Question

The table shows the prices (in dollars) for a sample of automobile batteries. The prices are classified according to battery type. At α=.05α = .05α = .05, is there enough evidence to conclude that at least one mean battery price is different from the others? (Assume the samples are random and independent, the populations are normally distributed, and the population variances are equal)

Battery Type A 110 100 125 90 120

Battery Type B 280 145 180 175 90

Battery Type C 140 125 85 140 80

P-value = ???

Answer 1

Solution:

We can use the excel ANOVA: Single Factor data analysis tool to find the answer to the given questions.
The excel steps are:

Enter the data in excel.

Click on Data > Data Analysis > ANOVA: Single Factor > OK

Input Range: Select the data range for all the data including labels

Mark Labels in the first row

Alpha = 0.05

Choose the output range and click OK.

The output is given below:

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Battery type A	5	545	109	205
Battery type B	5	870	174	4792.5
Battery type C	5	570	114	867.5
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	13083.33333	2	6541.666667	3.346121057	0.070003393	3.885293835
Within Groups	23460	12	1955
Total	36543.33333	14