Hi I need help with both QUESTION 17Find the margin of errorA survey found that 89% of a random sample of 1024 American adults approved of cloning endangered animals. Find the margin of error for this survey if we want90% confidence in our estimate of the percentage of American adults who approve of cloning endangered animals.O 4.85%O 16.5%O 1.10%O 1.61%O 1.92%QUESTION 18Interpret the confidence interval.A researcher wants to estimate the mean cholesterol level of people in his city A random sample of 21 people yields an average cholesterol level of 219, with amargin of error of +12. Assume the researcher used a confidence level of 90%.O The researcher can be 90% confident that the mean cholesterol level for people in his city is 219.90% of the people sampled have cholesterol levels between 207 and 231.O If we took many random samples of people in the city, about 9 out of 10 of them could produce a confidence interval of (207,231).OAhout 9 out of 10 people in the researcher's city have cholesterol levels between 207 and 231Click Save and Submit to save and submit. Click Save All Answers to save all answers.Sa0耳|WType here to searchDELLeLL

Name: Hi I need help with both QUESTION 17Find the margin of errorA survey f
Uploaded: 2024-03-20T07:06:27.000Z
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Description: Hi I need help with both QUESTION 17Find the margin of errorA survey found that 89% of a random sample of 1024 American adults approved of cloning endangered animals. Find the margin of error for this survey if we want90% confidence in our estimate o

Solution: Question 17) Given information n= 1024 Sample size p^=0.89 Sample proportion α=0.10 Level of significance The 90% margin of error for the percentage of Americans adults who approve of cloning endangered animals Margin of Error= Zα/2×p^×(1-p^)nAt α=0.10 Zα/2=Z0.05=1.645 From excelMargin of Error = 1.645×0.89×0.111024Margin of Error=1.645×0.09791024Margin of Error=1.645×0.0000956054Margin of Error=1.645×0.0097778Margin of Error= 0.0160844Margin of Error =0.0161Margin of Error=1.61% Converted into percentage by multiplying 100 Margin of Error = 1.61 % Option D is correct Question 18) Given n= 21 Sample size x=219 Sample meanMargin of Error =±12 The 90% confidence interval for population mean is x±Margin of Error(219±12)(219-12, 219+21)(207, 231) Interpretation of confidence interval If we took many random sample of people in the city , about 9 out of 10 of them could produce confidence interval of ( 207, 231) Option C is correct