A bank sampled its customers to determine the proportion of customers who use their debit card at least once each month. A sample of 50 customers found that only 12 use their debit card monthly. Find 95% and 99% confidence intervals for the proportion of customers who use their debit card monthly. Use the appropriate formula and verify your result using the Confidence Intervals workbook. Note: Interpret the results of interval in context. Use the Confidence Interval Templates to solve this problem.

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: Wildlife biologists inspect 200 deer taken by hunters and find 45 of them carrying Lyme disease…

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Q: I've asked 91 students on Monday whether they did their homework over the weekend. 20 of them said…

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Q: f n=450 and X = 360, construct a 99% confidence interval.

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Q: A survey of 366 Cook County residents is conducted, in which 253 of the respondents had shopped…

A: Given: n = 366 X = 253 Formula Used: Confidence Interval = p^ ± Z*p^(1-p^)n

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

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

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Q: In a survey of 2000 adults in a recent​ year, 708 made a New​ Year's resolution to eat healthier.…

A: Confidence interval for population proportion is given by, p^ +- z* √(pq/n) At 90% z = 1.645 At 95%…

Q: In a survey of 2113 adults in a recent​ year, 741 made a New​ Year's resolution to eat healthier.…

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Q: We wish to estimate what percent of adult residents in Ventura County like chocolate. Out of 500…

A: The sample proportion of the sample of size n can be defined as:  Here, X = Total number of success…

Q: Find the 95% confidence interval for the proportion of customers planning to buy the shoes, and…

A: Provided data is, Mary was successful and collected 600 responses in time. 195 of the surveyed…

Q: A sample of 500 nursing applicants included 80 men.  Determine the 95% confidence interval for the…

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Q: Suppose I wanted to know the proportion of students who are paying attention in class, so I give a…

A: Confidence Interval for Proportion A confidence interval is a statistical method that denotes a…

Q: m what their favorite color was. 23 people said their favorite color is blue. Using this sample,…

A: Sample proportion is the fraction of successful samples and total number of samples. Here,  Total…

Q: Jane

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

A: Given X=25 n=500 95% confidence interval

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

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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: A market researched for a consumer electronics company wants to study the television viewing habits…

A: Given: n=40x¯=15.3S=3.8 Degrees of freedom: df=n-1=40-1=39 The critical-value is obtained by using…

Q: Software testers have reported their testing results of your teams latest software release. Out of a…

A: Critical value: Consider α = 0.05. Using excel formula “=NORM.S.INV(1-(0.05/2))”, the critical value…

Q: ed 1294 people whether they performed any volunteer work during the past year. A total of 517 people…

A: The sample proportion is computed as follows,

Q: Translate the statement into a confidence interval. Approximate the level of confidence. In a survey…

A: The sample size is 1072, 67% say they are willing to get a second or part-time job to pay for their…

Q: If n=240 and X = 192, construct a 90% confidence interval. Enter your answer as an open-interval…

A: 1. The given values are n=240,x=192. The proportion of x is, p^=xn=192240=0.8 Computation of…

Q: Suppose that a health expert has claimed that 28% of college students smoke cigarettes. To…

A: Given 95% confidence interval for the percentage of college students who smoke is 19% to 24%. Given…

Q: To assess the proportion of the readers of a magazine being professional, a random sample of 1000…

A: Given that the random sample of 1000 readers is selected and 660 are found to be professionals. That…

Q: In a sample of 85 statistics students, 41 said that finding confidence intervals for population…

A: It is given that x is 41 and sample size (n) is 85. The confidence level is 0.99.

Q: In a survey of 3102 adults,1401 say they have started paying bills online in the last year.…

A: Obtain the 99% confidence interval for the true population proportion. The 99% confidence interval…

Q: A research institute poll asked respondents if they felt vulnerable to identify a theft. In the poll…

A: Given,n=1000x=501p^=xnp^=5011000=0.5011-p^=1-0.501=0.499α=1-0.99=0.01α2=0.005Z0.005=2.58    (from…

Q: 9. A survey of 280 homeless persons showed that 63 were veterans. Construct a 90% confidence…

A: In a survey of 280 homeless persons showed that 63 were veterans.

Q: A research institute asked students if they felt vulnerable to identity theft. In the poll, n=1047,…

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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: 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: The Challenger Disaster. In a letter to the editor that appeared in the February 23, 1987, issue of…

A: a.Probability that none of the “critically 1” items would fail is given by:

Q: What conditions are necessary in order to use the dependent samples t​-test for the mean of the…

A: Necessary conditions for dependent samples in t-test, and for the difference in the means of two…

Q: You want to determine if the proportion of people who will vote for the president next year has…

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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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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.