MATH 225N Week 6 Assignment 2020 - Chamberlain College of Nursing | Confidence Interval for Mean – Population Standard Deviation Known
Find the sample
... [Show More] size required to estimate a population mean with a given confidence level Question
The population standard deviation for the number of corn kernels on an ear of corn is 94 kernels. If we want to be 90%confident that the sample mean is within 17 kernels of the true population mean, what is the minimum sample size that should be taken?
z0.101.282 z0.051.645 z0.0251.960. z0.012.326. z0.0052.576
Use the table above for the z-score. Correct! You nailed it.
21 83
329
699
Answer Explanation
Correct answer: 83
The formula for sample size is
n=z2σ2EBM2
In this formula,
z=zα2=z0.05=1.645
because the confidence level is 90%. From the problem, we know that σ=94 and EBM=17. Therefore,
n=z2σ2EBM2=(1.645)2(94)2172≈82.74
Use n=83 to ensure that the sample size is large enough.
Find the sample size required to estimate a population mean with a given confidence level Question
Suppose the scores of a standardized test are normally distributed. If the population standard deviation is 2 points, what minimum sample size is needed to be 90% confident that the sample mean is within 1 point of the true population mean?
z0.101.282z0.051.645z0.0251.960z0.012.326z0.0052.576
Use the table above for the z-score, and be sure to round up to the nearest integer. Yes that's right. Keep it up! 11 test scores
Answer Explanation
Correct answers: • 11 test scores
The formula for sample size is
n=z2σ2EBM2
In this formula,
z=zα2=z0.05=1.645
because the confidence level is 90%. From the problem, we know that σ=2 and EBM=1. Therefore,
n=z2σ2EBM2=(1.645)2(2)212≈10.82
Use n=11 to ensure that the sample size is large enough.
Also, the sample size formula shown above is sometimes written using an alternate format of n=(zσE)2. In this formula, E is used to denote margin of error and the entire parentheses is raised to the exponent 2.
Therefore, the margin of error for the mean can be denoted by "EBM" or by "E". Either formula for the sample size can be used and these formulas are considered as equivalent.
Calculate and interpret the confidence interval for a population mean with a known standard deviation Question
The length, in words, of the essays written for a contest are normally distributed with a population standard deviation of 442 words and an unknown population mean. If a random sample of 24 essays is taken and results in a sample mean of 1330 words, find a 99% confidence interval for the population mean.
z0.10
z0.05
z0.025
z0.01
z0.005
1.282
1.645
1.960
2.326
2.576
You may use a calculator or the common z values above.
• Round the final answer to two decimal places. Perfect. Your hard work is paying off 😀 (1097.59, 1562.41)
Answer Explanation
Correct answers: • (1097.59, 1562.41)
Confidence intervals are written as
(pointestimate−marginof error,pointestimate+marginof error)
The point estimate is the sample mean, x¯, and the margin of error is
marginof error=(zα2)(σn‾√)
Substituting the given values σ=442, n=24, and zα2=2.576 for a confidence level of 99%, we have
marginof error=(2.576)(44224‾‾‾√)≈(2.576)(90.223)≈232.41
With x¯=1330 and a margin of error of 232.41, the confidence interval is
(1330−232.41,1330+232.41)(1097.59,1562.41).
So we estimate with 99% confidence that the true population mean is between 1097.59 and 1562.41 words.
Calculate and interpret the confidence interval for a population mean with a known standard deviation Question
The lengths, in inches, of adult corn snakes are normally distributed with a population standard deviation of 8 inches and an unknown population mean. A random sample of 25 snakes is taken and results in a sample mean of 58 inches.
Identify the parameters needed to calculate a confidence interval at the 99% confidence level. Then find the confidence interval.
z0.10z0.10
z0.05z0.05
z0.025z0.025
z0.01z0.01
z0.005z0.005
1.282
1.645
1.960
2.326
2.576
You may use a calculator or the common z values above.
• Round the final answer to two decimal places.
Correct answers:
• 158 • 28 • 325 • 42.576 • 553.88 • 662.12
Confidence intervals are written as
(point estimate−margin of error, point estimate+margin of error)
The point estimate is the sample mean, x¯, and the margin of error is
margino f error=(zα2)(σn‾√)
Substituting the given values σ=8, n=25, and zα2=2.576 for a confidence level of 99%, we have
margin of error=(2.576)(825‾‾‾√)≈(2.576)(1.6)≈4.12
With x¯=58 and a margin of error of 4.12, the confidence interval is
(58−4.12,58+4.12)(53.88,62.12).
