A survey found that 33 of 66 randomly selected women and 37 of 66 randomly selected men follow a regular exercise program. Find 95% confidence interval for the difference between the proportions of women and men who follow a regular exercise program. A. (-0.263, 0.702) B. (0.330, 0.670) C. (-0.231, 0.109) D. (0.298, 0.702)

Q: A 93.32% Cl is given as (2.5, 3.5). From this information, find a 95% confidence interval (yes, this…

A: The provided information is as follows:The 93.32% confidence interval is (2.5,3.5).

Q: Of 500 random people surveyed, 72 were smokers.  Find the 95% confidence interval. SE=0.0157

A:

Q: Use the Student's t distribution to find t, for a 0.95 confidence level when the sample is 27.

A: sample size=n=27 Degree of Freedom=n-1=27-1 =26 Confidence level is 0.95 Therefore the Level of…

Q: A confidence interval for the true proportional difference between the amount of seniors that have…

A: Ha: P1- P2= 0 H0:P1-P2 not equal to zero

Q: Lindsey selected a random sample of 25 tomatoes at the farmer's market. The 95% confidence interval…

A: When Lindsey selected a random sample of 25 tomatoes at the farmer's market, she was aiming to…

Q: 95% confidence interval for the true percent of workers in this city who have full health insurance…

A: The question is about confidence interval.Given :Randomly selected no. of workers ( n ) = 1700No. of…

Q: A  95% confidence interval for the  proportion of voters who say "they want change" is computed and…

A: For the population proportion, the confidence interval can be computed using the formulae: C.I=p⏞±E…

Q: main, the

A:

Q: For a confidence level of 80% with a sample size of 24, find the t-score.

A: From the provided information, Confidence level = 80% Level of significance (α) = 1 – 0.80 = 0.20…

Q: Beth wants to determine a 90% confidence interval for the proportion of high school students in a…

A: We have given that,  Population proportion (p) = 0.57, margin of error (E) = 0.04 and confidence…

Q: The population mean of a set of data is unknown. The sample mean is 44, and the error bound for the…

A:

Q: T.J. wants to estimate the percentage of college students who have taken out student loans. He…

A:

Q: Over half of all American teens (ages 12 to 17 years) have an online profile on at least one social…

A: Given data is appropriate for testing of hypothesis for single proportion . Because it is a large…

Q: Which percentile would be used to find the multiplier for 60% confidence? Select one: a. 20th…

A: Given find the multiplier for 60% confidence

Q: own residents were interviewed and it was found that 422 responded favorably to a question while 501…

A: given data, 422-favorably responded501-negetively responded 923- total  interviewed for positive…

Q: A survey of folk music fans yield this 95% confidence interval estimate of the proportion of fans…

A: Introduction:The 95% confidence interval for the requires proportion is (0.74, 0.86). So, lower…

Q: When 290 college students are randomly surveyed, it is found that 120 own a car. Find a 99%…

A: Given that, x=120,n=290 The sample proportion is, Critical values: By using the standard normal…

Q: A random sample of n1 = 262 people who live in a city were selected and 117 identified as a "dog…

A: Reviewing information, n1=262 x1 : No. of identified dog person in a city=117 n2=108 x2 : No. of…

Q: A recent study indicated that 24% of women over 55 are widows. A researcher wishes to estimate with…

A:

Q: Construct the confidence interval for the population mean . C=0.98, x=15.3, = 10.0, and n=75

A: The Confidence Interval formula is X  ±  Z s √(n) Where: X is the mean Z is the Z-value…

Q: you used the percentage of alcohol to predict wine quality. The data are stored in Vinho Verde. For…

A: a. To construct a 95% confidence interval estimate of the mean wine quality rating for all wines…

Q: . A survey of folk music fans yields this 95% confidence interval estimate of the proportion of fans…

A: The 95% confidence interval for proportion of fans who love music of David Wilcox is 0.74 to 0.86

Q: FIND t a n = 15 for the 98% confidence interval for the mean.

A: Provided information that, n=15 Alpha (α)=1-0.98=0.02 α/2=0.01

Q: A marketing research company is estimating which of two soft drinks college students prefer. A…

A:

Q: Use the confidence interval to find the margin of error and the sample mean. (12, 20) O E = 4, X =…

A: confidence interval for mean : (12,20)

Q: Determine tc for n = 15 and a confidence level of 99%.

A: From the given information, Sample size (n) =15, Confidence level (1-α)= 0.99 Thus, the t-critical…

Q: Construct the 99.8% confidence interval for the difference P1 - P2 when x, = 23, n, = 53, x2 = 15,…

A:

Q: ement below is the best explanation? ere is 90% confidence that 21% of the adult citizens of the…

A: Given n=2500 Pcap=0.21 ME=0.044 Alpha=1-0.90=0.10 90% confidence level

Q: (0.24, 0.56) is a confidence interval for the proportion of people who liked a new soda flavor.…

A: Made comparisons with confidence interval formula for proportion.

Q: Use the confidence interval to find the margin of error and the sample mean.   (.364,0.500)

A:

Q: In one city, 27.8% of the 9000 students surveyed said they eat at the school cafeteria. Find the 95%…

A: To find the 95% confidence interval for the proportion of students who said they eat at the school…

Q: Michelle, a student studying music, randomly sampled fellow music majors and asked whether they…

A: Given information- Confidence interval = (0.09, 0.26) We have to find margin of error. We know that-…

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.