Question

A researcher is interested in investigating whether living situation and pet ownership are dependent. The table below shows the results of a survey.

Frequencies of Living Situation and Pet

	Single	Family	Couple
Dog	90	90	78
Cat	102	79	93
Various	32	58	49
None	53	47	35

What can be concluded at the αα = 0.10 significance level?

1. What is the correct statistical test to use?
   • Goodness-of-Fit
   • Independence
   • Paired t-test
   • Homogeneity
2. What are the null and alternative hypotheses?

H0:

• The distribution of pets is the same for each living situation.
• The distribution of pets is not the same for each living situation.
• Pet ownership and living situation are independent.
• Pet ownership and living situation are dependent.

H1:

• Pet ownership and living situation are dependent.
• The distribution of pets is not the same for each living situation.
• The distribution of pets is the same for each living situation.
• Pet ownership and living situation are independent.

3. The test-statistic for this data = _____ (Please show your answer to three decimal places.)  

4. The p-value for this sample = _____ (Please show your answer to four decimal places.)  

5. The p-value is? (greater than, less than or equal to) α

6.

1. Based on this, we should
   • reject the null
   • fail to reject the null
   • accept the null
2. Thus, the final conclusion is...
   • There is sufficient evidence to conclude that pet ownership and living situation are independent.
   • There is sufficient evidence to conclude that the distribution of pets is not the same for each living situation.
   • There is insufficient evidence to conclude that pet ownership and living situation are dependent.
   • There is sufficient evidence to conclude that pet ownership and living situation are dependent.
   • There is insufficient evidence to conclude that the distribution of pets is not the same for each living situation.

Answer 1

		Observed Frequencies:
	Single	Family	Couple	TOTAL
Dog	90	90	78	258
Cat	102	79	93	274
Various	32	58	49	139
None	53	47	35	135
TOTAL	277	274	255	806
E1=	RT*CT/GT =	258*227/806
		88.6674938
O=	OBSERVED FRQ
E =	EXPECTED FREQ
EXPECTED  frequencies
	Single	Family	Couple	TOTAL
Dog	88.6674938	87.70719603	81.62531017	258
Cat	94.1662531	93.14640199	86.68734491	274
Various	47.77047146	47.25310174	43.9764268	139
None	46.39578164	45.89330025	42.71091811	135
TOTAL	277	274	255	806
	Level =	0.1
	CALCULATION TABLE
	O	E	O-E	(O-E)2/E
Dog S	90	88.6674938	1.332506203	0.02002507
Cat S	102	94.1662531	7.833746898	0.651694088
Various S	32	47.77047146	-15.77047146	5.206307633
None S	53	46.39578164	6.604218362	0.940079004
Dog   F	90	87.70719603	2.29280397	0.0599375
Cat  F	79	93.14640199	-14.14640199	2.14845324
Various   F	58	47.25310174	10.74689826	2.444195577
None  F	47	45.89330025	1.106699752	0.02668765
Dog  C	78	81.62531017	-3.625310174	0.161014688
Cat  C	93	86.68734491	6.312655087	0.459693561
Various  C	49	43.9764268	5.023573201	0.573859441
None  C	35	42.71091811	-7.710918114	1.392109109
SUM =	806	806	0	14.08405656
What is the correct statistical test to use?
Independence
H0:	Pet ownership and living situation are independent
H1:	Pet ownership and living situation are dependent.
3. The test-statistic for this data =	14.084
4. The p-value for this sample =	0.028712
5. The p-value is less than α
	0.028712 < 0.10
Based on this, we should
reject the null
Thus, the final conclusion is..
There is sufficient evidence to conclude that pet ownership and living situation are dependent.