Question

Part (a)
using the data set below on the account balances of customers at a bank’s four locations.

Using that data set and an α of 0.05, test the null hypothesis that the mean account balances

are equal in the four towns using a one-way ANOVA in Excel.

using Excel file include the output as part of the answer.

Part (b)   

Do you reject the null hypothesis or not? Indicate on which part of the Excel output you base your decision.

Part (c)
Assuming that an acquaintance of yours has never heard of ANOVA, explain to him what the decision in part (b),

i.e., “rejecting H₀” or “not rejecting H₀” means in this context at a level that can be understood by a high school senior.

Customers' Checking Account Balances by Location

City 1	City 2	City 3	City 4
789	748	1831	1756
2051	1501	740	2125
765	1886	1554	1995
1645	1593	137	1526
1266	1474	2276	1746
2138	1913	2144	1616
1487	1218	1053	1958
1622	1006	1120	1675
1169	343	1838	1885
2215	1494	1735	2204
167	580	1326	2409
2557	1320	1790	1338
634	1784	32	2076
1326	1044	1455	2375
1790	890	1913	1125
32	1708	1218	1326
1455	1913	1006	1790
	1218	343	32
	1006	1494	1455
	343	580

Answer 1

Solution:

part(a)

Our null hypothesis is that the mean account balances are equal in the four towns.

The alternative hypothesis is that the mean account balances are not equal in the four towns.

We perform One-Way ANOVA and obtain the following results:

Descriptives
City
	N	Mean	Std. Deviation	Std. Error	95% Confidence Interval for Mean		Minimum	Maximum
					Lower Bound	Upper Bound
City 1	17	1359.29	709.179	172.001	994.67	1723.92	32	2557
City 2	20	1249.10	498.057	111.369	1016.00	1482.20	343	1913
City 3	20	1279.25	651.091	145.588	974.53	1583.97	32	2276
City 4	19	1705.89	538.340	123.504	1446.42	1965.37	32	2409
Total	76	1395.88	617.338	70.813	1254.81	1536.95	32	2557

ANOVA
City
	Sum of Squares	df	Mean Square	F	Sig.
Between Groups	2551767.065	3	850589.022	2.353	.079
Within Groups	26031162.869	72	361543.929
Total	28582929.934	75

The excel output is the following:

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	2551767.065	3	850589	2.352657	0.07928	2.731807
Within Groups	26031162.87	72	361544
Total	28582929.93	75

Part(b).

So, here we find the p-value is = 0.07928.

Since p-value = 0.07928>0.05, we fail to reject the null hypothesis that there is a significant difference between the mean of cities.

Part(c)

Here we fail to reject the null hypothesis H0 which states that the mean account balances are equal in the four towns. The data do not provide enough evidence against the alternative hypothesis that the mean account balances are not equal in the four towns. Which means there may or may not be a significant difference between the mean of the four towns.