Question

3. 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?

2. What statistic would you use to analyze this?

3. At a 0.01 level of significance, what would be your decision rule?

4. From your analysis, is there a difference between drying times?

5. 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)

please show all the work

Answer 1

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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