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An activity for students to learn about hypothesis testing on population means. The activity involves working in teams to complete exercises and assessing team performance. Examples of hypothesis tests for mean yields of corn and accidents at an intersection, and includes instructions for calculating test statistics and determining critical values or p-values. Students are expected to use their textbook and calculator for the exercises.
Typology: Exams
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Statistical Applications ACTIVITY 4: Hypothesis Tests on the Mean
Why
The predominant (but far from the only) mode of decision-making in statistics is theātest of significanceā. The mechanics vary between situations, but the underlying logic is the same. we need to establish the basic ideas ā using tests on a population mean.
LEARNING OBJECTIVES
CRITERIA
RESOURCES
PLAN
MODELS
I Population - corn fields fertilized with the new fertilizer. Variable X = corn yield (bu/A). Test is on mean. Test is : H 0 : μ = 142 Ha : μ > 142 II Test statistic is t = (^) sĀÆxxā/^142 ān with df = n ā 1 III Critical value approach: Reject H 0 if sample t > t. 05 = 2. 132 (with df = 4)[table p. 920] IV t = (^1496). 38.^6 /āā^1425 = 2. 66 V Reject H 0 and support Ha
VI The sample does give evidence at the .05 level that the mean yield is increased by the fertilizer.
Using the p-value approach, weād have:
III t = (^1496). 38.^6 /āā^1425 = 2. 66 df = 4
IV p = P (t > 2 .66), approximating from the table [p. 920] p is between .05 and .025 (since t is between t.05 = 2. 132 and t.025 = 2. 776 ) [Calculator/ computer gives p =. 028 ]
Steps V, VI are the same.
I Population: weeks [note our sample is of twenty weeks]. Variable: X = number of accidents at this intersection (with the new conditions) Parameter: Mean. Test is: H 0 : μ = 2. 2 Ha : μ 6 = 2. 2 II Statistic t = (^) sĀÆxxā/^2 ā.^2 n df = n ā 1 III To test at .05 level: Reject H 0 if sample t > t. 025 or if sample t < āt. 025 -from table, with df = 19, t. 025 = 2. 093 IV sample t = 21 ..^124 /āā^220.^2 = ā. 26 V We do not reject H 0 the sample does not provide evidence at the .05 level that the mean number of accidents has changed.
Calculating the p-value, from the tabel we can only tell that t does not even reach t. 20 , so we know that p > 2(.20) =. 40. Calculator gives p =. 80. Both methods are saying that this amount of change is easily explained as usual variability - no evidence of a change in the basic situation.
EXERCISE Show all the setup - identify population, variables,etc. Write your conclusion in words.
READING ASSIGNMENT (in preparation for next class) Read Chapter 9 (9.6ā9.7) - Decision-making and types of errors
SKILL EXERCISES:Use your calculator or Minitab for number -crunching [Minitab will carry out hypoth- esis tests when you have actual data to work with] but you have to write the hypotheses and conclusion. p. 358 #20, p.363 #28, 29, p.368(proportions)#40, 41