The confidence interval with mean value 2 and margin error is 1.3 is
Q: The mean of the numbers of chocolate chips in a sample of 40 Chips Ahoy regular cookies is 23.95 and…
A: From the provided information, Sample size (n) = 40 Sample mean (x̄) = 23.95 Sample standard…
Q: . How large a sample would the researcher need to take to estimate the mean blood cholesterol to…
A: Given : Margin of error, ME = 3 population standard deviation, σ = 30…
Q: mple of 400 items is found to have mean of 82 and standard deviation is 18. Find 95% confidence…
A: Given: Sample mean x¯=82 Sample standard deviation (s) = 18 Sample size (n) = 400 Confidence…
Q: Describe the Interpretation of Confidence Intervals for μ1 – μ2.
A: The formula for Confidence interval for mean difference is given by,
Q: Determine the point estimate of the population mean and margin of error for the confidence interval…
A: To find: 1. Point estimate for population mean 2. Margin of error for population mean
Q: A thread manufacturer wants to estimate the mean tensile strength of a particular type of thread. A…
A: Given : Sample Size, n = 15 Sample Mean, x̄ = 7.64 sample standard…
Q: A random sample of 49 lunch customers was selected at a restaurant. The average amount of time the…
A: Given: n= 49.0 , x̄= 33.0 σ= 10.0 At 95.0 % confidence level , z score= 1.96
Q: A confidence interval is desired for the true mean stray-load loss (Watts) for a certain type of…
A: Solution: Given information: n= 25 sample size x = 53.6 Sample mean σ= 3.3 Population standard…
Q: Data on 4500 college graduates show that the mean time required to graduate with a bachelor's degree…
A: Given data: n = 4500 Mean(x-bar) = 6.35 years Standard deviation =1.73 years Confidence level = 95%
Q: Find the sample size for some finite and infinite population when the percentage of 4300 population…
A: It is given that, Population size, N=4300 Level of significance= 0.01. So. Zα = 1.28 Percentage of…
Q: A sample of 36 men has a mean weight of 175 pounds with a sample standard deviation of 13 pounds.…
A: It is given that sample mean is 175 and standard deviation is 13.
Q: The mean height for a sample of n = 81 women is x = 63.5 inches, and the standard deviation is s =…
A:
Q: researcher was interested in the mean corn yeild for farmers in Illinois. She took an SRS of 28…
A: Given that Margin of error =E =3.5 Population standard deviation =13.25 95% confidence level.
Q: For the one-mean t-interval procedure, express the formula for the endpoints of a confidence…
A: The formula for a mean t-interval procedure is given as follows:
Q: A researcher collected sample data for 14 middle-aged women. The sample had a mean serum cholesterol…
A: The question is about confidence interval.Given :Randomly selected no. of middle aged women ( n ) =…
Q: The average measured verbal IQ (IQV) scores of children is 100. In a sample of 20 children, the…
A: The objective of this question is to calculate the 98% confidence interval for the true mean IQV of…
Q: A researcher collected sample data for 18 middle-aged women. The sample had a mean serum cholesterol…
A:
Q: A fitness center is interested in finding a 90% confidence interval for the mean number of days per…
A: Given Information: Sample size (n) = 241 Sample mean (x¯) = 2.3 Standard deviation (s) = 2.9 To…
Q: The number of cars sold annually by used car salespeople is normally distributed with a standard…
A:
Q: A researcher collected sample data for 11 middle-aged women. The sample had a mean serum cholesterol…
A: Given datan=11mean=190.4standard deviation=6.6confidence interval=95%
Q: A longer confidence interval on the population mean will have better estimation of the unknown…
A: Here use basic definition of confidence interval
Q: trol manager at a light bulb factory needs to estimate the mean life of a large shipment of light…
A:
Q: researcher wants to estimate the mean blood cholesterol level of young men ages 15-25 with a 90.05%…
A: Formula for the sample size needed to estimate a population mean is
Q: The president of a large university wishes to estimate the average of the students presently…
A: Given Information: Standard deviation of students enrolled,σ=2.2 years. Mean of students enrolled,x¯…
Q: The blood alcohol concentration (BAC) level in a sample of 18 blood specimens is tested. This sample…
A:
Q: Type only the upper interval by using type your answer to the nearest hundredths place.
A:
Q: The average BMI (Body Mass Index) of 50 people from a certain town is 24 kg/sq m. Find the 90%…
A: Given information: The average BMI for 50 people from a certain town is 24 kg/sq m Population…
Q: The researcher recorded the amount of time each bird spent in the plain chamber during a 60-minute…
A: Sample Mean , x¯ = 54.6 min Sum of squares = SS= 3508 Sample size, n = 14 Sample standard deviation,…
Q: Bob samples 30 one-pound bags of celery hearts from the produce section at Kroger's. He weighs each…
A:
Q: A researcher collected sample data for 16 middle-aged women. The sample had a mean serum cholesterol…
A: Given , Sample size , n = 16Sample mean , 191.7standard deviation , s = 8confidence interval = 95%…
Q: The distribution of study times for first year students follows a Normal distribution with standard…
A: Introduction :- We have to find, 88.58% confidence interval for population mean Margin of error.
