2. Compute the margin of error of p given the following: Confidence level 97%, p 0.85, n 120.

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A: Given, n=339 x=236

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A: The 95% confidence interval for the proportion of cars on the road in the Country G that have…

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Q: Find the sample size needed to estimate the percentage of adults who can wiggle their ears. Use a…

A: a)

Q: Suppose we wanted to estimate the average amount of sleep (in hours) a college student gets a night.…

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Q: Use the sample data and confidence level given below to complete parts (a) through (d). In a study…

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Q: n a survey of 3117 ​adults, 1473 say they have started paying bills online in the last year.…

A: Given problem Given that   In a survey of 3117 ​adults, 1473 say they have started paying bills…

Q: A sociologist wants to construct a 94% confidence interval for the proportion of teenagers who uses…

A: Given, p=0.23 E=0.02 z value at 94% confidence is Zc=1.88 n=?

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Q: a survey of 3088 adults aged 57 through 85​ years, it was found that 87.1​% of them used at least…

A: The formula for computing confidence interval is,  n is the sample size and P is the proportion.

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Q: Out of 400 people sampled, 32 had kids. Based on this, construct a 95% confidence interval for the…

A: Solution: Let X be the number of people had kids and n be the sample number of people. From the…

Q: Use the sample data and confidence level given below to complete parts (a) through (d). In a study…

A: a.

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A: Sample taken for ages 19 to 25 2253 men and 2629 women random sample In which 986 men and 923 women…

Q: If a 90% confidence interval for the mean is found to be (1.25,4.75) using a sample of n=100, find…

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Q: 829 adult were surveyed in one city, and 51% of them are opposed to the use of the photo-cop for…

A: Given data: n = 829 p = 0.51

Q: na survey of 2587 adults in a recent year, 1365 say they have made a New Year's resolution.…

A: Given that  Sample size n =2587 Favorable cases x =1365 Sample proportion p^=x/n  =1365/2587 =0.5276

Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 500…

A:

Q: ) Find the 99% confidence interval (CI) and margin of error (ME) used to estimate the population…

A: Given data,n=175p=72%=0.72z-value at 99% confidence is Zc=2.576

Q: Construct the indicated confidence interval for the difference between population proportions p1-p2.…

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Q: Find the sample size needed to estimate the percentage of adults who can wiggle their ears. Use a…

A: margin of error(E)=0.05Assume sample…

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A: Required conditions: n1>30 n2>30 n1*p1^>=10 n1*(1-p1^)>=10 n2*p2>=10 n2*p2^>=10

Q: Find the sample size needed to estimate the percentage of adults who can wiggle their ears. Use a…

A: GIVEN DATA,margin of error =1%=0.01CI=0.95α=1-0.95=0.05Zc=Zα/2=Z0.05/2=Z0.025=1.96 [ from excel…

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Q: An interactive poll found that 397 of 2,391 adults aged 18 or older have at least one tattoo. A.…

A: Given data: Sample size (n) = 2391 x = 397 adults To construct: 1.95 % confidence interval 2.98%…

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Q: Insurance companies are interested in knowing the population percent of drivers who always buckle up…

A: Givenx=316sample size(n)=391Confidence level=95%p^=xn=316391≅0.8082

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A: Given that, x=1447,n=3643 The sample proportion is,

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A: We have given that,  Sample size (n) = 36 and favorable cases         (X) = 12 Then,  We Will find…

Q: 7. A recent survey of 500 women indicated that 29% of women over age 55 in the study were widows.…

A: Givensample size(n)=500p^=29%=0.29confidence level=90%

Q: A surprising number of young adults (ages 19 to 25) still live in their parents' homes. A random…

A: Required conditions: n1>30 n2>30 n1*p1^>=10 n1*(1-p1^)>=10 n2*p2>=10 n2*p2^>=10

Q: f. A sample of 60 Grade 10 students' ages was obtained to estimate the mean lot all Grade 9…

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Q: In a random sample of 250 high school students, it was found that 85.2% (213) of them later…

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Q: Construct a 95% confidence interval for the population proportion of students who graduate within…

A: The sample size is n = 400 And the number of students who graduate within six years, X = 108 The…

Q: National statistics show that 23% of men (sample 1) smoke and 18.5% of women (sample 2) do. A random…

A: We need to construct the 95%95\%95% confidence interval for the difference between population…

Q: 2.) A survey of 200 college students revealed that 160 of them eat dessert daily. Construct a 95%…

A: