Question

2,-/12 pointsDevoreSta19 10.Е.027 Although tea is the world's most widely consumed beverage after water, little is known about its nutritional value. Folacin is the only B vitamin present in any significant amount in tea, and recent advances in assay methods have made accurate determination of folacin content feasible. Consider the accompanying data on folacin content for randomly selected specimens of the four leading brands of green tea. 1: 7.8 6.3 6.6 8.7 9.0 10.1 9.6 2: 5.8 7.5 9.8 6.2 8.4 3: 6.7 7.6 5.1 7.4 5.3 6.2 4: 6.5 7.2 8.0 4.6 5.0 3.9 Does this data suggest that true average folacin content is the same for all brands? (a) Carry out a test using α-0.05. State the appropriate hypotheses Ha: at least two μ's are unequal Ha: all four μ's are equal Ha: at least two uis are equal Ha: all four i's are unequal Find the value of the test statistic in this test. (Round your answer to two decimal places.) What can be said about the P-value? O P-value 0.100 0.050 < P-value 0.100 0.010 < P-value 0.050 0.001 < P-value 0.010 O P-value < 0.001 State the conclusion in the problem context. O Fail to reject Ho. There is not sufficient evidence that at least two of the tea blends have different folacin content. Fail to reject Ho- There is sufficient evidence that at least two of the tea blends have different folacin content O Reject Ho There is not sufficient evidence that at least two of the tea blends have different folacin content

Answer 1

	sampl1	sampl2	sample3	sample 4
count, ni =	7	5	6	6
mean , x̅ i =	8.30	7.54	6.38	5.87
std. dev., si =	1.46	1.63	1.05	1.61
sample variances, si^2 =	2.120	2.668	1.094	2.591
total sum	58.1	37.7	38.3	35.2	169.3	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =				7.05
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	1.552100694	0.24	0.450017361	1.410
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	10.86470486	1.18	2.700104167	8.461	23.20592
SS(within ) = SSW = Σ(n-1)s² =	12.72	10.67	5.468333333	12.953	41.81367

no. of treatment , k =   4
df between = k-1 =    3
N = Σn =   24
df within = N-k =   20
  
mean square between groups , MSB = SSB/k-1 =    7.735
  
mean square within groups , MSW = SSW/N-k =    2.091
  
F-stat = MSB/MSW =    3.700
P value =   0.029

anova table
	SS	df	MS	F	p-value	F-critical
Between:	23.206	3	7.735	3.70	0.0288	3.0984
Within:	41.814	20	2.091
Total:	65.020	23
					α =	0.05

a)

option a)

F-stat=3.70

0.010<p-value <0.050

conclusion : option d) . reject Ho, there is sufficient evidence.............

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b)

the normal probability plot is linear ,so normality assumption is plausible

the equal varianceassum[tion is plausible ..................

so, answer is option a) and c)

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c)

	sampl1	sampl2	sample3	sample 4
count, ni =	7	5	6	6
mean , x̅ i =	8.30	7.54	6.38	5.87

Tukey Kramer test

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

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

population mean difference		critical value			result
µ1-µ2	0.760	w12=2.37			means are not different
µ1-µ3	1.917	w13=2.25			means are not different
µ1-µ4	2.433	w14=2.25			means are different
µ2-µ3	1.157	w23=2.45			means are not different
µ2-µ4	1.673	w24=2.45			means are not different
µ3-µ4	0.517	w34=2.34			means are not different

brand 1 and brand 4 is significantly different