The management of an retirement community would like to establish a new “no-pet”
ID: 3275981 • Letter: T
Question
The management of an retirement community would like to establish a new “no-pet” policy for their senior residents but they have seen a report suggesting that keeping a dog as a pet may impact the number of doctor’s visits made by the dog owners. A psychologist was therefore hired to conduct a small study on the residents in this community to examine the potential effect of dog ownership on a senior resident’s number visits to the doctor. The psychologist recruited 10 seniors who has one dog as a pet (and no other pet animals) and 12 seniors with similar demographics with no dog or any other type of pet. The data are listed in the table below. The psychologist is not predicting a particular direction of the potential differences between the two groups and she sets the alpha level at .10 for the hypothesis test.
Owning a dog
Not owning a dog
Subject ID #
# of visits to doctor
Subject ID #
# of visits to doctor
1
5
11
10
2
4
12
8
3
9
13
9
4
5
14
9
5
7
15
13
6
6
16
7
7
7
17
6
8
8
18
12
9
10
19
10
10
9
20
8
21
11
22
5
Calculate 2/1 and 2/2(estimated variance for population 1 and variance for population 2) (4 points total: 2 points for each variance. 1 for work and 1 for result)
Calculate the pooled variance for the two populations (2 points total: 1 point for work, 1 point for result)
Use the pooled variance to calculate the variance for sampling distribution 1 and the variance for sampling distribution 2 (2 points total: 1 for work and 1 for result)
Calculate standard deviation of the comparison distribution (2 points total: 1 for work and 1 for result)
Owning a dog
Not owning a dog
Subject ID #
# of visits to doctor
Subject ID #
# of visits to doctor
1
5
11
10
2
4
12
8
3
9
13
9
4
5
14
9
5
7
15
13
6
6
16
7
7
7
17
6
8
8
18
12
9
10
19
10
10
9
20
8
21
11
22
5
Explanation / Answer
Estimated variance for population 1 21 = (xi -x)2 /N
where x= xi /n = 70/10 = 7
21 = (xi - 7)2 /10 = 4
similarly,
Estimated variance for population 2 22= (xi -x)2 /N
where x= xi /n = 108/12 = 7
22 = (xi - 9)2 /12 = 5.64
Pooled Variance 2p= [(n1 -1) 12 + (n2 -1) 22 ]/ (n1 + n2 -2) = [9 * 4 + 11 * 5.64]/ 20 = 4.9
standard deviation of the comparison distribution s = sqrt (2p) = 2.2136
test statistic
t = (7 - 9) / sqrt[4.9 * (1/10 + 1/12) ] = 2.11
for dF = 20 and alpha = 0.10
critical value of t = 1.325
so we can reject the null hypothesis and can say that there is potential effects of dogs on number of doctors visit.
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