2.) A survey of 200 college students revealed that 160 of them eat dessert daily. Construct a 95% confidence interval for the population proportion of college students who eat dessert daily. Can we claim that 75% eart dessert daily?

Q: If n=380 and p^(p-hat) =0.12, find the margin of error at a 95% confidence level Give your answer…

A: Given data: n = 380  p^(p-hat) =0.12, To find : The margin of error at a 95% confidence level

Q: If n = 340 and p^=0.92, find the margin of error at a 90% confidence level E= Give your answer to…

A: Given,n=340p^=0.921-p^=1-0.92=0.08α=1-0.90=0.1α2=0.05Z0.05=1.645

Q: If a retailer would like to estimate the proportion of their customers who bought an item after…

A: Given data,     P=58%=0.58     E=7%=0.07     z-value at 95% confidence is  zc =1.98     n=?

Q: 12. In a sample of 600 Canadians, 265 say they would rather retire in the US than in Canada.…

A:

Q: Suppose we want to estimate the proportion of teenagers (aged 13-18) who are lactose intolerant. If…

A: Given information Error = 4% = 0.04 Confidence level = 0.95

Q: If n=550 and ˆpp^ (p-hat) =0.25, find the margin of error at a 99% confidence level Give your…

A: The margin of error = The margin of error (ME) expresses to the maximum expected differences between…

Q: 2. A nutritionist wants to know the population proportion of Grade 1 learners who eat vegetables.…

A: We have given thatn = 1200x = 200p̂ = x/n = 200/1200 = 0.167Significance level (α) = 1 - 0.95 = 0.05

Q: Of 118 randomly selected adults, 34 were found to have high blood pressure. Construct a 95%…

A: According to the given information, we have Total number of adults = 118 34 were found to have high…

Q: In a study of 200 accidents that required treatment in an emergency room, 80 occurred at work. Find…

A: Obtain the 90% confidence interval for the true proportion of accidents that occurred at work. The…

Q: Suppose in a recent poll of 1250 randomly selected Californians who took vacations, 132 of them…

A:

Q: Suppose we want to estimate the proportion of teenagers (aged 13-18) who are lactose intolerant. If…

A: Given information Error = 2% = 0.02 Confidence level = 0.95

Q: In a Gallup Poll of 1005 individuals, 452 thought they were better off financially than a year ago.…

A: Given; p=452n=1005 95% confidence interval is given by- p^±z×p^(1-p^)n true proportion is given as-…

Q: Suppose we want to estimate the proportion of teenagers (aged 13-18) who are lactose intolerant. If…

A:

Q: If n=240 and ˆpp^ (p-hat) =0.91, find the margin of error at a 90% confidence level Give your…

A:

Q: If n=590 and p=0.93, construct a 90% confidence interval.

A: Express the expression for the confidence interval. C.I=p±Z*p(1-p)n Here, Z* is the critical value .…

Q: If n=210 and ˆpp^ (p-hat) =0.86, find the margin of error at a 90% confidence level Give your…

A:

Q: If n = 560 and ˆpp^ (p-hat) = 0.25, construct a 99% confidence interval.

A: Calculations: x=140 and N= 560

Q: 1. The Hartford Insurance Company wants to estimate the percentage of drivers who change tapes or…

A: Please see the attachment..

Q: In a survey of 3473 adults aged 57 through 85​ years, it was found that 84.6% of them used at least…

A:

Q: The following table shows the results of a survey in which 1017 adults from the United States, 1060…

A: (1)

Q: A random sample of 600 voters in Centreville indicated that 90 of them believed their congressional…

A: Formula for confidence interval for population proportion : CI =p^ ± E  Where ,  p^ is sample…

Q: It was observed that 30 of the 100 randomly selected students smoked. Find the confidence interval…

A: It is given that, 30 out of 100 randomly selected students smoked i.e. proportion of the student who…

Q: If n=270 and p (p-hat) =0.93, find the margin of error at a 90% confidence level Give your answer to…

A:

Q: 9. A survey found that out of 900 workers, 150 said they were interrupted by frank phone calls. Find…

A:

Q: In a sample of 1901 patients who underwent a certain type of surgery, 22% experienced complications.…

A:

