A researcher uses a matched-subjects design to investigate whether single people who own pets are generally happier than singles without pets. A mood inventory questionnaire is administered to a group of 20 to 29-year old non-pet owners and a similar age group of pet owners.  The pet owners are matched one to one with the non-pet owners for income, number of close friendships, and general health.  The data are as follows: MATCHED PAIR NON-PET OWNER PET OWNER A 12 14 B 8 7 C 10 13 D 9 9 E 7 13 F 10 12   Is there a significant difference in the mood scores for non-pet owners versus pet owners? Test with α = 0.05 for two tails. Construct the 95% confidence interval to estimate the size of the mean difference in mood between the population of pet owners and the population of non-pet owners. (You should find that a mean difference of µ = 0 is an acceptable value, which is consistent with the conclusion from the hypothesis test.)

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
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ISBN:9780079039897
Author:Carter
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Chapter4: Equations Of Linear Functions
Section: Chapter Questions
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A researcher uses a matched-subjects design to investigate whether single people who own pets are generally happier than singles without pets. A mood inventory questionnaire is administered to a group of 20 to 29-year old non-pet owners and a similar age group of pet owners.  The pet owners are matched one to one with the non-pet owners for income, number of close friendships, and general health.  The data are as follows:

MATCHED PAIR

NON-PET OWNER

PET OWNER

A

12

14

B

8

7

C

10

13

D

9

9

E

7

13

F

10

12

 

  • Is there a significant difference in the mood scores for non-pet owners versus pet owners? Test with α = 0.05 for two tails.
  • Construct the 95% confidence interval to estimate the size of the mean difference in mood between the population of pet owners and the population of non-pet owners. (You should find that a mean difference of µ = 0 is an acceptable value, which is consistent with the conclusion from the hypothesis test.)
  • a) 1. H0: µ = 0; HA: µ ≠ 0.
  • α = 0.05, two-sided test -> 0.025 per critical area; D.F. = 6 – 1 = 5. ->Critical t value = 2.57

PAIR

D

MD = (2 – 1 + 3 + 0 + 6 + 2)/6 = 2

s2D = [(2 – 2)2 + (2 – (–1))2 + (2 – 3)2 + (2 – 0)2 + (2 – 6)2 + (2 – 2)2]/(6 – 1) = 6

sMD = SQRT(6/6) = 1

TEST T VALUE = (2 – 0)/1 = 2

A

12 – 14 = 2

B

8 – 7 = -1

C

10 – 13 = 3

D

9 – 9 = 0

E

7 – 13 = 6

F

10  – 12 = 2

  1.  

 

 

 

 

  1. 4)  |2| < |2.57| -> FAIL TO REJECT H0.
  2. LOWER BOUND OF C.I. -> 2 – 2.57*1 = –0.57

    UPPER BOUND OF C.I. -> 2 + 2.57*1 = 4.57

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