b) It will be wider c) It will not change Determine the margin of error. 11. In a survey of 500 residents, 300 were opposed to the use of photo-cops for issuing traffic tickets. Find the margin of error that corresponds to a 95% confidence interval for the proportion of residents that are opposed to the use of photo-cops for issuing traffic tickets. a) 0.03 b) 0.4 c) 0.05 d) 0.6 e) 0.04

Q: A 95% confidence interval for the proportion of cars on the road in the Germany that have automatic…

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

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

A: Margin of error E=4%=0.04C.I=95%=0.95

Q: 4. According to a recent poll of eight social networking sites, the averagemonthly visitor counnt…

A: In question, Given that in a poll of 8 networking sites the average monthly visitor count was 12.1…

Q: A survey was conducted that asked 1011 people how many books they had read in the past year. Results…

A: Givensample size(n)=1011Mean(x)=14.7standard deviation(s)=16.6confidence level=95%

Q: The Centers for Disease Control reported the percentage of people 18 years of age and older who…

A: a) The sample size can be estimated as follows: n=ze2p(1-p)where, z = z scoree = errorp =…

Q: In 2011 the U.S. Department of Transportation reported with 95% confidence that the average length…

A: Given    alpha=0.05, Xbar=9.72, MOE=0.22

Q: 4) A simple random sample is to be taken of 572 business majors in a college to estimate the…

A: When we don't have any prior information about population proportion, we generally take it as 0.50.…

Q: The NCAA is interested in estimating the difference in mean number of daily training hours for men…

A:

Q: 5. A researcher surveyed of 300 young adults ages 18-29 found that on average, those surveyed sent…

A: 5. Given that, n=300x¯=67.7s=23.6 Confidence level=90% Significance level=1-0.90-0.10…

Q: A recommendation for the amount of vitamin C that a person should have per day is 75 mg. A…

A: given that

Q: The Centers for Disease Control reported the percentage of people 18 years of age and older who…

A: Obtain the sample size of a preliminary estimate for p with 0.26. The formula of sample size is,…

Q: 7. Find the sample size, n, needed to estimate the percentage of adults who have consulted fortune…

A: Given, Margin of error=0.09 Confidence level=98% p^=0.15

Q: The U.S. Department of Transportation, National Highway Traffic Safety Administration, reported that…

A: From the provided information, Sample size (n) = 54 Out of which 35 involved an intoxicated driver.…

Q: Use the sample data and confidence level to construct the confidence interval estimate of the…

A: n=1200 x=900 CI=99% α=1-0.99=0.01

Q: Increasing the sample mean would cause a confidence interval to (a) shift left (b) shift right (c)…

A:

Q: 7. In a survey of 800 people, mean amount of money they spend on movies per month is $48. Using z=…

A: Sample size, n = 800 Mean = 48 Standard deviation = 1.5 Z = 1.96

Q: 13. A sample of the reading scores of 35 fifth-graders has a mean of 82. The standard deviation of…

A:

Q: ind a 95% confidence interval for the difference between the two proportions.   b) Is there a…

A: Let p1 and p2 be the two population proportions of people who said that it is serious.Given…

Q: Suppose that we wanted to estimate the true average number of eggs a queen bee lays with 99%…

A:

Q: 1.Community college students conduct a survey at their college. They ask "Do you plan to transfer to…

A: We have given that,  Favorable cases (X) = 80 and sample size (N) = 100 Then,  We will find the 94%…

Q: You recently sent out a survey to determine if the percentage of adults who use social media has…

A: It is given that sample size (n) is 2,461 and x is 1,702.

