A senior manager, responsible for a group of 120 junior executives, is interested in the total amount of time per week spent by these people in internal meetings. A random sample of 35 of these executives was asked to keep diary records during the next week. When the results were analyzed, it was found that these sample members spent a total of 143 hours in internal meetings. The sample standard deviation was 3.1 hours. Find a 90% confidence interval for the total number of hours spent in internal meetings by all 120 junior executives in the week.

Q: researcher is conducting a study to examine what factors influence older adult's Binge Drinking…

A: Given that  Sample size n =1666 Sample mean =2.62 Population standard deviation =11.52

Q: Several years ago, 44% of parents who had children in grades K-12 were satisfied with the quality of…

A: The claim is that 44% of parents who had children in grades K-12 were satisfied with the quality of…

Q: A research center survey of 2,384 adults found that 1,868 had bought something online. Of these…

A: From the provided information,Sample size (n) = 2384From which 1247 are weekly online…

Q: In a random sample of 100 customers, the items that customers purchased were tracked. It was found…

A: Given,sample size(n)=100sample proportion(p^)=0.171-p^=1-0.17=0.83α=1-0.90=0.1α2=0.05Z0.05=1.645

Q: Insurance companies are interested in knowing the population percent of drivers who always buckle up…

A: It is given that, x is 311 and n is 400.

Q: knowing

A: Given: Total number of drivers = 383 Drivers who buckle up = 297 Proportion of drivers, p =…

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: Insurance companies are interested in knowing the population percent of drivers who always buckle up…

A: Solution

Q: A surprising number of young adults (ages 19 to 25) still live in their parents' homes. A random…

A: Sample taken for ages 19 to 25 2253 men and 2629 women random sample In which 986 men and 923 women…

Q: The Centers for Disease Control and Prevention reported a survey of randomly selected Americans age…

A: Confidence interval for Proportions : The following is the formula for confidence interval of the…

Q: In a survey, 45 people were asked how much they spent on their last partners birthday gift. The…

A: Given information- Sample size, n = 45 Significance level, α = 0.10 Sample mean, x-bar = $90 Sample…

Q: Of 127 randomly selected adults, 31 were found to have high blood pressure. Construct a 95%…

A: Given that, Of 127 randomly selected adults, 31 were found to have high blood pressure.  X = 31 N =…

Q: A poll of 827 students at Alpha state college found that 61% of those polled preferred a quarter…

A: It is given that the proportion of the students who preferred a quarter system to a trimester…

Q: A recent survey showed that from a sample of 500 packages delivered by a Postal Service, 480 were…

A:

Q: A recent survey showed that from a sample of 500 packages delivered by a Postal Service, 480 were…

A: Given data X = 480 Sample size (n) = 500

Q: In a survey of 1037 US adults ages 65 and over, 643 were concerned about getting the flu. A 99%…

A: Given,n=1037x=643p^=xnp^=6431037=0.6201-p^=1-0.620=0.380α=1-0.99=0.01α2=0.005Z0.005=2.58    (from…

Q: In a survey of 612 randomly selected female college students, 141 report that they had been sexually…

A: Given Information:n=612x=141c=95%=0.95

Q: In a survey of 2255 randomly selected US adults (age 18 or older), 1787 of them use the Internet…

A: We want to calculate the 90% confidence interval for each of the following proportions

Q: A random variance survey of 495 adults found that 59 had diabetes. which of the following is a 98%…

A:

Q: In a survey with human resources managers, 71% of respondents thought they should hire new staff…

A: Given information: Denote the population proportion of all respondents who thought that they should…

Q: Construct a 95% confidence interval for the ,ean number of hours of watching streaming videos…

A: given: mean=18.5       standard deviation = 3.2 we need to construct 95% CI for mean number of hours

Q: A surprising number of young adults (ages 19 to 25) still live in their parents' homes. A random…

A: Required conditions: n1>30 n2>30 n1*p1^>=10 n1*(1-p1^)>=10 n2*p2>=10 n2*p2^>=10

Q: a newspaper article stated that the us supreme court received 812 letters from around the country on…

A: To find the 95% confidence interval for the true proportion of all American adults who feel that…

Q: A recent study showed that among 725 randomly selected subjects who completed 4 years of college,…

A: We have given that,  Favorable cases (X) = 139 and Sample size    (n) = 725 Then,  We will find 95%…

Q: A sample of 3048 randomly chosen people found that 341 of them were left-handed. Construct a 99%…

A: Sample proportion=p^=xn=3413048Critical value for 99% confidence interval is=z=2.576n=sample…

Q: A surprising number of young adults (ages 19 to 25) still live in their parents' homes. A random…

A:

Q: Several years ago, 44% of parents who had children in grades K-12 were satisfied with the quality of…

A: We have given that 44% of parents who had children in grades K-12 were satisfied with the quality of…

Q: A surprising number of young adults (ages 19 to 25) still live in their parents' homes. A random…

A:

Q: Out of 100 people sampled, 44 have lied to a family member in the past month. Based on this,…

A:

Q: Out of 300 people sampled, 279 had kids. Based on this, construct a 90% confidence interval for the…

A:

Q: Insurance companies are interested in knowing the population percent of drivers who always buckle up…

A: Givenx=316sample size(n)=391Confidence level=95%p^=xn=316391≅0.8082

Q: Suppose a marketing company computed a 96% confidence interval for the true proportion of customers…

A: The objective of the question is to understand the meaning of the given 96% confidence interval for…

Q: A survey of credit card holders revealed that Americans carried an average credit card balance of…

A: M1=$3900M2=$3300n1=400n2=450σ1=$880σ2=$810

Q: Insurance companies are interested in knowing the popul drivers who always buckle up before riding…

A: Given data,No of people surveyed n=387who claims to buckle up is x=311Construct a 93 % CI?

Q: A surprising number of young adults (ages 19 to 25) still live in their parents' homes. A random…

A: Required conditions: n1>30 n2>30 n1*p1^>=10 n1*(1-p1^)>=10 n2*p2>=10 n2*p2^>=10

Q: A survey showed 980 of 1500 adult respondents believe in global warming. Construct a 95% confidence…

A:

Q: The manager at Bell’s Gym wants to know the proportion of members who use the Cardio Machines. A…

A: DETAILED SOLUTION IS PROVIDED BELOWExplanation:

Q: company manufactures a certain over-the-counter drug. The company samples 80 pills and finds that…

A: We have given that  Sample size n = 80 Sample mean = 325.5 Standard deviation s=10.3 90%confidence…

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.