Question

In: Statistics and Probability

A researcher hypothesizes that students’ performance in a test is better in situations where the surrounding...

A researcher hypothesizes that students’ performance in a test is better in situations where the surrounding noise is kept lowest. She collects data on her students under three different levels of noise. A higher score indicates better performance. Her data follow. SHOW ALL YOUR WORK.

High Noise

Mod Noise

Low Noise

3

4

6

G =

5

6

8

ΣX2 =

393

4

4

6

N =

3

2

7

k =

2

3

8

T = 17

T = 19

T =

M = 3.4

M = 3.8

M =

SS = 5.2

SS = 8.8

SS =

n = 5

n = 5

n =

  1. State the hypotheses
  1. Calculate the F ratio and complete the Summary Table.

Source

SS

df

MS

F

Between

Within

Total

  1. What is the critical value of F? What decision can you make? What does that indicate?
  1. Calculate Turkey’s HSD and determine which pairwise comparisons are statistically significant.

Solutions

Expert Solution

treatment A1 A2 A3
count, ni = 5 5 5
mean , x̅ i = 3.400 3.80 7.000
std. dev., si = 1.140 1.483 1.000
sample variances, si^2 = 1.300 2.200 1.000
total sum 17 19 35 71 (grand sum)
grand mean , x̅̅ = Σni*x̅i/Σni =   4.73
square of deviation of sample mean from grand mean,( x̅ - x̅̅)² 1.778 0.871 5.138
TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² = 8.889 4.356 25.689 38.93333333
SS(within ) = SSW = Σ(n-1)s² = 5.200 8.800 4.000 18.0000

no. of treatment , k =   3
df between = k-1 =    2
N = Σn =   15
df within = N-k =   12
  
mean square between groups , MSB = SSB/k-1 =    19.4667
  
mean square within groups , MSW = SSW/N-k =    1.5000
  
F-stat = MSB/MSW =    12.9778

anova table
SS df MS F p-value F-critical
Between: 38.9333 2 19.467 12.978 0.0010 3.885
Within: 18.0000 12 1.500
Total: 56.9333 14
α = 0.05

critical F = 3.885

reject the null hypothesis

conclusion :    there is enough evidence of significant mean difference among three treatments

==========

tukey's test

Level of significance 0.05
no. of treatments,k 3
DF error =N-k= 12
MSE = 1.5000
q-statistic value(α,k,N-k) 3.7728

critical value = q*√(MSE/2*(1/ni+1/nj)) = 2.07

population mean difference critical value result
µ1-µ2 0.40 2.07 means are not different
µ1-µ3 3.600 2.07 means are different
µ2-µ3 3.20 2.07 means are different

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