Calculate and interpret the confidence interval for a population mean with a known standard deviation Question The lengths, in inches, of adult corn snakes are normally distributed with a population standard deviation of 8 inches and an unknown population mean. A random sample of 25 snakes is taken and results in a sample mean of 58 inches. What is the correct interpretation of the confidence interval? Select the correct answer below:
We can estimate with 99% confidence that the true population mean length of adult corn snakes is between 53.88 and 62.12 inches. We can estimate with 99% confidence that the sample mean length of adult corn snakes is between 53.88 and 62.12 inches. We can estimate that 99% of adult corn snakes will have a length that is between 53.88 and 62.12 inches. Correct! You nailed it. We can estimate with 99% confidence that the true population mean length of adult corn snakes is between 53.88and 62.12 inches.
We can estimate with 99% confidence that the sample mean length of adult corn snakes is between 53.88 and 62.12inches.
We can estimate that 99% of adult corn snakes will have a length that is between 53.88 and 62.12 inches.
Calculate and interpret the confidence interval for a population mean with a known standard deviation Question
The weights, in pounds, of dogs in a city are normally distributed with a population standard deviation of 2 pounds and an unknown population mean. A random sample of 16 dogs is taken and results in a sample mean of 28 pounds.
Identify the parameters needed to calculate a confidence interval at the 90% confidence level. Then find the confidence interval.
z0.10
z0.05
z0.025
z0.01
z0.005
1.282
1.645
1.960
2.326
2.576
You may use a calculator or the common z values above.
• Round the final answer to two decimal places.
Correct answers:
• 128 • 22 • 316 • 41.645 • 527.18 • 628.82
Confidence intervals are written as
(pointestimate−marginof error,pointestimate+marginof error)
The point estimate is the sample mean, x¯, and the margin of error is
marginof error=(zα2)(σn‾√)
Substituting the given values σ=2, n=16, and zα2=1.645 for a confidence level of 90%, we have
marginof error=(1.645)(216‾‾‾√)≈(1.645)(0.5)≈0.82
With x¯=28 and a margin of error of 0.82, the confidence interval is
(28−0.82,28+0.82)(27.18,28.82) Calculate and interpret the confidence interval for a population mean with a known standard deviation Question The weights, in pounds, of dogs in a city are normally distributed with a population standard deviation of 2 pounds and an unknown population mean. A random sample of 16 dogs is taken and results in a sample mean of 28 pounds. What is the correct interpretation of the confidence interval? Select the correct answer below:
We can estimate that 90% of the dogs in the city have a weight that lies between 27.18 and 28.82 pounds. We can estimate with 90% confidence that the sample mean weight of dogs in the city is between 27.18 and 28.82pounds. We can estimate with 90% confidence that the true population mean weight of dogs in the city is between 27.18and 28.82 pounds. Correct! You nailed it.
We can estimate that 90% of the dogs in the city have a weight that lies between 27.18 and 28.82 pounds.
We can estimate with 90% confidence that the sample mean weight of dogs in the city is between 27.18 and 28.82pounds. We can estimate with 90% confidence that the true population mean weight of dogs in the city is between 27.18and 28.82 pounds.
Answer Explanation
Correct answer: We can estimate with 90% confidence that the true population mean weight of dogs in the city is between 27.18 and 28.82 pounds.
Once a confidence interval is calculated, the interpretation should clearly state the confidence level (CL), explain what population parameter is being estimated, and state the confidence interval.
We can estimate with 90% confidence that the true population mean weight of dogs in the city is between 27.18 and 28.82pounds.
Calculate and interpret the confidence interval for a population mean with a known standard deviation Question
Suppose heights of dogs, in inches, in a city are normally distributed and have a known population standard deviation of 7inches and an unknown population mean. A random
sample of 15 dogs is taken and gives a sample mean of 34 inches. Find the confidence interval for the population mean with a 99% confidence level.
z0.10
z0.05
z0.025
z0.01
z0.005
1.282
1.645
1.960
2.326
2.576
You may use a calculator or the common z values above.
• Round the final answer to two decimal places. MY answer wrong (28.62, 39.38)
Answer Explanation
Correct answers: • (29.34, 38.66)
Confidence intervals are written as
(pointestimate−marginof error,pointestimate+marginof error)
The point estimate is the sample mean, x¯, and the margin of error is
marginof error=(zα2)(σn‾√)
Substituting the given values σ=7, n=15, and zα2=2.576 for a confidence level of 99%, we have
marginof error=(2.576)(715‾‾‾√)≈(2.576)(1.807)≈4.66
With x¯=34 and a margin of error of 4.66, the confidence interval is
(34−4.66,34+4.66)(29.34,38.66).
So we estimate with 99% confidence that the true population mean is between 29.34 and 38.66 inches.
