High School
SAT
ACT
University
MCAT
LSAT
DAT

Confidence Interval for a Mean

Related Topics

Wize University Statistics Textbook > Estimating with Confidence Intervals

Margin of Error and Sample Size

2 Activities

Wize University Statistics Textbook > Estimating with Confidence Intervals

Confidence Interval for a Mean

3 Activities

Which statement is true about a 95% confidence interval [188,212]?
More Margin of Error and Sample Size Questions:
Margin of Error and Sample Size
Approximately 70% of laptops in our sample have components that are made in China. How many laptops should you sample to estimate the true proportion with 95% confidence and a margin of error of 0.075?
Margin of Error and Sample Size
In a sample random sample of 20, the average is 18 and the standard deviation is 1.25. Let m1m_1 be the margin of error for a 95% confidence interval. Let m2m_2 be the margin of error for a 90% confidence interval. Calculate the value of m1−m2m_1-m_2
Margin of Error and Sample Size
Previous studies have shown that approximately 70% of laptops have components that
are made in Cina. How many laptops should you sample to estimate the true proportion with 95% confidence and a margin of error of 0.09?
Margin of Error and Sample Size
You wish to estimate the proportion of all laptops in Kanata that have components made in Cina. In a random sample of 12 laptops, 10 have components made in Cina.
The margin of error for a 95% confidence interval for the proportion of laptops made in Cina is
Margin of error and sample size
Professor Karl has two large statistics classes. He teaches one class in the morning and one in the evening. He samples students in both classes to assess his teaching performance. He uses their latest quiz marks for his assessment.
Here is some summarized statistics:
Assuming that the variances of the class quiz marks are equal, find the margin of error for a 95% confidence interval for the difference in the two means
Margin of Error and Sample Size
In a sample random sample of 20, the average is 18 and the standard deviation is 1.25. Let m1m_1 be the margin of error for a 95% confidence interval. Let m2m_2 be the margin of error for a 90% confidence interval. Calculate the value of m1−m2m_1-m_2
Margin of Error and Sample Size
A research group wants to estimate the number of transactions people make on their credit cards each month. The standard deviation is known to be 42 transactions. They want to construct a 95% confidence interval to estimate the mean number of transactions for October 2015 to within no more than 8 transactions.
Determine the sample size required.
What sample size is required if they want to use a 99% confidence interval?
Margin og Error and Sample Size
A statistician wishes to compute a 90% confidence interval for the average number of calories a typical adult consumes. He wishes to have a margin of error of no more than 10 calories. What is the minimum sample size he can use? The population standard deviation is 75.
Margin of Error
Increasing the sample size will decrease the margin of error while keeping the same level of confidence.
Margin of Error and Sample Size
A research group wants to estimate the number of transactions people make on their credit cards each month. The standard deviation is known to be 42 transactions. They want to construct a 95% confidence interval to estimate the mean number of transactions for October 2015 to within no more than 8 transactions.
Margin of Error and Sample Size
A statistician wishes to compute a 90% confidence interval for the average number of calories a typical adult consumes. He wishes to have a margin of error of no more than 10 calories. What is the minimum sample size he can use? The population standard deviation is 75.
Margin of Error and Sample Size
True or False?
1) Increasing n will increase the margin of error?
2) Increasing Zα2Z_{\frac{α}{2}} will increase the margin of error?
Margin of Error and Sample Size
In predicting the mean length of a species of snakes, how many times as many observations do we need to sample to cut the margin of error by a half without changing the level of confidence?
Statistical errors
For the following statements, select true or false. Enter T for True, F for False. (Use capital letters: T, F)
1. The mean weight of five cats is 20 lbs. We take away one that at weights 28 lbs. The mean weight of the remaining cats is 18 lbs.
T
2. The 95% confidence interval is [$800, $1,200]. The sample mean must be $1,000.
T
More Confidence Interval for a Mean Questions:
Confidence Interval for a Mean
A 90% confidence interval for the mean number of pairs of shoes Justine purchases every few years is (5, 17).
Confidence Interval for a Mean
You wish to test if being a member of a union increases hourly wages, in dollars, for the average construction worker. A sample of 18 construction workers' earnings were recorded prior to and after forming a union.
The results are: mean difference = −4.135-4.135 standard deviation of differences = 7.197.19
Compute a 95% confidence interval for the mean difference in the hourly wages
Confidence Interval for a Mean
You wish to test Ho:μ=29   Ha:μ≠29H_o:\mu=29\ \ \ H_a:\mu\ne29
A 90% confidence interval for the population mean is (22, 28)\left(22,\ 28\right)
Which of the following is NOT true?
Confidence Interval for a Mean
