In a survey of 2383 U.S. adults, 1073 think there should be more government regulation of banks.             Find the point estimate for the population proportion of U.S. Adults who think there should be more regulation of banks.              Construct a 90% confidence interval for the population proportion.             Find the minimum sample size needed to be sure at the 99% confidence level that the estimate is within 3 percentage points of the population proportion.

Q: In a survey of 2591 adults in a recent​ year 1414, say they have made a New​ Year's resolution.…

A: From the provided information,Confidence level = 90% and 95%

Q: In a survey of 3001 ​adults, 1486 say they have started paying bills online in the last year.…

A: Assuming that the number of people paying bills online has a binomial distribution with n=3001, and…

Q: In a survey of 2224 adults in a recent​ year, 1280 say they have made a New​ Year's resolution.…

A: Obtain the 90% confidence interval for the population proportion. The 90% confidence interval for…

Q: In a survey of 2865 ​adults, 1413 say they have started paying bills online in the last year.…

A:

Q: In a survey of 2549 adults in a recent​ year, 1270 say they have made a New​ Year's resolution.…

A:

Q: In a survey of 2195 adults in a recent​ year, 1454 say they have made a New​ Year's resolution.…

A:

Q: In a survey of 3478 adults, 1462 say they have started paying bills online in the last year.…

A: From the provided information,Confidence level = 99%

Q: In a survey of 2703 ​adults, 1438 say they have started paying bills online in the last year.…

A: We have given that, Number of favorable cases(X) = 1438,   Sample size(n) = 2703 Sample proportion…

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: a survey of 3447 adults, 1477 say they have started paying bills online in the last year.…

A:

Q: In a survey of 3185 adults, 1455 say they have started paying bills online in the last year.…

A:

Q: In a survey of 2282​ adults, 749 say they believe in UFOs. Construct a 90% confidence interval for…

A:

Q: In a survey of 2405 adults in a recent​ year, 1217 say they have made a New Year's resolution.…

A: Given: n=2405 x=1217

Q: When 302 college students are randomly selected and surveyed, it is found that 122 own a car. What…

A: E have to find point estimate with with confidance interval..

Q: In a survey of 2542 adults in a recent​ year, 1487 say they have made a New​ Year's resolution.…

A: Given,  n = 2542 x = 1487 proportion p  = 1487/2542                       = 0.5850

Q: In a survey of 50,00 people, 80% indicated that they listened to their favorite radio station…

A: Hern= 5000 P^ = 0.80 We have to construct 95% ci for p

Q: in a survey of 2042 adults in a recent​ year, 715 made a New​ Year's resolution to eat healthier.…

A:

Q: In a survey of 2407 adults in a recent​ year,1468 say they have made a New​ Year's resolution.…

A: Given that,  Sample size, n= 2407 And X= 1468 We know that proportion, p=X/n So p= 1468/2407 p=…

Q: In a survey of 2108 adults in a recent​ year, 1305 say they have made a New​ Year's resolution.…

A:

Q: In a survey of 2000 adults in a recent​ year 1404, say they have made a New​ Year's resolution.…

A: n=2000x=1404

Q: 1000 randomly selected Americans were asked if they believed the minimum wage should be raised. 600…

A: Givensample size(n)=1000x=600sample proportion(p^)=xn=6001000=0.6confidence level=95%

Q: a survey of 3409 adults, 1445 say they have started paying bills online in the last year.  Construct…

A: The confidence interval = 99%The faovarble cases [x] = 1445The sample size [n] = 3409

Q: According to a poll, 77% of California adults (389 out of 506 surveyed) feel that education is one…

A: Given that the number of favorable cases Favorable Cases XX = 389389 Sample Size NN = 506506…

Q: In a survey of 2693 ​adults, 1495 say they have started paying bills online in the last year.…

A: Given that, In a survey of 2693 ​adults, 1495 say they have started paying bills online in the last…

Q: In a survey of 3292 adults, 1470 say they have started paying bills online in the last year.…

A: Using Excel formula “=NORM.S.INV(1-(0.01/2))”, the critical value for 99% confidence level is…

Q: A researcher wanted to compare the percentages of men and women who play the lottery.  They randomly…

A:

Q: In a survey of 2646 ​adults, 1401 say they have started paying bills online in the last year.…

A: Number of adults (n)  = 2646 Number of adults who say they have started paying bills online in the…

Q: n a survey of 2847 ​adults, 1408 say they have started paying bills online in the last year.…

A: Given that  Sample size (n) =2847 People started paying bills online (x) = 1408 Proportion (p) =…

Q: An internet service provider sampled 540 customers and found that 55 of them experienced an…

A:

Q: If a 90% , a 95% and a 99 % confidence interval are found for a population ean, describe how the 3…

A: Solution :  A 90%, a 95% and a 99 % confidence interval are found for a population mean.

Q: In a survey of 3342 adults, 1489 say they have started paying bills online in the last year.…

A: Given Information: To construct a 99% confidence interval for the population proportion and…

Q: In a survey of 2417 adults in a recent​ year, 1458 say they have made a New​ Year's resolution.…

A:

Q: A random survey of 36 airline passengers indicates that twelve have never flown before. Find a 95%…

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

Q: In a survey of 2497 adults in a recent​ year, 1383 say they have made a New​ Year's resolution.…

A: The sample size (n) is 2497 and the value of x is 1383.

Q: n a survey of 3196 adults, 1405 say they have started paying bills online in the last year.…

A: Given data, n=3196 x=1405 p=x/n=1405/3196 =0.4396 z-value at 99% confidence is Zc=2.576

Q: In a survey of 2588 ​adults, 1466 say they have started paying bills online in the last year.…

A: Construct 99% confidence interval for population proportion:The sample of 2,588 adults among that…

Q: 700 randomly selected Americans were asked if they believed the minimum wage should be raised. 350…

A: Given that  p = 700 x = 350 Confidence level(c) = 90% = 0.90 Significance level (α) = 1-c = 1-0.90=…

Q: Suppose we wish to estimate the percentage of all taxpayers that want the current tax system…

A: The random variable current system follows binomial distribution. The sample size is 1310. The…

Q: In a survey of 2554 adults in a recent​ year, 1357 say they have made a New​ Year's resolution.…

A:

Q: In a survey of 2480 golfers, 372 said they were left-handed. Construct a 92% confidence interval for…

A: GIVEN DATA : Total number of golfers are 2480, Number of left handed golfer are 372        TO FIND :…

Q: In a random sample of 230 men , 18 or older , 142 were married. Construct and explain a 90%…

A: Solution: Let X be the number of married men 18 or older and n be the sample number of men. From the…

Q: According to a telephone survey of adults, 86% of homeowners said that owning a home was a good…

A: From the given information, According to a telephone survey of adults, 86% of homeowners said that…

Q: Out of 42738 freshmen entering the State University system,  27298 of them need to take remedial…

A:

Q: A Gallup poll of 1487 adults showed that 43% of the respondents have Facebook pages.  Construct a…

A: Confidence Interval: In statistics, we can say a range of values of the specified population…

Q: In a random sample of 230 men, 18 or older, 142 are married. Construct and explain a 98% confidence…

A: Given Data Sample Size, n = 230 Number of successes, x = 142.0 Significance level,…

Q: In a survey of 3038 ​adults, 1448 say they have started paying bills online in the last year.…

A: Confidence Interval:The form of the confidence interval is,CI = Point estimate ± Margin of error.The…

Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.