900 students were surveyed and had an average GPA of 2.7 with a standard deviation of 0.4. Calculate the margin of error for a 95% confidence level:
Q: In a random sample of 60 computers, the mean repair cost was $150 and assume that the population…
A: Given: In a random sample of 60 computers, the mean repair cost was $150 the population…
Q: random sample of ten people, the mean driving distance to work was 23.2 miles and the standard…
A: Given Sample mean=x̄=23.2, s=7.4,n=10
Q: Anystate Auto Insurance Company took a random sample of 380 insurance claims paid out during a…
A: Obtain the 90% confidence interval for the mean claim payment. The 90% confidence interval for the…
Q: Is there a difference in the average age that people get married in the US and its close neighbor,…
A: Obtain the 95% confidence interval for the mean rating. The 95% confidence interval for the mean…
Q: the standard deviation of the 16 observations was 0.95 inches (again, the sample mean was 17.2).…
A: Given : Standard deviation of the 16 observations was 0.95 inches & the sample mean was 17.2.…
Q: A random sample of 35 high-school students averaged 7.3 hours of sleep per night. Assume a…
A: Obtain the 95% confidence interval for the population mean. The 95% confidence interval for the…
Q: he average weight of 20 randomly selected cars was 3500 pounds. The sample standard deviation was…
A: given data, n=20df=n-1=20-1=19x¯=3500 pounds s=315 pounds CI=0.90α=1-0.90=0.10
Q: An automotive engineer wants to estimate the cost of repairing a car that experiences a 25 MPH…
A:
Q: In a random sample of seven people, the mean driving distance to work was 19.7 miles and the st and…
A: From provided information, random sample of 7 people is considered. The mean distance to work was…
Q: Find the value of za/2 that corresponds to a confidence level of 89.48%.
A: Confidence level: C=0.8948
Q: Find the best point estimate of the population mean. b. Construct the 95% confidence interval of the…
A: It is given that Sample size n = 35 Sample mean = 16.6 Population SD = 2.8 The critical values of Z…
Q: At 95% confidence, what is the margin of error? b. Construct the 95% confidence interval for…
A:
Q: Find the margin of error for the following; 95% confidence, sample mean is 32.5, standard deviation…
A:
Q: An economist would like to estimate the average household income in Okechobee county. The mean…
A:
Q: 300 high school students were asked how many hours of TV they watch per day. The mean was 2 hours,…
A:
Q: A runner wants to know her average mile running time. She records a sample of 25 running times. The…
A: Solution: Let X be the mile running time. From the given information, x-bar=6.5 minutes, s=0.7…
Q: The average salary of 64 employee was 5400 and the sample standard deviation was 270. Find the 90%…
A:
Q: given a standard deviation of + or - 0.023m at the 90% confidence level, what is the 68% error level…
A: The confidence level of the data is 90%. The standard deviation of the data with error percentage is…
Q: The mean time taken to design a house plan by 40 architects was found to be 19 hours with a standard…
A: Step 1:Given data :Sample size n=40degrees of freedom df=40−1=39Sample mean xˉ=19Sample standard…
Q: In a sample of 31 students, the mean scholarship amount received was $562, with a standard deviation…
A: Given data,n=31x=562s=17.04df=n-1=30α=0.10t-critical value at α=0.10 and df=30 is tc=1.697
Q: In a random sample of five people, the mean driving distance to work was 22.5 miles and the standard…
A:
Q: In a random sample of twelve cell phones, the mean full retail price was $566.50 and the standard…
A:
Q: use the t-distribution to find the margin of error and construct a 90% confidence interval for the…
A: Given that Sample size (n) = 12 Mean full retail price (x¯)= $597 Standard deviation (s) = $168…
Q: Calculate the lower endpoint of a two-sided 95% confidence interval for the population mean using…
A: given data sample size (n) = 41sample mean ( x¯ ) = 11.92population standard deviation (σ) = 1.8195%…
Q: for a sample of 150 that has a mean of 6540 and a standard deviation of .89, calculate the 95%…
A: It is given that mean and standard deviation are 6,540 and 0.89, respectively.
Q: You measure 46 textbooks' weights, and find they have a mean weight of 34 ounces. Assume the…
A: given data population sd(σ) = 14.3n = 46x¯ = 3495% ci for mean (μ) = ?