We have an SRS of 100 people. If about 95% of the data is between 193 and 393, find the 95% confidence interval.
Confidence Interval for a Mean
The 95% confidence interval is [$800, $1,200]. The sample mean must be $1,000.
Confidence interval for a mean
According to a sample of 100 apartment suites in Twin Pines, the sample mean rent is $970. The population standard deviation is $540. Construct a 99% confidence interval for the true population mean rent in Twin Pines.
(a) The mayor of Twin Pines says that the mean rent is only $700. Is this plausible?
(b) Helen lives in Twin Pines and says her rent is only $700. Is this plausible?
Confidence Interval for a Mean
A large software development firm recently relocated its facilities. Top management has encouraged their professional employees to engage in local service activities. Based on a random sample of 24 professionals, the sample mean is 16.75 hours. The population standard deviation is 2.4 hours.
A 95% confidence interval for the true mean number of hours of volunteer time is closest to:
Confidence interval for a mean
Kevin is the quality assurance manager for Gossip Girl dolls that talk gibberish until the AA batteries run out. The commercial for the toy says: “Over 300 hours of non-stop yapping!” He randomly selects 36 dolls and finds that they talk non-stop for about 275 hours on average until their batteries are drained. The number of hours is normally distributed with a standard deviation of 25 hours.
A bored and angry mother sues the company for false advertisement. Gio, the Marketing Director, defends himself in court by saying that the toys' average number of hours is not significantly different from 300. Use the confidence interval calculated in Question 9 to determine the conclusion of the two-tailed hypothesis test at the 5% level of significance.
Confidence Interval for a Mean
Kevin is the quality assurance manager for Gossip Girl dolls that talk gibberish until the AA batteries run out. The commercial for the toy says: “Over 300 hours of non-stop yapping!” He randomly selects 36 dolls and finds that they talk non-stop for about 275 hours on average until their batteries are drained. The number of hours are normally distributed with a standard deviation of 25 hours.
Construct a 95% confidence interval for the hours of non-stop yapping.
Confidence interval for a mean
A 90% confidence interval for the average length of a pregnancy is 33 to 38 weeks.
The correct conclusion is
Confidence Interval for a Mean
A sample of 61 households that have dogs reveal that an average of $140 is spent on dog food per month. The standard deviation of the sample is $115 per month.
Construct a 90% confidence interval for the monthly mean amount spent on dog food.
Confidence Interval
(v) Which of the following is/are true? (Check all that applies.)
Confidence intervals
A large software development firm recently relocated its facilities. Top management has encouraged their professional employees to engage in local service activities. Based on a random sample of 24 professionals, the sample mean is 16.75 hours and the standard deviation is 2.4 hours.
A 95% confidence interval for the true mean number of hours of volunteer time is closest to:
Confidence intervals
Based on a random sample size of 20 people with full-time jobs, the average time they spend in the shower each morning is 16 minutes with a standard deviation of 5 minutes. Construct a 95% confidence interval.
Population sampling
There are 600 Sephora locations in the United States. Christine randomly selected 40 stores, and then she randomly selected 100 customers from each of those stores. Less than 1% of her sample identified as heterosexual males. Based on her sample, the average spending at Sephora is $45. In addition, 340 customers in her sample revealed that they have VIB Rouge status, which the highest tier status that is reached when a customer spends $1,000 in a calendar year.
Which of the following is/are true? (Check all that applies.)
Population Sampling: Confidence Interval
Bob, a second-year student at Twin Pines University, is not very good at statistics. He wishes to infer about the average starting salary of recent graduates at Twin Pines University. His sample included 12 friends, all recent graduates, and it revealed a mean of $55,000. His 90% confidence interval is [39,116, 80,884].
Bob reports: "We conclude that 90% of recent graduates at Twin Pines University earn between $39,116 and $80,884."
List everything that is wrong about his statement. (Check all that applies.)
Statistical errors
For the following statements, select true or false. Enter T for True, F for False. (Use capital letters: T, F)
1. The mean weight of five cats is 20 lbs. We take away one that at weights 28 lbs. The mean weight of the remaining cats is 18 lbs.
T
2. The 95% confidence interval is [$800, $1,200]. The sample mean must be $1,000.
T
Confidence interval with t-distribution
A sample of 61 households that have dogs reveal that an average of $140 is spent on dog food per month with a standard deviation of $115 per month.
Confidence Interval for a Mean
A t distribution is used instead of the normal z distribution when
Confidence Interval for a Mean
A 90% confidence interval for the average length of a pregnancy is 33 to 38 weeks.
The correct conclusion is