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Hypothesis-Null & Alternative,P-value for F-test;2-Way ANOVA

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Hypothesis - Null & Alternative, p-value for F-test

(ANOVA) - How to determine significance difference in means

1) An investor wants to know if a difference exists, at a significance level of 0.05, between home prices in three subdivisions. Independent samples of homes from each of the three subdivisions - 18 homes in total - are obtained and their prices are recorded. The analysis of variance results for comparing these prices is started below.
SOURCE SUM OF SQUARES D.F. MEAN SQUARES F-stat
Treatment 157.44
______
_________
________
Error 253.50
______
_________
Total 410.94
______

a) What are the null and alternative hypotheses that need to be tested?
b) Complete the blanks in the table.
c) What is the estimate of the P-value for this F test?
d) Explain the conclusion for the hypothesis test

2) A company wishes to test new medicine expected to lower cholesterol levels. 10 patients are randomly selected and pretested. The results are listed below, the patients are placed on the drug for 6 months, after which their cholesterol levels were tested again.

Test the company's claim that the drug lowers cholesterol levels, using alpha = 0.01. Assume that the distribution is normally distributed.

Subject 1 2 3 4 5 6 7 8 9 10
Before 195 225 202 195 175 250 235 268 190 240
After 180 220 210 175 170 250 205 250 190 225

3) These measurements were taken from hole diameters of metal drilled on your equipment that need to meet precise specifications. Three tools are being used for the drilling and the drilling is done during three different shifts. Assume the diameters are approximately normally distributed and the requirements for equal variances are met. Use the following Minitab two-way ANOVA output to answer the questions.

a) Is there a significant difference in the means for the three shifts and the three tools, if yes, explain
b) If there are significant differences in the means, which pairwise means differ? Explain

Minitab output for Question 11
Two-way ANOVA: Diameter versus Shift, Tool

Source DF SS MS F P
Shift 2 0.0001561 0.0000780 25.39 0.000
Tool 2 0.0003116 0.0001558 50.69 0.000
Interaction 4 0.0000179 0.0000045 1.46 0.256
Error 18 0.0000553 0.0000031
Total 26 0.0005410

S = 0.001753 R-Sq = 89.77% R-Sq(adj) = 85.23%

Individual 95% CIs For Mean Based on Pooled StDev
Tool Mean ---------+---------+---------+---------+
1 25.0176 (---*---)
2 25.0094 (---*---)
3 25.0151 (---*---)
---------+---------+---------+---------+
25.0110 25.0140 25.0170 25.0200

Tukey 95.0% Simultaneous Confidence Intervals
Response Variable Diameter
All Pairwise Comparisons among Levels of Tool
Tool = 1 subtracted from:

Tool Lower Center Upper +---------+---------+---------+------
2 -0.01022 -0.008111 -0.006001 (---*---)
3 -0.00455 -0.002444 -0.000335 (---*---)
+---------+---------+---------+------
-0.0100 -0.0050 0.0000 0.0050

Tool = 2 subtracted from:

Tool Lower Center Upper +---------+---------+---------+------
3 0.003557 0.005667 0.007776 (---*----)
+---------+---------+---------+------
-0.0100 -0.0050 0.0000 0.0050

Tukey 95.0% Simultaneous Confidence Intervals
Response Variable Diameter
All Pairwise Comparisons among Levels of Shift
Shift = 1 subtracted from:

Shift Lower Center Upper ------+---------+---------+---------+
2 0.003557 0.005667 0.007776 (----*----)
3 -0.000665 0.001444 0.003554 (-----*----)
------+---------+---------+---------+
-0.0040 0.0000 0.0040 0.0080

Shift = 2 subtracted from:

Shift Lower Center Upper ------+---------+---------+---------+
3 -0.006332 -0.004222 -0.002112 (----*-----)

------+---------+---------+---------+
-0.0040 0.0000 0.0040 0.0080

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Solution Summary

A hypothesis-null and alternative p-values are determined. Complete, Neat and Step-by-step Solutions are provided in the attached file.

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