Q: handles from the shipment should be inspected to estimate p to within 0.07 with 95% confidence?…

A: It is given that  Margin of error, E = 0.07 Confidence level = 95%

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: B. Construct a 95% confidence interval estimate of 4-4, the difference between the mean weight of…

A: Given Information: B) 95% confidence interval estimate of difference between the mean weight of…

Q: The Centers for Disease Control reported the percentage of people 18 years of age and older who…

A: It is given that margin of error (E) is 0.02 and the confidence level is 95%.Here,

Q: Americans are stressed out! A psychologist would like to estimate the proportion of Americans that…

A: We have given that p = 58.3% = 0.583 q = 1 - p = 0.417 n = 705 Significance level (α) = 1 - 0.96 =…

Q: In 2010, the proportion of Texans without health insurance was 0.11. To update this information a…

A:

Q: The psychologist can be 95% confident that the interval from to includes the unknown population mean…

A: Given: n=250, M=2.0, s=4.2 The estimated standard error is given by, sM=sn=4.2250=0.2656 Hence, the…

Q: In a recent study of 35 ninth graders, the mean number of hours per week they played video games was…

A: Given data, n=35Mean x=16.6sd    σ=2.8

Q: The NCAA is interested in estimating the difference in mean number of daily training hours for men…

A: Solution: Given information: n1=10  Sample size of men n2= 10 Sample size of women x1= 2.7 hours…

Q: 7. In 2005 the New Jersey Adult Tobacco Survey found that 22.5% of 464 blacks surveyed said that…

A:

Q: A marketing firm wanted to estimate the true proportion of protected homes today that have a home…

A: Given Data: n=280p∧=0.69   As "n" is large enough we will use normal approximation to calculate…

Q: e) Find the critical value used to construct a 80% Confidence Interval for the mean. Only for this…

A: Solution: e. From the given information, the population standard deviation is σ=23.8 is known. The…

Q: (b) Suppose the researcher wants to estimate the mean with a margin of error of no more than $20…

A: A researcher likes to estimate the mean amount of money the typical American donates to his or her…

Q: A polling agency wants to compare support for the president's foreign policies between men and…

A: The confidence interval is (-0.05 to -0.025).

Q: A recent GSS reported that the 600 females had a mean of 7.5 close friends (a=8,4) and the 575 male…

A: The null and alternative hypotheses are   The test statistic is given by   p-value = 0.0131…

Q: 18. Obesity Obesity is defined as a body mass index (BMI) of 30 kg/m² or more. A 95% confidence…

A: Given Data : Lower limit = 0.224 upper limit = 0.235

Q: a 95% confidence interval for p the proportion.

A: Here Given  Sample Size=n=200 Sample Proportion = p^=0.22

Q: Assume the managers have requested estimates of the mean attitudes towards each mall with a margin…

A: M.E= 0.05 Springdale Mean -3.81 Springdale STDEV - 1.06 West Mall Mean - 3.41 West Mall STDEV - 1.18…

Q: A tv sports commenter wants to estimat th proportion of citizens who "follow professional football."…

A:

Q: 3. A random of families in Solano St. Intramuros, resulted in the following family sizes: 5, 6, 7,…

A: Given data  observations : 5,6,7,4,3,8,2,9 we have to construct 95% ci for mean.  sample size (n) =…

Q: 5. Asample of 350 engineering applicant included 140 women. Find the 95% confidence interval of the…

A: Given: Sample size, n=350 No. of women, x=140

Q: 3. A sample of 25 students from a Faculty has been obtained to estimate the average grade of the…

A: Solution: Given information: n= 25 Sample size  x=3.7 Sample mean σ=1.3  Population mean α=0.05…

Q: A survey was conducted to find out mean expenditure of students in a week. Data collected from a…

A: Givensample size(n)=35Mean(x)=80standard deviation(s)=15confidence level=99%

Q: 4) Out of the 440 people surveyed, a 110 said they listen to Spotify at least one a month. With the…

A:

Q: 6. The hospital would also like to estimate the Blue Shield as their insurance provider. A ra used…

A:

Q: a) What proportion of workers answered yes to the survey? b) What was the margin of error?

A: Here, the 95% confidence interval for the proportion of workers that answered 'yes' is (0.62, 0.73).…