Question

According to the 2018 general social survey (gss), 60% of all u. s. households have at least one pet. a veterinarian in ann arbor decided to conduct a survey to assess if over 60% of all ann arbor households have at least one pet. a 10% significance level was selected to be used. the random sample of 280 ann arbor households, resulted in 182 stating they currently have at least one pet. these results are turned over to you to perform the appropriate test. 1) state the appropriate hypotheses to be tested with a complete definition of the parameter of interest. (the significance level has been selected to be 10%). 2) check the necessary assumption(s) (you may assume that the selected sample is a random sample). 3) perform the appropriate test (include all supporting computations for the test statistic and p-value) 4) give your decision and provide a conclusion in context. your answer needs to be organized well and use labels for your steps 1, 2, 3, and 4. question 8 answer no answer entered.

Answer 1

Explanation:

1. the claim here is that over 60 percents of the households living in ann harbor have pets.

null hypothesis;

h0: p≤0.60

alternative:

h1: p>0.60

2.

n is the size of the sample = 280

n(p) = 280 x 0.60 = 168 greater than 10

n(1-p) = 280(1-0.6) = 280 x 0.4 = 112 greater than 10

we use the normal dist since each of the above are greater than 10

3. n = 280

x = 182

x/n = 182/280

= 0.65

we get the test statistic z

z =
`(0.65-0.6)/(√(0.6*(1-0.6)/280) )`

=
`(0.05)/(√(0.000857) )`

=
`(0.05)/(0.02927)`

≈
`1.71`

we get the p value = 1-p(z<1.71)

= 1-.9562

this gives 0.0438

4. we can see that 0.0438 < 0.10

so we conclude that over 60% of these households have at least one pet since we rejected h0.