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15 points and will give brainliest to excellent answer

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Problem 1

Answers:

Percentage of patients that were dogs = 46%

Standard Error = 0.07048404074682

Margin of error for 90% confidence interval = 0.11594624702851

Margin of error for 95% confidence interval = 0.13814871986377

Round the decimal values however you need them

To get the first answer, you add up the numbers given (7,4,5,5,2) and divide that over 50. So 7+4+5+5+2 = 23 which leads to 23/50 = 0.46 = 46%; therefore phat = 0.46 is the sample proportion of dogs.

Use the SE (standard error) formula given to you with phat = 0.46 and n = 50 to get SE = 0.07048404074682

The critical z value at 90% confidence is 1.645; this value is found in your Z table (back of your stats textbook). Multiply the SE value by 1.645 to get 0.07048404074682*1.645 = 0.11594624702851

Also found in your textbook is 1.960 which is the z critical value at 95% confidence. Multiply this with the SE value to get 0.07048404074682*1.960 = 0.13814871986377

Problem 2

Choice B) picking balls from a bin; the 60 randomly selected get chosen for the first bus, while the remaining 60 go to the second bus

Choice A is fairly vague on what the lower and upper boundaries are. What is the smallest number allowed? What about the largest? This isn't clear so it's possible that we could end up with more positive numbers than negative (eg: if we had an interval -10 < x < 110). So choice A is false. A similar issue shows up with choice D.

Choice B is true. Assuming the selection process is random and not biased, then each ball is equally likely for each selection. The fact that the balls are colored seems to be extra info which I'm not sure why your teacher threw that in there.

Choice C is false because choice B is true

Choice D is false for similar reasons as choice A. It's not clear where we start and where we end. If we had the interval 2 < x < 6 then x could take on the values {3, 4, 5} and we see that picking an odd number is twice as likely than picking an even. In this example, there is bias.

Problem 3

Choice B) Roll a die; each number corresponds to a different class

Choice A is false because choice B being the answer contradicts it

Choice B is true: there are 6 sides on the die, and each side is equally likely to be landed on, so each class is equally likely

Choice C is false because 2*2*2 = 8 represents the number of combos you can have when you flip three coins (one combo being HTH for heads tail heads) but there are 6 classes, not 8

Choice D is false because while we want 6 regions on the spinner. Each region must have the same area; otherwise, one class is weighted heavier than the others making it more likely you select that particular class.

its kind of blurry. can you put a clearer pic?

step-by-step explanation: