Question

In: Statistics and Probability

Four different paints are advertised to have the same drying times. To verify the manufacturer’s claim,...

  1. Four different paints are advertised to have the same drying times. To verify the manufacturer’s claim, seven samples were tested for each of the paints. The time in minutes until the paint was dry enough for a second coat to be applied was recorded. Below are the results which may be imported into MSExcel for analysis: (Assume the populations are normally distributed, the populations are independent and the population variances are equal)

Paint 1

Paint 2

Paint 3

Paint 4

120

120

117

128

112

130

122

131

121

121

123

131

118

126

115

129

118

126

123

127

121

114

126

126

118

117

126

137

  1. Write the Null and Alternative hypothesis to test whether there is a difference in dry time between the samples of each paint?
  1. What statistic would you use to analyze this

  1. At a 0.01 level of significance, what would be your decision rule?
  1. From your analysis, is there a difference between drying times

  1. If there was a difference in dry times how would you determine which paint has the different drying time? (i.e. what formula would you use and what is your decision criteria?) (this is not asking for a calculation)

Solutions

Expert Solution

A)

Ho: µ1=µ2=µ3=µ4
H1: not all means are equal

.....

B)

F stat will be used

treatment G1 G2 G3 G4
count, ni = 7 7 7 7
mean , x̅ i = 118.286 122.00 121.71 129.86
std. dev., si = 3.1 5.6 4.2 3.7
sample variances, si^2 = 9.571 31.667 17.905 13.476
total sum 828 854 852 909 3443 (grand sum)
grand mean , x̅̅ = Σni*x̅i/Σni =   122.96
( x̅ - x̅̅ )² 21.889 0.930 1.563 47.511
TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² = 153.223 6.509 10.938 332.580 503.25
SS(within ) = SSW = Σ(n-1)s² = 57.429 190.000 107.429 80.857 435.7143

no. of treatment , k = 4
df between = k-1 = 3
N = Σn = 28
df within = N-k = 24
  
mean square between groups , MSB = SSB/k-1 = 503.25/3= 167.7500
mean square within groups , MSW = SSW/N-k = 435.7143/24= 18.1548
  
F-stat = MSB/MSW = 167.75/18.1548= 9.24
  
P value = 0.0003

anova table
SS df MS F p-value
Between: 503.3 3 167.8 9.24 0.000
Within: 435.7 24 18.2
Total: 939.0 27

F =9.24

................

c)

α =    0.01

F-critical = 4.718

so, f stat should be greater than 4.718

d)



Decision:  f stat > critical , reject null hypothesis

so, there is difference between drying times

....

e)

tukey kramer rule:

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

if absolute difference of means > critical value,means are significnantly different ,otherwise not

........................

Please let me know in case of any doubt.

Thanks in advance!


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