Suppose a veterinarian wants to estimate the difference between the proportion o
ID: 3319001 • Letter: S
Question
Suppose a veterinarian wants to estimate the difference between the proportion of cat owners who are single and the proportion of dog owners who are single. Of the pet owners that visit any of the veterinarian clinics in her city regularly, she identifies 3872 pet owners that exclusively have cats and 4108 pet owners that exclusively have dogs. From this list of pet owners, she surveys 158 randomly selected cat owners and 139 randomly selected dog owners and asks each of them if they are single or married. Her findings are summarized in the table.
Calculate the upper and lower limits (bounds) for a large sample 90% zconfidence interval for the difference in two population proportions, p1p2. Give each of your answers with three decimal places of precision.
Then, complete the following sentence to state the interpretation of the confidence interval.
The that the
who are single is between and
.
Answer Bank
0.068
difference in the proportions of all cat and dog owners
probability is 90%
proportion of dog owners
mean number of cat owners
0.333
0.092
0.097
0.866
0.068
veterinarian can be 90% confident
difference in the mean numbers of cat and dog owners
0.463
0.043
0.048
0.089
mean number of dog owners
difference in the proportions of cat and dog owners in the samples
0.039
0.415
0.107
0.156
proportion of cat owners
0.335
0.118
0.680
0.118
lower limit =
upper limit =
Population Populationdescription Sample
size Number
of successes Sample
proportion 1 cat owners n1=158 x1=63 p 1=0.39873 2 dog owners n2=139 x2=52 p 2=0.37410
Explanation / Answer
Sample X N Sample p
1 63 158 0.398734
2 52 139 0.374101
Difference = p (1) - p (2)
Estimate for difference: 0.0246335
90% CI for difference: (-0.0684412, 0.117708)
Test for difference = 0 (vs 0): Z = 0.43 P-Value = 0.664
Fisher’s exact test: P-Value = 0.721
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