Q: A city planner wants to estimate the average monthly residential water usage in the city. He…
A:
Q: The 95% Confidence Interval for the weights (in pounds) of a special type of rock has found to be…
A: Given, The 95% Confidence Interval for the weights (in pounds) of a special type of rock has found…
Q: 20 years old women’s weighs have a population standard deviation σ = 20 lb. The distribution is…
A: The objective of this question is to calculate the margin of error and the confidence interval for…
Q: A researcher collected sample data for 10 middle-aged women. The sample had a mean serum cholesterol…
A: Given , Sample size , n = 10 Sample mean , 189.6standard deviation , s = 9confidence interval = 95%…
Q: CONFIDENCE INTERVALS FOR ONE POPULATION MEAN A study is conducted on the amount of time Alzheimer's…
A: Given n=36, x=35 and s=3, Then the standard error isS.E.=sn=336=36=0.5
Q: A researcher wants to estimate the mean blood cholesterol level of young men ages 15-25 with a…
A:
Q: Find a 95% confidence interval for the mean time of Visayas Team athletes for them to finish a…
A: It is given that Sample size n = 12 Sample mean M = 98 Sample SD s = 1.5 The critical value of t at…
Q: ne distance a basketball player runs during a ga r assessing the performance of the players. 73
A: to find 95% confidence interval for true population mean
Q: The age when smokers first start from previous studies is normally distributed with a population…
A:
Q: A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 185…
A:
Q: The researcher recorded the amount of time each bird spent in the plain chamber during a 60-minute…
A: It is given that mean and sum of squares are 44.62 and 206, respectively. The sample size (n) is 10.
Q: the 95% confidence interval for estimating the population mean μ If sample mean X = 50, sample size…
A: We have given that Sample size = 60 Sample mean = 50 Population standard deviation = 10. 95%…
Q: The 82.9% confidence interval estimate for μμ was (25.81, 31.35). Find the point estimate and the…
A: The confidence interval for population mean is (25.81, 31.35). Let M denote the point estimate and E…
The confidence interval with
Step by step
Solved in 2 steps
- To calculate 95% confidence interval we must multiply the standard error by 3, then add and subtract the result to the best estimate. add and subtract the standard error to the best estimate. multiply the standard error in y-intercept by 2, then add and subtract the result to the slope of the graph. multiply the standard error by 2, then add and subtract the result to the best estimate.A researcher collected sample data for 19 middle-aged women. The sample had a mean serum cholesterol level (measured in milligrams per one hundred milliliters) of 192.4, with a standard deviation of 8.2. Assuming that serum cholesterol levels for middle-aged women are normally distributed, find a 90% confidence interval for the mean serum cholesterol level of all women in this age group. Give the lower limit and upper limit of the 90% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.) Lower limit: Upper limit:A researcher collected sample data for 19 middle-aged women. The sample had a mean serum cholesterol level (measured in milligrams per one hundred milliliters) of 190.3, with a standard deviation of 9.5. Assuming that serum cholesterol levels for middle-aged women are normally distributed, find a 95% confidence interval for the mean serum cholesterol level of all women in this age group. Give the lower limit and upper limit of the 95% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.) Lower limit= Upper limit=
- An article claims that babies spend twice as much time in REM (rapid eye movement) sleep than adults. This implies that babies have an average of 6 hours of REM sleep per night. A doctor would like to test the hypotheses = 6 versus ≠ 6 where = the true mean amount of REM sleep per night for babies. A 95% confidence interval for the true mean amount of REM sleep per night for babies is (5.45, 6.25) hours. Based on the confidence interval, what conclusion would you make for a test of these hypotheses? The 95% confidence interval include as a plausible value, so we H0. We convincing evidence that the true mean amount of REM sleep per night for babies 6 hours.The 99% confidence interval on a population mean computed with a sample of size 36 and sample mean 1.4 is computed knowing that the standard deviation of this population is 0.3. The margin of error for this confidence interval is aboutFind the level of the confidence interval that has the value Za/2 = 1.28. (Round the final answer to three decimal places.)
- A city planner wants to estimate the average monthly residential water usage in the city. He selected a random sample of 42 households from the city, which gave the mean water usage to be 3409.30 gallons over a one-month period. Based on earlier data, the population standard deviation of the monthly residential water usage in this city is 387.90 gallons. Make a 97% confidence interval for the average monthly residential water usage for all households in this city.Suppose that a jewelry store tracked the amount of emeralds they sold each week to more accurately estimate how many emeralds to keep in stock. At the end of one year, the store used the weekly sales information to construct a 95% confidence interval for the mean number of emeralds sold per week. The confidence interval they calculated was (3.25,4.51). Use the confidence interval to find the point estimate for the mean and the margin of error. Give your answers precise to two decimal places.A researcher collected sample data for 15 middle-aged women. The sample had a mean serum cholesterol level (measured in milligrams per one hundred milliliters) of 192.7, with a standard deviation of 6. Assuming that serum cholesterol levels for middle-aged women are normally distributed, find a 99% confidence interval for the mean serum cholesterol level of all women in this age group. Give the lower limit and upper limit of the 99% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.) Lower limit: Upper limit: X Ś
- A researcher collected sample data for 19 middle-aged women. The sample had a mean serum cholesterol level (measured in milligrams per one hundred milliliters) of 191.5, with a standard deviation of 6.5. Assuming that serum cholesterol levels for middle-aged women are normally distributed, find a 95% confidence interval for the mean serum cholesterol level of all women in this age group. Give the lower limit and upper limit of the 95% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. lower limit ____ Upper Limit ____A researcher collected sample data for 20 middle-aged women. The sample had a mean serum cholesterol level (measured in milligrams per one hundred milliliters) of 190.4, with a standard deviation of 9.7. Assuming that serum cholesterol levels for middle-aged women are normally distributed, find a 99% confidence interval for the mean serum cholesterol level of all women in this age group. Give the lower limit and upper limit of the 99% confidence interval.Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place Lower limit= Upper limit=If n=550 and p' (p-prime) = 0.19, construct a 90% confidence interval.Give your answers to three decimals.