Q: If n=200 and ˆpp^ (p-hat) =0.39, find the margin of error at a 99% confidence level

A: Find z-critical value:…

Q: If n=340 and p (p-hat) =0.23, find the margin of error at a 95% confidence level Give your answer to…

A: Given,n=340p^=0.23α=1-0.95=0.05α2=0.025Z0.025=1.96

Q: If n=530 and ˆpp^ (p-hat) =0.51, find the margin of error at a 90% confidence level Give your…

A:

Q: If n=270 and ˆpp^ (p-hat) =0.59, find the margin of error at a 99% confidence level Give your…

A: Given  n=270, pcap=0.59, alpha=0.01 Zcritical=2.578 at alpha=0.01 We want to find margin of error

Q: If n=110 and ˆpp^ (p-hat) =0.71, find the margin of error at a 90% confidence level Give your…

A: Sample size n = 110 Sample proportion p^ = 0.71 Confidence level = 90% Formula Used: Margin of error…

Q: If 64% of a sample of 550 people leaving a shopping mall claims to have spent over $25 create and…

A: Given,   Sample size = 550 Sample proportion = 0.64   Critical value: Using z-table, the z-critical…

Q: If n=490 and ˆpp^ (p-hat) =0.46, find the margin of error at a 99% confidence level.Give your answer…

A:

Q: If n=190 and ˆpp^ (p-hat) =0.29, find the margin of error at a 90% confidence level Give your…

A:

Q: If n=530 and ˆpp^ (p-hat) =0.75, find the margin of error at a 90% confidence level Give your…

A: Givenn=530p^=0.75confidence level=90%

Q: In a random sample of 184 college students, 97 had part-time jobs. Find the margin of error for the…

A: Given : Sample Size, n = 184 Number of successes, x = 97 confidence level,c=0.99

Q: If n=580 and ˆpp^ (p-hat) =0.59, find the margin of error at a 95% confidence level Give your…

A: Obtain the margin of error for the 95% confidence interval. The margin of error for the 95%…

Q: If n=490 and ˆpp^ (p-hat) =0.54, find the margin of error at a 90% confidence level Give your…

A:

Q: 1. In a poll of 2,500 likely voters, 966 say that the Philippines spend too little on education.…

A:

Q: The table shows the results of a survey in which separate samples of 400 adults each from the East,…

A:

Q: If n = 520 and p^ = 0.85, construct a 99% confidence interval

A: Here, n = 520 and p hat = 0.85.

Q: The following data represents the monthly cost for telephone service for a random sample of 8…

A: The data for a random sample of 8 households in Wilmington is provided which represents the monthly…

Q: If n=570 and ˆpp^ (p-hat) =0.89, find the margin of error at a 95% confidence levelGive your answer…

A: It is given that the value of n is 570, p hat is 0.89 and the confidence level is 0.95.Here,

Q: (c) A 99% confidence interval from a sample size of 27 observations. 3.054

A: It is given that the sample size is 27 and the confidence level is 0.99.

Q: Suppose you want to construct an 80% confidence interval for the proportion of businesses that…

A: Given:Margin of Error (E)=3.5%=0.035   For Confidence level 80%, the significance level (α) is…

Q: Identify the at-score for a 99% confidence interval if the sample size is 70.

A: givensample size(n)=70α=1-0.99=0.01

Q: If n=130 and ˆpp^ (p-hat) =0.19, find the margin of error at a 95% confidence level Give your…

A:

Q: Construct a 99% confidence interval for the true proportion of all car license applications that…

A: Given Information:x=25n=155c=99%=0.99

Q: 1. Construct a 90% confidence interval for the proportion of employees requiring size 3.

A: here given  A civil engineering firm issues every new employee with a safety helmet. Four different…

Q: 2. A survey found that out of 900 workers, 150 said that they were interrupted by frank phone calls.…

A: In question, We have given the number of favorable cases 150 and sample size 900 then we'll find 95%…

Q: If n=360 and ˆp(p-hat) =0.69, find the margin of error at a 95% confidence level

A:

2-Proportion Test

In a 2-proportion test, two separate proportions are measured at the same time, and their similarity is evaluated. It is also known as a 2-proportion z-test, and a comparison of two different proportions is done in this method.