A new rocket launching system is considering short range launches. The existing system is p= 0.78 as the probability of a successful launch. A sample of 45 experimental launches is made with the new system. Construct a 95% confidence interval for p
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A new rocket launching system is considering short range launches. The existing system is p= 0.78 as the probability of a successful launch. A sample of 45 experimental launches is made with the new system. Construct a 95% confidence interval for p.
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- Suppose a data set of 4,000 observations (n - 4,000) was analyzed using OLS to examine the factors influencing students' college grade point averages. The regression results are as follows, with standard errors in parentheses: colgpa-1.550 +0.0010 sat-0.03 hspere-0.05 hsize +-0.007 haize (0.00) (0.00006) (0.0007) (0.01) (0.000) where colgpa- GPA in college sat score on the SAT exam hspere -percentile in high school graduating class haize size of graduating class, in hundreds 11-4,000 R-0.254 R²-0.253 -0.45 The predicted college GPA for someone who scored 1,120 on the SAT and graduated in the 27th percentile of his or her 700-person high school graduating class, is Suppose you would like to construct a 95% confidence interval around the expected college GPA for someone with aat 1,120, hapere 27, and haize 7. colga 1853+0.0010 sat (0,01) (0.0000) where colgpa-GPA in college sat0 sat-1,120 0.03 Asperc0-0.05 haize0+ 0.007 haize (0) (0000)If n=13, ¯xx¯(x-bar)=37, and s=9, construct a confidence interval at a 80% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place.A high school is running a campaign against the over-use of technology in teens. The committee running the campaign decides to look at the difference in social media usage between teens and adults. They take a random sample of 200 teens in their city (Group 1) and find that 85% of them use social media, and then take another random sample of 180 adults in their city (Group 2) and find that 55% of them use social media. Find a 90% confidence interval for the difference in proportions. Enter the confidence interval
- A simple random sample of 1000 people age 18 or over is taken in a large town. It turns out that 223 people in the sample are newspaper readers. Calculate a 90%-confidence interval for the percentage of people (age 18 and over) in that town who read newspapers. Choose the closest answer. a. (19.7%, 24.9%) b. (21.5%, 23.0%) c. (21.0%, 23.5%) d. (20.1%, 24.5%)if n=20, x=43, and s=9, construct a confidence interval at a 98% confidence level, the data came from a normally distributed population.Construct a 95% confidence interval for p1 - p2 for a survey that finds 30% of 240 males and 41% of 200 females are opposed to the death penalty. Group of answer choices a.(-0.200, -0.021) b.(-1.532, 1.342) c.(-1.324, 1.512) d.(-0.561, 0.651)
- A retail store is considering adding a new product line to its inventory. The store wants to know whether the new product line will be popular with customers. To determine this, they take a random sample of customers and ask if they would be interested in buying the new product line. Of the 125 people they asked, 70 were interested in this new product line. The manager is interested in creating a 90 % confidence interval regarding the population parameter.A survey of 2450 golfers showed that 281 of them are left-handed. Construct a 90% confidence interval for the proportion of golfers that are left-handed. Question 10 options: (0.104, 0.125) (0.369, 0.451) (0.100, 0.130) (0.203, 0.293)A beverage company would like to determine if their drink is preferred to a rival company’s drink. They perform an experiment with 150 randomly selected individuals and have them taste test the two different drinks in a controlled environment so that each person's preference is independent of others. A 90% confidence interval for the proportion preferring their drink is (0.453, 0.587). Which of the following is the correct interpretation of this confidence interval? 90% of all consumers prefer this beverage company’s drink to their rival’s between 45.3% and 58.7% of the time. We’re 90% confident that between 45.3% and 58.7% of all consumers would prefer this beverage company’s drink to their rival’s drink. We’re 90% confident that there’s a 45.3% to 58.7% chance that a randomly selected person would prefer this beverage company’s drink to their rival’s drink. We’re 90% confident that the proportion of people in the sample who prefer this beverage company’s drink over their…
- If n=14, ¯xx¯(x-bar)=34, and s=4, construct a confidence interval at a 80% confidence level. Assume the data came from a normally distributed population.In a study of treatments for very painful "cluster" headaches,... Find the 90% confidence interval using technology.If n=17, ¯xx¯(x-bar)=50, and s=20, construct a confidence interval at a 98% confidence level. Assume the data came from a normally distributed population.
