A veterinarian wants to know if pit bulls or golden retrievers have a higher incidence of tooth decay at the age of three. The vet surveys 120 three-year-old pit bulls and finds 30 of them have tooth decay. The vet then surveys 160 three-year-old golden retrievers and finds 32 of them have tooth decay. Number the population of pit bulls and golden retrievers by 1 and 2, respectively. Calculate a 90% confidence interval for the difference in the population proportion of pit bulls and golden retrievers that have tooth decay. Which of the following is correct? A) [-0.0492, 0.1492]
B) [-0.0332, 0.1332]
C) [0.0199. 0.0801]
D) [0.0428, 0.0572]

B) [-0.0332,0.1332]

Step-by-step explanation: Confidence Interval is an interval where we can be a percentage sure the true mean is.

The confidence interval for a difference in population proportion is calculated following these steps:

First, let's find population proportion for each population:

Second, calculate standard deviation for each proportion:

Now, we calculate standard error for difference:

SE = 0.0505

The z-score for a 90% CI is 1.645.

Then, confidence interval is

± z-score.SE

±

0.05 ± 0.0831

The limits of this interval are:

inferior: 0.05 - 0.0831 =

superior: 0.05 + 0.0831 = 0.1332

The 90% confidence interval for the difference in the population proportion of pit pulls and golden retrievers is .

there is no question content here, but you for the free points! : d

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)