Calculate and interpret the confidence interval for a population mean with a known standard deviation
Calculating Confidence Intervals With a Known Standard Deviation
Calculate and interpret the confidence interval for a population mean with a known standard deviation Question
The germination periods, in days, for grass seed are normally distributed with a population standard deviation of 5 days and an unknown population mean. If a random sample of 17 types of grass seed is taken and results in a sample mean of 52days, find a 80% confidence interval for the population mean.
z0.10
z0.05
z0.025
z0.01
z0.005
1.282
1.645
1.960
2.326
2.576
You may use a calculator or the common z values above.
Select the correct answer below:
(50.45,53.55)
(50.01,53.99)
(49.85,54.15)
(49.62,54.38)
(49.18,54.82)
(48.88,55.12)
Correct answer: (50.45,53.55)
Confidence intervals are written as (x¯−EBM,x¯+EBM), so we need the sample mean, x¯, and the EBM. We know x¯=52. We can use the formula to find the error bound: EBM=(zα2)(σn‾√)=(1.282)(517‾‾‾√)≈(1.282)(1.213)≈1.55 So, the error bound (EBM) is 1.55. So we can write this confidence interval as: (52−1.55,52+1.55) or (50.45,53.55). Using a calculator: Press STAT and arrow over to TESTS. Arrow down to 7: ZInterval, and press ENTER. Arrow to Stats, and press ENTER. Arrow down and enter 5 for σ (standard deviation), 52 for x¯ (sample mean), 17 for n (sample size), and 0.8 for the C-Level (confidence level). Arrow down to Calculate, and press ENTER. So we estimate with 80% confidence that the true population mean is between 50.45 and 53.55 days.
Calculate and interpret the confidence interval for a population mean with a known standard deviation Question
Suppose scores of a standardized test are normally distributed and have a known population standard deviation of 6 points and an unknown population mean. A random sample of 22 scores is taken and gives a sample mean of 92 points.
Identify the parameters needed to calculate a confidence interval at the 98% confidence level. Then find the confidence interval.
z0.10
z0.05
z0.025
z0.01
z0.005
1.282
1.645
1.960
2.326
2.576
You may use a calculator or the common z values above.
• Round the final answer to two decimal places, if necessary. • Yes that's right. Keep it up!
•
• 92
• 6
• 22
• 2.326
• (89.02, 94.98) Calculate and interpret the confidence interval for a population mean with a known standard deviation Question • Suppose scores of a standardized test are normally distributed and have a known population standard deviation of 6 points and an unknown population mean. A random sample of 22 scores is taken and gives a sample mean of 92 points. • What is the correct interpretation of the 98% confidence interval? Select the correct answer below: • We can estimate that 98% of the time the test is taken, a student scores between 89.02 and 94.98 points. • We can estimate with 98% confidence that the true population mean score is between 89.02 and 94.98 points. • We can estimate with 98% confidence that the sample mean score is between 89.02 and 94.98 points. Yes that's right. Keep it up!
Once a confidence interval is calculated, the interpretation should clearly state the confidence level (CL), explain what population parameter is being estimated, and state the confidence interval.
We can estimate with 98% confidence that the true population mean score is between 89.02 and 94.98 points.
Find the sample size required to estimate a population mean with a given confidence level Question
Suppose the heights of seasonal pine saplings are normally distributed. If the population standard deviation is 14millimeters, what minimum sample size is needed to be 95% confident that the sample mean is within 4 millimeters of the true population mean?
z0.101.282z0.051.645z0.0251.960z0.012.326z0.0052.576
Use the table above for the z-score, and be sure to round up to the nearest integer. Well done! You got it right. 48 pine saplings
Answer Explanation
Correct answers: • 48 pine saplings
The formula for sample size is
n=z2σ2EBM2
In this formula,
z=zα2=z0.025=1.96
because the confidence level is 95%. From the problem, we know that σ=14 and EBM=4. Therefore,
n=z2σ2EBM2=(1.96)2(14)242≈47.06
Use n=48 to ensure that the sample size is large enough.
Also, the sample size formula shown above is sometimes written using an alternate format of n=(zσE)2. In this formula, E is used to denote margin of error and the entire parentheses is raised to the exponent 2.
Therefore, the margin of error for the mean can be denoted by "EBM" or by "E". Either formula for the sample size can be used and these formulas are considered as equivalent.
Find the sample size required to estimate a population mean with a given confidence level Question
The population standard deviation for the scores of a standardized test is 4 points. If we want to be 90% confident that the sample mean is within 1 point of the true population mean, what is the minimum sample size that should be taken?
z0.101.282 z0.051.645 z0.0251.960 z0.012.326 z0.0052.576
Use the table above for the z-score, and be sure to round up to the nearest integer.
Provide your answer below:
__44_test scores
Great work! That's correct. [Show Less]