Question

Please round to 3 decimal places

Consider the accompanying data collected for a two-way ANOVA Click the icon to view the data table a) Using a b) Using c) Using a 0.025, are the Factor B means different? = 0.025, is there significant interaction between Factors A and B? a = 0.025, are the Factor A means different? a) Using a 0.025, is there significant interaction between Factors A and B? Identify the hypotheses for the interaction between Factors A and B. Choose the correct answer below B do interact, H,: Factor A and B do not interact OA. Ha: Factor A and O B. Ho HA#Hg. H: PA PB O C. H: Factor A and B do not interact, H,: Factor A and B do interact OD. Ho:PAHg. H: HA PB Find the p-value for the interaction between Factors A and B. (Round to three decimal places as needed) p-value Draw the appropriate conclusion for the interaction between Factors A and B. Choose the correct answer below. OA Reject the null hypothesis. There is insufficient evidence to conclude that the means differ. O B. Do not reject the null hypothesis. There is insufficient evidence to conclude that the means differ. O C. Do not reject the null hypothesis. O D. Reject the null hypothesis. There is sufficient evidence to conclude that Factors A and B interact b) Using a 0.025, are the Factor A means different? There is insufficient evidence to conclude that Factors A and B interact Identify the hypotheses to test for Factor A. Choose the correct answer below OA. Ho: PAPB H: PA #HB O B. Ho: HA1HA2 " HA3, H,: Not all Factor A means are equal O C. Ho: HA1#HA2 # PA3 . H : PA1 HA2 HA3 O D. Ho: HA1 PA2HA3 . H : PA1 HA2 HA3 Find the p-value for Factor A. p-value (Round to three decimal places as needed.) Draw the appropriate conclusion for Factor A. Choose the correct answer below. O A. Reject the null O B. Do not reject the null hypothesis. There is insufficient evidence to conclude that not all Factor A means are equal. O c. Reject the null hypothesis. There is sufficient evidence to conclude that not all Factor A means are equal. O D. It is inappropriate to draw a conclusion from this test because the Factors A and B interact. hypothesis. There is insufficient evidence to conclude that the means differ.

Answer 1

excel output-

Anova: Two-Factor With Replication
SUMMARY	level 1	level2	level3	Total
level1
Count	3	3	3	9
Sum	28	41	62	131
Average	9.333333	13.66667	20.66667	14.55556
Variance	26.33333	22.33333	94.33333	60.27778
level2
Count	3	3	3	9
Sum	53	74	93	220
Average	17.66667	24.66667	31	24.44444
Variance	74.33333	41.33333	37	71.52778
level3
Count	3	3	3	9
Sum	95	107	106	308
Average	31.66667	35.66667	35.33333	34.22222
Variance	72.33333	44.33333	21.33333	38.19444
Total
Count	9	9	9
Sum	176	222	261
Average	19.55556	24.66667	29
Variance	138.7778	117.75	80.75
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Sample	1740.519	2	870.2593	18.06072	4.98E-05	3.554557
Columns	402.2963	2	201.1481	4.174481	0.032403	3.554557
Interaction	90.37037	4	22.59259	0.46887	0.757862	2.927744
Within	867.3333	18	48.18519
Total	3100.519	26

a)
Ho: factors A and B do not interact
H1: factor A and B do interact

p value=0.758
do not reject Ho, there is insufficient evidence to conclude that factors A and B interact

b)

Ho: µA1=µA2=µA3; H1: not all factor A means are equal
p value for factor A=0.032
so,
do not reject null hypothesis, there is insufficient evidence to conclude that not all factor A means are equal

c)

Ho: µB1=µB2=µB3; H1: not all factor B means are equal
p value for factor B=0.000 <α
So,
reject null hypothesis, there is sufficient evidence to conclude that not all factor B means are equal