Question

The Great Eagles of the Misty Mountains wish to estimate the mean number of orcs per...

The Great Eagles of the Misty Mountains wish to estimate the mean number of orcs per raiding party. To do so, they measure the number of orcs in 13 raiding parties. They find an average of 72 orcs per party with a standard deviation of 5.7 orcs in their sample. Compute the 90% confidence interval about the mean number of orcs per raiding party. Answer with the Upper Bound only.

Round to 2 decimal places, as needed.

Homework Answers

Answer #1

Solution :

sample size = n = 13

Degrees of freedom = df = n - 1 = 12

t ,df = 1.356

Margin of error = E = t/2,df * (s /n)

= 1.356 * (5.7 / 13)

Margin of error = E = 2.14

The 95% confidence interval estimate of the population mean is,

Upper Bound =   + E

= 72 + 2.14

= 74.14

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
You wish to estimate the mean number of travel days per year for salespeople. The mean...
You wish to estimate the mean number of travel days per year for salespeople. The mean of a small pilot study was 150 days, with a standard deviation of 42 days. If you want to estimate the population mean within 6 days, how many salespeople should you sample? Use the 99% confidence level. (Use z Distribution Table.) (Round up your answer to the next whole number.)
Researchers studying trees in the Tobacco Root Mountains wish to estimate the proportion of trees that...
Researchers studying trees in the Tobacco Root Mountains wish to estimate the proportion of trees that have been infested by bark beetles. They want to randomly sample trees throughout the forest to estimate this proportion with 90% confidence and a margin of error of 5%. 1.How many trees should they sample? (Hint: use pb = .5) Is my thinking correct: n=p^*q^*Z*/ME^2 where p^=o.5, q^=0.5, Z*=1.645 ME^2=0.0025 so 164.5 so answer 165?
For these two questions In order to estimate the number of calls to expect at a...
For these two questions In order to estimate the number of calls to expect at a new suicide hotline, volunteers contact a random sample of 35 similar hotlines across the nation and find that the sample mean is 42.0 calls per month and the sample standard deviation is 5.0 calls per month. Assume that the population standard deviation is unknown. Find the error of estimate for a 90% confidence interval. Group of answer choices 1.39 1.08 1.43 1.10 Assume that...
Consider a population having a standard deviation equal to 9.94. We wish to estimate the mean...
Consider a population having a standard deviation equal to 9.94. We wish to estimate the mean of this population. (a) How large a random sample is needed to construct a 95% confidence interval for the mean of this population with a margin of error equal to 1? (Round your answer to the next whole number.) The random sample is             units. (b) Suppose that we now take a random sample of the size we have determined in part a. If we...
You want to estimate the mean number of chocolate chips in cookies. The average number of...
You want to estimate the mean number of chocolate chips in cookies. The average number of chocolate chips per cookie in a sample of 40 cookies was 23.95, with a sample standard deviation of 2.55. (A) Find a 90% confidence interval for the mean number of chocolate chips using the t-distribution. (B) Find a 90% confidence interval for the mean number of chocolate chips using the standard normal distribution (for this question, assume the population standard deviation is equal to...
A survey of 20 randomly selected adult men showed that the mean time they spend per...
A survey of 20 randomly selected adult men showed that the mean time they spend per week watching sports on television is 9.34 hours with a standard deviation of 1.34 hours. Assuming that the time spent per week watching sports on television by all adult men is (approximately) normally distributed, construct a 90 % confidence interval for the population mean,   μ . Round your answers to two decimal places. Lower bound: Enter your answer; confidence interval, lower bound Upper bound:...
The owner of Britten’s Egg Farm wants to estimate the mean number of eggs produced per...
The owner of Britten’s Egg Farm wants to estimate the mean number of eggs produced per chicken. A sample of 19 chickens shows they produced an average of 24 eggs per month with a standard deviation of 4 eggs per month. (Use t Distribution Table.) a-1. What is the value of the population mean? It is unknown. 24 4 a-2. What is the best estimate of this value? For a 90% confidence interval, what is the value of t? (Round...
In order to approximate the number of text messages sent daily by Swedish teenagers, Neu Star...
In order to approximate the number of text messages sent daily by Swedish teenagers, Neu Star Communications took a random sample of Swedish youth, finding the number of text messages they sent on a given day in 2018: (use 2 digits after decimal point) Hint: Σ X = 984 Σ X2 = 80784 n=12 Point Estimate for population mean is Point estimate for population standard deviation is Find 90% confidence interval for the population mean number of calls received by...
The owner of Britten's Egg Farm wants to estimate the mean number of eggs laid per...
The owner of Britten's Egg Farm wants to estimate the mean number of eggs laid per chicken. A sample of 14 chickens shows they laid an average of 17 eggs per month with a standard deviation of 4 eggs per month.   a-1. What is the value of the population mean? (Click to select)It is unknown.417.00 a-2. What is the best estimate of this value? b-1. Explain why we need to use the t distribution. (Click to select)Use the t distribution...
Explain why if a random sample of 50 provides a sample mean of 31 with a...
Explain why if a random sample of 50 provides a sample mean of 31 with a standard deviation of s=14. The upper bound of a 90% confidence interval estimate of the population mean is 34.32.
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT