Question

3. The following data are weights of food (in kilograms) consumed per day by adult deer collected at different times of the year.

Month Weight of food consumed (kg)

Feb. 4.7, 4.9, 5.0, 4.8, 4.7

Mar. 4.6, 4.4, 4.3, 4.4, 4.1, 4.2

Oct. 4.8, 4.7, 4.6, 4.4, 4.7, 4.8

Dec. 4.9, 5.2, 5.4, 5.1, 5.6

1. (a) Is there any difference in the food consumption for all the months studied?

2. (b) If food consumption is not the same for all the months, which ones are different (you must show the test used and results for interpretation)?

Answer 1

Solve using Excel, go to Data, select Data Analysis, choose Anova: single factor and groupp by rows

a) H0: µ1 = µ2 = µ3 = µ4, There is no difference in the food consumption for all the months studied

H1: There is a difference in the food consumption for all the months studied

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Feb	5	24.100	4.820	0.017
Mar	6	26.000	4.333	0.031
Oct	6	28.000	4.667	0.023
Dec	5	26.200	5.240	0.073
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	2.307	3	0.769	22.084	0.000	3.160
Within Groups	0.627	18	0.035
Total	2.933	21

Since p-value is less than 0.05, we reject the null hypotesis and conclude that there is a difference in the food consumption for all the months studied.

b) To detect difference in pairs, conduct t-tests on Excel

using Excel, go to Data, select Data Analysis, choose t-Test: Two-Sample Assuming Unequal Variances

For (Feb, Mar)

H0: µ1 = µ2

H1: µ1 ≠ µ2

t-Test: Two-Sample Assuming Unequal Variances
	Feb	Mar
Mean	4.82	4.333333
Variance	0.017	0.030667
Observations	5	6
Hypothesized Mean Difference	0
df	9
t Stat	5.275194
P(T<=t) one-tail	0.000255
t Critical one-tail	1.833113
P(T<=t) two-tail	0.00051
t Critical two-tail	2.262157

Since p-value = 0.00051 is more than 0.05, we reject the null hypothessis and conclude that µ1 ≠ µ2

For (Feb, Oct),

H0: µ1 = µ3

H1: µ1 ≠ µ3

t-Test: Two-Sample Assuming Unequal Variances
	Feb	Oct
Mean	4.82	4.666667
Variance	0.017	0.022667
Observations	5	6
Hypothesized Mean Difference	0
df	9
t Stat	1.809846
P(T<=t) one-tail	0.05188
t Critical one-tail	1.833113
P(T<=t) two-tail	0.10376
t Critical two-tail	2.262157

Since p-value = 0.10376 is more than 0.05, we do not reject the null hypothessis and conclude that µ1 = µ3

For (Feb, Mar)

H0: µ1 = µ4

H1: µ1 ≠ µ4

t-Test: Two-Sample Assuming Unequal Variances
	Feb	Dec
Mean	4.82	5.24
Variance	0.017	0.073
Observations	5	5
Hypothesized Mean Difference	0
df	6
t Stat	-3.1305
P(T<=t) one-tail	0.010156
t Critical one-tail	1.94318
P(T<=t) two-tail	0.020312
t Critical two-tail	2.446912

Since p-value = 0.020312 is less than 0.05, we reject the null hypothessis and conclude that µ1 ≠ µ4

For (Mar, Oct),

H0: µ2 = µ3

H1: µ2 ≠ µ3

t-Test: Two-Sample Assuming Unequal Variances
	Mar	Oct
Mean	4.333333	4.666667
Variance	0.030667	0.022667
Observations	6	6
Hypothesized Mean Difference	0
df	10
t Stat	-3.53553
P(T<=t) one-tail	0.002698
t Critical one-tail	1.812461
P(T<=t) two-tail	0.005397
t Critical two-tail	2.228139

Since p-value = 0.005397 is more than 0.05, we do not reject the null hypothessis and conclude that µ2 = µ3

For (Mar, Dec)

H0: µ2 = µ4

H1: µ2 ≠ µ4

t-Test: Two-Sample Assuming Unequal Variances
	Mar	Dec
Mean	4.333333	5.24
Variance	0.030667	0.073
Observations	6	5
Hypothesized Mean Difference	0
df	7
t Stat	-6.45791
P(T<=t) one-tail	0.000174
t Critical one-tail	1.894579
P(T<=t) two-tail	0.000348
t Critical two-tail	2.364624

Since p-value = 0.00003 is less than 0.05, we reject the null hypothessis and conclude that µ2 ≠ µ4

For (Mar, Oct),

H0: µ2 = µ3

H1: µ2 ≠ µ3

t-Test: Two-Sample Assuming Unequal Variances
	Mar	Oct
Mean	4.333333	4.666667
Variance	0.030667	0.022667
Observations	6	6
Hypothesized Mean Difference	0
df	10
t Stat	-3.53553
P(T<=t) one-tail	0.002698
t Critical one-tail	1.812461
P(T<=t) two-tail	0.005397
t Critical two-tail	2.228139

Since p-value = 0.005397 is more than 0.05, we do not reject the null hypothessis and conclude that µ2 = µ3

For (Oct, Dec)

H0: µ3 = µ4

H1: µ3 ≠ µ4

t-Test: Two-Sample Assuming Unequal Variances
	Oct	Dec
Mean	4.666667	5.24
Variance	0.022667	0.073
Observations	6	5
Hypothesized Mean Difference	0
df	6
t Stat	-4.22922
P(T<=t) one-tail	0.002752
t Critical one-tail	1.94318
P(T<=t) two-tail	0.005505
t Critical two-tail	2.446912

Since p-value = 0.0005 is less than 0.05, we reject the null hypothessis and conclude that µ3 ≠ µ4