Q: Kennedy High School reported that the mean score on an end of semester physics exam wass 508 with a…
A: Given, Mean score μ=508Standard deviation σ=121Sample size n=200
Q: In a sample of 32 students, the mean scholarship amount received was $887, with a standard deviation…
A: Given information: n=32x¯=887s=27.4 df=n-1=32-1=31 Confidence level is 90% Significance level is…
Q: ten cell phones, the mean full retail price was $406.50 and the standard deviation was $174.00.…
A: Given that Sample size n =10 Sample mean =406.50 Sample standard deviation =174.00
Q: In a random sample of six people, the mean driving distance to work was 24.3 miles and the…
A:
Q: 300 high school students were asked how many hours of TV they watch per day. The mean was 2 hours,…
A: It is given that the confidence level is 0.90.
Q: for a sample of 57 that has a mean of 6200 and a standard deviation of 254, calculate the 95%…
A:
Q: or a sample of 121 that has a mean of 6540 and a standard deviation of 780, calculate the 90%…
A:
Q: a. b. Find the best point estimate of the population mean. Construct and interpret the 85%…
A: here given , sample size = n= 20 sample mean = $48.60 sample standard deviation = $3.53 here…
Q: In a random sample of 60 refrigerators, the mean repair cost was $157.08. Assume the population…
A: Given that,The sample size is refrigerators.The sample mean is The population standard deviation is…
Q: Construct a 98% confidence interval for the population mean, μ. Assume the population has a normal…
A: In this scenario, it is assumed that the population has normal distribution. So, it is appropriate…
Q: A sample of 22 public high school students spent an average of 7 hours studying per week. Assuming…
A:
Q: In a random sample of seven cell phones, the mean full retail price was $402.50 and the standard…
A:
Q: In a random sample of eight cell phones, the mean full retail price was $431.50 and the standard…
A:
Q: 300 high school students were asked how many hours.of TV they watch per day. The mean was 2 hours,…
A: Given that: Mean=2 Standard deviation=SD=σ= 0.5 Confidence level= 90% n=sample size=n=300 So…
Q: A local bank needs information concerning the checking account balances of its customers. A random…
A:
900 students were surveyed and had an average GPA of 2.7 with a standard deviation of 0.4. Calculate the margin of error for a 95% confidence level:
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 1 images
- You measure 25 randomly selected textbooks' weights, and find they have a mean weight of 73 ounces. Assume the population standard deviation is 5.8 ounces. Based on this, construct a 95% confidence interval for the true population mean textbook weight.A group of 49 randomly selected students has a mean age of 22.4 years. Assume the population standard deviation is 3.8. Construct a 90% confidence interval for the population mean. Include a statement interpreting your answer.The standard deviation, S, for the annual medical cost of 24 households was found to equal $1689 find a 95% confidence interval for the standard deviation of the annual medical cost for all households. Assume medical costs are normally distributed.
- A study by the International Coffee Organisation found that persons aged 30 – 59 drank an average of 0.57 cups of coffee per day. Assuming that the sample size is 200 with a sample standard deviation of 0.4 cups, construct a 95% confidence interval for the population mean cup of coffee. Interpret your answer.In a recent sample of 89 used cars sales costs, the sample mean was $6,225 with a standard deviation of $3,151. Assume the underlying distribution is approximately normal. Which distribution should you use for this problem? Construct a 95% confidence interval for the population mean cost of a used car.In a random sample of 17 people, the mean commute time to work was 30.2 minutes and the standard deviation was 7.1 minutes. Assume the population is normally distributed and use a t- distribution to construct a 95% confidence interval for the population mean y. What is the margin of error of y? Interpret the results.
- Ohio is surveying the price of nitrogen, which is necessary for many manufacturing processes. The state surveyed 60 stores across the state; the mean price is $0.31 per pound, and the standard deviation is $0.05. The confidence level is 95%. What is the margin of error? O 0.0125 O 0.0127 O 0.0129 O 0.0131In a sample of 100 individuals, the average height is 170 cm and the standard deviation is 15 cm. Using a 95% confidence level, what is the margin of error for the mean height of the population?Construct a 90% confidence interval for the population mean, μ . Assume the population has a normal distribution. A sample of 15 randomly selected students has a grade point average of 2.81 with a standard deviation of 0.63 .
- A survey of 73 randomly selected homeowners finds that they spend a mean of $74 per month on home maintenance. Construct a 95% confidence interval for the mean amount of money spent per month on home maintenance by all homeowners. Assume that the population standard deviation is $16 per month. Round to the nearest cent.In a sample of 45 students, the mean scholarship amount received was $854, with a standard deviation of $17.25. Contruct a 90% confidence interval for the mean scholarship award for all students. Round your answer to the nearest cent.A statistician selected sample of 16 accounts reviewable and determined the mean of the sample to be $5,000 with a standard deviation of $400. She reported that the sample information indicated the mean of the population ranges from $4,739.80 to $5,260.20. She did not report what confidence coefficient she has used. Based on the above information, determine the confidence coefficient that was used.
