Question

1.

Sales (in thousands of dollars) over 5 days at 4 different stores are recorded below.

Store 1	Store 2	Store 3	Store 4
10	20	13	30
15	20	17	25
10	25	15	30
20	15	20	35
20	20	14	30

Test the claim that there is a significant difference among the mean amount of sales at the 4 stores, using α=0.05. Round you answer to 3 decimal places.

Group of answer choices

p-value=0.000, evidence support claim

p-value=0.578, evidence support claim

p-value=3.796, evidence not support claim

p-value=0.578, evidence not support claim

p-value=0.000, evidence not support claim

2.

We randomly sample 200 adults and ask whether they like plain, peanuts or almond M&M's. The result is summarized according to gender. Test the claim that the preference of M&M's is independent of gender. Use α=0.01. Round your answer to 3 decimal places.

	Plain	Peanuts	Almond
Male	40	50	20
Female	50	20	20

Group of answer choices

p-value=0.237, evidence support claim

p-value=0.002, evidence not support claim

p-value=0.293, evidence support claim

p-value=0.237, evidence not support claim

p-value=0.002, evidence support claim

3.

Mr. Arnold teaches an algebra course and he gave 2 tests to the 29 students in the class. Below is a summary of how the 29 students performed on those 2 tests:

Test 1: Average=86.66, standard deviation=8.78

Test 2: Average=66.83, standard deviation=17.67

Test 1- Test 2: Average= 18.83, standard deviation=14.87

Assuming that the data are normal, we want to test the claim that students' average score from Test 1 is higher than Test 2. Use a 0.05 level of significance. Round your answer to 3 decimal places.

Group of answer choices

p-value=1.000, evidence not support claim

p-value=0.000, evidence not support claim

p-value=1.041, evidence support claim

p-value=0.000, evidence support claim

p-value=1.000, evidence support claim

Answer 1

1)

Applying one way ANOVA from excel: data-data analysis: one way ANOVA:

Source of Variation	SS	df	MS	F	P-value
Between Groups	712.4	3	237.4667	16.46216	0.0000
Within Groups	230.8	16	14.425
Total	943.2	19

p-value=0.000, evidence support claim

2)

Applying chi square test of independence:

Expected	Ei=row total*column total/grand total	c	d	e	Total
	a	49.50	38.50	22.00	110
	b	40.50	31.50	18.00	90
	total	90	70	40	200
chi square    χ2	=(Oi-Ei)2/Ei	c	d	e	Total
	a	1.823	3.435	0.182	5.4401
	b	2.228	4.198	0.222	6.6490
	total	4.0516	7.6335	0.4040	12.0891
test statistic X2 =		12.089
p value =		0.0024

p-value=0.002, evidence not support claim

3)

population mean μ=	0
sample mean 'x̄=	18.830
sample size    n=	29.00
std deviation s=	14.870
std error ='sx=s/√n=	2.7613
test stat t ='(x-μ)*√n/sx=	6.819
p value      =	0.0000

p-value=0.000, evidence support claim