Pulse rate is an important measure of the fitness of a person's cardiovascular system. The mean pulse rate for all U.S. adult males is approximately 72 heart beats per minute. A random sample of 21 U.S. male adults who jog at least 15 miles per week had a mean pulse rate of 52.6 beats per minute and a standard deviation of 3.22
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What value would you expect to find for the standard deviation of the sample mean? Determine the minimum number of subjects to be polled for a survey with a margin of error of six percentage points and a confidence level of 95%. How many of the 200 confidence interval estimate should you expect to actually contain the population mean M? What is the standard deviation of this distribution?
16) If the random variable z is the standard normal distribution, then z(0.2324) = 23) Determine the minimum number of subjects to be polled for a survey with a margin of error of six percentage points and a confidence level of 95% a.253 b.267 c.278 d.306 e.none of these 32)
See attached file. a. Determine the .95 confidence interval, in thousands of kilowatt-hours, for the mean of all six-room homes. b. Determine the .95 prediction interval, in thousands of kilowatt-hours, for a particular six-room home. Number of Kilowatt-Hours Number of Kilowatt-Hours Rooms (thousands) Rooms (thous
Aggregate with a low measre of thermal conductivity is desirable for paving. Two types of aggregate are randomly sampled and tested fir thermal conductivity, with 25 observations from X (lower cost type) with mean 0.485 and standard deviation 0.18, and 25 observations from Y (higher cost type) with mean 0.372 and standard deviat
A. Test is the difference in lifetime of vehicles exist between the two manufacturers at a=0.01. Please see attached.
Obtain a 95% CI for o^2. Please see attached.
44. The amount of lateral expansion (mils) was determined for a sample of n = 9 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s = 2.81 mils. Assuming normality, derive a 95% CI for σ and for σ^2.
2. Each of the following is a confidence interval for μ = true average (i.e., population mean) resonance frequency (Hz) for all tennis rackets of a certain type: (114.4. 115.6) (114.1,115.9) a. What is the value of the sample mean resonance frequency? b. Both intervals were calculated from the same sample data. The confi
1. The owner of Britten's Egg Farm wants to estimate the mean number of eggs laid per chicken. A sample of 20 chickens shows they laid an average of 20 eggs per month with a standard deviation of 2 eggs per month. a. What is the value of the population mean? What is the best estimate of this value? b. Explain why we need to
Normal distribution: estimation of proportions, confidence interval, margin of error, minimum sample size
1) Assume that the population proportions are to be estimated from the samples described. Find the margin of error and the 95% confidence interval. Sample size = 256, sample proportion = 0.6 2) Importance: A Phone survey of 500 people revealed that 99% of those surveyed ranked good relationships as important or very import
Question: Give an estimate of the sample size needed to obtain the specified margin of error for a 95% confidence interval. Margin of error = 0.01, standard deviation = 0.25 Scenario: Researchers want to determine the mean number of hours students spend watching TV. A margin of error of 0.25 hour is desired. Past studies
1) The distribution of sample mean for samples of 500 homes is normal with a mean of 2.64 and a standard deviation of 0.06. Suppose you select a random sample of n=500 homes and determine that the mean number of people per home for this sample id 2.55. How many standard deviations is the sample mean of the sampling distribution?
Basic Business Statistics: 17 Multiple choice questions on Sampling distribution, Normal Distribution, t-distribution, confidence interval estimate, standard error of the mean, finite population correction factor, sample size, degrees of freedom, one-sided confidence interval estimate
Please see attached. 1. For air travelers, one of the biggest complaints is of the waiting time between when the airplane taxis away from the terminal until the flight takes off. This waiting time is known to have a skewed-right distribution with a mean of 10 minutes and a standard deviation of 8 minutes. Suppose 100 flights ha
In the spring of 2002, residents of the United States were quite worried about the possibility of further terrorists attacks. To gauge the public's sentiment on this topic, the Gallup Poll asked 1002 U.S. adults the following question during May 20-22, 2002. How likely is it that there will be further acts of terrorism in the
1) Does the M&M/Mars Corporation use the same proportion of red candies in its plain and peanut varieties? A random sample of 56 plains M&Ms contained 12 red candies, and another random sample of 32 peanut M&Ms contained 8 red candies. a) Construct a 95% confidence interval for the difference in the proportions of red candies f
1. Why are confidence intervals useful? 2. You and a colleague conducted a study on grocery totals for shoppers in the State of Michigan. Your estimated grocery totals at CI 95%: ($78, $98). In writing the report, your colleague stated: "There is a 95% chance that the true value of µ will fall between $78 and $98. a.
Please help answer the following statistics question. Provide a detailed explanation with that is at least 300 words. What is the difference between a point estimate and a confidence interval?
Would publishing a margin of error with survey results be beneficial?
1) In an attempt to compare the starting salaries of college graduates majoring in education and social sciences, random samples of 50 recent college graduates in each major were selected and the attached information was obtained. (see attachment for data table) a) Find a point estimate for the difference in the average sta
What does a confidence interval tell us and why is it importance to use a confidence interval?
Question: Do you have university education? Yes=45% (n=450) No=60% (n=800) To find out whether 45% is significantly different from 60%, I use this: =SQRT (45*(100-45)*1.96*1.96/450) answer=4.59 Then I use =SQRT(+60*(100-60)*1.96*1.96/800) answer=3.40 Then I calculate the total margin of error
A poll of 1400 randomly selected students in grades 6 through 8 was conducted and found that 30% enjoy playing sports. Would confidence in the results increase if the sample size were 3200 instead of 1400? Why or why not?
1. A parcel service randomly selected 48 packages it received. The sample has a mean weight of 18.6 pounds. Assume that σ = 3.4 pounds. Construct a 90% confidence interval for the true mean weight, μ, of all packages received by the parcel service. 2. A phone company wants to estimate the mean duration of local
I have some trouble understanding the concept first and it will be really helpful if I can see the problem worked out. Here is example question from the textbook. Please help me. The data studies the effect of dietary calcium on blood pressure. In the experiment, one group of black male adults received calcium supplements f
Dr. Patton is a Professor of English. Recently he counted the number of misspelled words in a group of student essays. For his class of 40 students, the mean number of misspelled words was 6.05 and the standard deviation 2.44 per essay. Construct a 95% confidence interval for the mean number of misspelled words in the populatio
When each student in a random sample of students from a particular university was asked to tap a pencil.......
PiggyBank wants to estimate the mean dollars that each card holder will spend each month. It would like to be within plus or minus $10 of the true mean with a 98% confidence level. The standard deviation is thought to be $500. How many card holders should be sampled? After you've determined how many card holders should be sample
A bank wants to estimate the mean dollars that each card holder will spend each month. It would like to be within plus or minus $10 of the true mean with a 98% confidence level. The standard deviation is thought to be $500. How many card holders should be sampled? After you've determined how many card holders should be sampl
The mean and standard deviation of the sample of 65 customer satisfaction ratings are x-bar = 42.95 and s = 2.6424. If we let "mu" denote the mean of all possible customer satisfaction ratings: a) Calculate 95 and 99 percent confidence intervals for "mu". b) Using the 95 percent confidence interval, can we be 95% confident
Express the Confidence Interval in the form of xbar +-E Given Information: Z interval: (291.67, 543.11) xbar = 417.39 n=98 Choose correct confidence interval: a) 417.39 +- 291.67 b) 417.39+-125.72 Why?