Question

Suppose samples of six different brands of diet or imitation margarine were analyzed to determine the level of physiologically active polyunsaturated fatty acids (PAPUFA, in percent), resulting in the data shown in the accompanying table. Imperial 14.1 13.6 14.4 14.3 Parkay 12.8 12.5 13.4 13.0 12.3 Blue Bonnet 13.5 13.4 14.1 14.3 Chiffon 13.2 12.7 12.6 14.1 Mazola 16.8 17.2 16.4 17.3 18.0 17.2 18.7 18.4 Fleischmann's 18.1 (a) Test for differences among the true average PAPUFA percentages for the different brands. Use a = 0.05. Calculate the test statistic. (Round your answer to two decimal places.) What can be said about the P-value for this test? OP-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 What can you conclude? O Reject Ho. There is not convincing evidence that the mean PAPUFA percentages for the six brands are not all equal. Reject Ho. There is convincing evidence that the mean PAPUFA percentages for the six brands are not all equal. O Fail to reject My. There is convincing evidence that the mean PAPUFA percentages for the six brands are not all equal. O Fail to reject My. There is not convincing evidence that the mean PAPUFA percentages for the six brands are not all equal. (b) Use the T-K procedure to compute 95% simultaneous confidence intervals for all differences between means. (Round your answers to three decimal places.) Imperial and Parkay Imperial and Blue Bonnet Imperial and Chiffon Imperial and Mazola Imperial and Fleischmann's Parkay and Blue Bonnet Parkay and Chiffon Parkay and Mazola Parkay and Fleischmann's Blue Bonnet and Chiffon Blue Bonnet and Mazola Blue Bonnet and Fleischmann's Chiffon and Mazola Chiffon and Fleischmann's Mazola and Fleischmann's

Answer 1

(A)

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ₁ = μ₂ = μ₃ = μ₄ = μ₅ = μ₆

Ha: Not all means are equal

The above hypotheses will be tested using an F-ratio for a One-Way ANOVA.

(2) Rejection Region

Based on the information provided, the significance level is \alpha = 0.05α=0.05, and the degrees of freedom are df_1 = 5df1​=5 and df_2 = 5df2​=5, therefore, the rejection region for this F-test is R = \{F: F &gt; F_c = 2.711\}R={F:F>Fc​=2.711}.

(3) Test Statistics

The following table is obtained:

	Group 1	Group 2	Group 3	Group 4	Group 5	Group 6
	14.1	12.8	13.5	13.2	16.8	18.1
	13.6	12.5	13.4	12.7	17.2	17.2
	14.4	13.4	14.1	12.6	16.4	18.7
	14.3	13.0	14.3	14.1	17.3	18.4
		12.3			18.0
Sum =	56.4	64	55.3	52.6	85.7	72.4
Average =	14.1	12.8	13.825	13.15	17.14	18.1
\sum_i X_{ij}^2 =∑i​Xij2​=	795.62	819.94	765.11	693.1	1470.33	1311.7
St. Dev. =	0.356	0.43	0.443	0.686	0.598	0.648
SS =	0.37999999999988	0.74000000000001	0.58750000000009	1.4099999999999	1.4319999999998	1.2599999999998
n =	4	5	4	4	5	4

The total sample size is N = 26N=26. Therefore, the total degrees of freedom are:

dftotal​=26−1=25

Also, the between-groups degrees of freedom are dfbetween​=6−1=5, and the within-groups degrees of freedom are:

dfwithin​=dftotal​−dfbetween​=25−5=20

First, we need to compute the total sum of values and the grand mean. The following is obtained

∑​Xij​=56.4+64+55.3+52.6+85.7+72.4=386.4

Also, the sum of squared values is

∑​Xij2​=795.62+819.94+765.11+693.1+1470.33+1311.7=5855.8

Based on the above calculations, the total sum of squares is computed as follows

$SS_{total} = \sum_{i,j} X_{ij}^2 - \frac{1}{N} \left(\sum_{i,j} X_{ij}\right)^2 = 5855.8 - \frac{ 386.4^2}{ 26} = 113.302$

The within sum of squares is computed as shown in the calculation below:

$SS_{within} = \sum SS_{within groups} = 0.37999999999988+0.74000000000001+0.58750000000009+1.4099999999999+1.4319999999998+1.2599999999998 = 5.809$

M Sbetween S Sbetween dfbetween 107.492 5 = 21.498

M Swithin S Swithin dfurithin 5.809 20 = 0.29

$F = \frac{MS_{between}}{MS_{within}} = \frac{ 21.498}{ 0.29} = 74.011$

F=74.011

(4) Decision about the null hypothesis

Since it is observed that F=74.011>Fc​=2.711, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0(option(D) p<0.001) and since p=0<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that not all 6 population means are equal, at the α=0.05 significance level.

option(B) is correct.

(B)

A=Imperial D=Chiffon

B=Parkey E=Mazola

C=Blue Bonnet F=Fleischmaan's

Tukey HSD results

treatments pair	Tukey HSD Q statistic	Tukey HSD p-value	Tukey HSD inferfence
A vs B	5.0851	0.0192747	* p<0.05
A vs C	1.0205	0.8999947	insignificant
A vs D	3.5253	0.1733858	insignificant
A vs E	11.8913	0.0010053	** p<0.01
A vs F	14.8435	0.0010053	** p<0.01
B vs C	4.0094	0.0922096	insignificant
B vs D	1.3691	0.8999947	insignificant
B vs E	18.0061	0.0010053	** p<0.01
B vs F	20.7315	0.0010053	** p<0.01
C vs D	2.5048	0.5038367	insignificant
C vs E	12.9670	0.0010053	** p<0.01
C vs F	15.8640	0.0010053	** p<0.01
D vs E	15.6073	0.0010053	** p<0.01
D vs F	18.3688	0.0010053	** p<0.01
E vs F	3.7551	0.1292520	insignificant

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