Question

vv==v=v=a. report the descriptive statistics for each sample b. report and analyze the ANOVA test for the three cities

Florida	New York	North Carolina
13	14	10
12	9	12
17	15	15
17	12	18
20	16	12
21	24	14
16	18	17
14	14	8
13	15	14
17	17	16
12	20	18
9	11	17
12	23	19
15	19	15
16	17	13
15	14	14
13	9	11
10	14	12
11	13	13
17	11	11

Answer 1

Solve using Excel:

a) Go to Data, select Data Analysis, choose Descriptive Statistics and group by columns.

Florida		New York		North Carolina
Mean	14.50	Mean	15.25	Mean	13.95
Standard Error	0.71	Standard Error	0.92	Standard Error	0.66
Median	14.50	Median	14.50	Median	14.00
Mode	17.00	Mode	14.00	Mode	12.00
Standard Deviation	3.17	Standard Deviation	4.13	Standard Deviation	2.95
Sample Variance	10.05	Sample Variance	17.04	Sample Variance	8.68
Kurtosis	-0.34	Kurtosis	-0.03	Kurtosis	-0.59
Skewness	0.28	Skewness	0.53	Skewness	-0.04
Range	12	Range	15	Range	11
Minimum	9	Minimum	9	Minimum	8
Maximum	21	Maximum	24	Maximum	19
Sum	290	Sum	305	Sum	279
Count	20	Count	20	Count	20

b) Using Excel. go to Data, select Data Analysis, choose Anova: Single Factor.

SUMMARY
Groups	Count	Sum	Average	Variance
Florida	20	290	14.500	10.053
New York	20	305	15.250	17.039
North Carolina	20	279	13.950	8.682
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	17.033	2	8.517	0.714	0.494	3.159
Within Groups	679.700	57	11.925
Total	696.733	59

H₀: μ₁ = μ₂ = μ₃, There is no significant difference in means of the three cities

H₁: At least one μ_i, There is a significant difference in means of the three cities

Test statistic: F = 0.714

Level of significance = 0.05

Critical value: F crit = 3.159

Since test statistic is less than critical value, we do not reject the null hypothesis.

So, μ₁ = μ₂ = μ₃, there is no significant difference in means of the three cities.