want to compare the means between more than 2 independent groups.
a) The null hypothesis can be stated as Ho: There is NO significant difference between the means of aluminum content found in three countries.
The alternate hypothesis can be stated as: Ha: There is A significant difference between means of at least one pair of the aluminum content found in three countries.
To test the hypothesis,
df= degrees of freedom
As the value of sig value id 0.502 which is greater than 0.05 (95% Confidence Interval), we conclude that we fail to reject the null hypothesis. Therefore, there is NO significant difference between the means of aluminum content found in three countries.
b)
The null hypothesis can be stated as Ho: There is NO significant difference between the means of aluminum content found in each of the five years.
The alternate hypothesis can be stated as: Ha: There is A significant difference between means of at least one pair of the aluminum content found in each of the five years.
To test the hypothesis,
df= degrees of freedom
As the value of sig value id 0.045 which is greater than 0.05 (95% Confidence Interval), we conclude that we reject the null hypothesis. Therefore, here is A significant difference between means of at least one pair of the aluminum content found in each of the five years.
Therefore, the year of data collection mattered in this condition.
c) In order to apply other tools apart fro ANOVA, the best alternative tool would be to apply Independent sample t-test for each of the pair of the countries and look at the values of significance and compare them with 0.05. If the value is greater than 0.05, it concludes that there is no significant difference between the average of the aluminum content between the two countries.
Therefore applying independent sample t-test for each of the three pairs.
Between Lycoming and Cambria
Ho: There is No significant difference between the average aluminum content between the two countries.
Ha: There is A significant difference between the average aluminum content between the two countries.
The value of t calculated at 8 degrees of freedom is -1.120 and the value of significance is 0.295 which is greater than 0.05. This concludes that we fail to reject our null hypothesis.
Between Center and Cambria
Ho: There is No significant difference between the average aluminum content between the two countries.
Ha: There is A significant difference between the average aluminum content between the two countries.
The value of t calculated at 8 degrees of freedom is -0.543 and the value of significance is 0.602 which is greater than 0.05. This concludes that we fail to reject our null hypothesis.
Between Lycoming and Center
Ho: There is No significant difference between the average aluminum content between the two countries.
Ha: There is A significant difference between the average aluminum content between the two countries.
The value of t calculated at 8 degrees of freedom is -0.701 and the value of significance is 0.503 which is greater than 0.05. This concludes that we fail to reject our null hypothesis.
Therefore, we conclude that the mean scores of aluminum content is same for all the three countries. The results are similar to what we derived in part (A).
ANOVA Table from SPSS Sum of Squares df Mean Square F Sig. Between Groups 103.600 2 51.800 0.731 0.502 Within Groups 850.400 12 70.867 Total 954.000 14Related Questions
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