Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $30,000 and $45,000. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. a. What is the planning value for the population standard deviation? How large a sample should be taken if the desired margin of error isb.$500? c.$200? d.$100? e.Would you recommend trying to obtain the $100 margin of error? Explain.

Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $30,000 and $45,000. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. a. What is the planning value for the population standard deviation? How large a sample should be taken if the desired margin of error isb.$500? c.$200? d.$100? e.Would you recommend trying to obtain the $100 margin of error? Explain.

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

See similar textbooks

Related questions

Q: Find the sample size need to find the Population Mean if the margin of error is $105, the confidence…

A: Solution: From the given information, confidence level is 0.95, σ=315 and E=105.

Q: a. What is the planning value for the population standard deviation? population mean annual starting…

A: Here given annual starting salary for college graduate with degree in business administration are…

Q: You are given the sample mean and the population standard deviation. Use this information to…

A: sample size(n)=35Mean()=126.00population standard deviation()=$15.20confidence level=90%

Q: After surgery, patients remain in the hospital for a few days to be recover. Historical data from a…

A: Given,sample size(n)=50sample mean(x¯)=6.2sample standard deviation(s)=7.4

Q: You are conducting a study to see how much high school students spend on back to school shopping.…

A: The sample size can be calculated as: n=z×σE2 Here, z = desired confidence level E = The allowable…

Q: The average daily points for my fantasy baseball team are {4.1, 5.7, 3.8, 4.7, 4.7, 5.5, 4.1, 2.7,…

A: Solution: Note: Hello! As you have posted more than 3 sub parts, we are answering the first 3…

Q: You want to obtain a sample to estimate how much parents spend on their kid's birthday parties.…

Q: You are given the sample mean and the population standard deviation. Use this information to…

A: Givensample size(n)=50Mean(x)=129.00standard deviation(σ)=18.20

Q: You are given the sample mean and the population standard deviation. Use this information to…

A: Mean()=108.78standard deviation()=10.26sample size(n)=36

Q: You are given the sample mean and the population standard deviation. Use this information to…

A: GivenMean(x)=145.00standard deviation(σ)=16.60sample size(n)=50

Q: The average income in a certain area is assumed to be approximately $35,000. Assume the standard…

Q: You are given the sample mean and the population standard deviation. Use this information to…

A: To construct the confidence intervals, we use the following formulas: For a 90% confidence…

Q: In the 1992 presidential election, Alaska's 40 election districts averaged 2053 votes per district…

A: “Since you have posted a question with multiple sub-parts, we will solve first three subparts for…

Q: You are given the sample mean and the population standard deviation. Use this information to…

Q: convenient, use technology to construct the confidence intervals. A random sample of 50 home theater…

A: For the calculation, Minitab is used, We are given a random sample of size 50 with sample mean 126…

Q: You work for a soft-drink company in the quality control division. You are interested in the…

A: Given data…

Q: You are given the sample mean and the population standard deviation. Use this information to…

Q: The mean work week for engineers in a start-up company is believed to be about 56 hours. A newly…

Q: If the age distribution is a bellshaped distribution, what percentage of the students are less than…

A: Obtain the percentage of the students are less than 23 years old. The percentage of the students are…

Q: a. Determine the sampling distribution of the sample mean for samples of size 64. The mean of the…

A: GivenMean(μ)=$51.5 thousand = $51500standard deviation(σ)=$7.6 thousand = $7600

Q: In a random sample of seven people, the mean driving distance to work was 24.1 miles and the…

Q: Scores on an aptitude test are normally distributed with a mean of 1000 and standard deviation of…

A: It is give that the value of mean is 1000 and standard deviation is 120. If 1,590,000 student take…

Q: You are given the sample mean and the population standard deviation. Use this information to…

Q: If the weights of newborn babies are normally distributed with a mean of 2950 grams and a standard…

A: Given, X~Nμ,σ where μ=2950 grams and σ=340 gramsfind P(x<2610)=?

Q: nstruct a 90% contidence interval for the population mea e 90% confidence interval is ( 139.97…

A: As the population standard deviation is known, we will use z distribution. The critical value for…

Q: random sample of 12 second-year university students enrolled in a business statistics course was…

A: Given that Sample size n =12 Population mean =60 Population standard deviation =8 Level of…

Q: You are given the sample mean and the population standard deviation. Use this information to…

A: From the given information, let X be the price of a certain stock. Total number of business days n =…

Q: You are given the sample mean and the population standard deviation. Use this information to…

A: For the 48 sampled business days (i.e., the sample size, n=48), the sample mean closing price (in $)…

Q: This past yeal, through (e). A Click the icon to view the data table. (d) Explain why a large sample…

A: Given dataThis past year an organisation conducted a surveyThe results of the survey are shown

Q: ) Find the sample standard deviation: b) Find the lower and upper χ2χ2 critical values at 80%…

A: here use given sample data for calculating sample standard deviation confidence level = 80%

Q: A market surveyor wishes to know how many energy drinks teenagers drink each week. They want to…

A: Population standard deviation()=1.3Margin of error(E)=0.06

Q: IQ scores have a bell-shaped (mound shaped) distribution with a mean of 100 and a standard deviation…

A: Note: Hi there! Thank you for posting the question. As there are multiple different questions,…

Q: Internet providers: In a survey of 992 homeowners with high-speed Internet, the average monthly cost…

A: X follows a Normal distribution with mean = 65.8 and S.D = 11.14 Then z=(x-mean)/SD =…

Q: You are given the sample mean and the population standard deviation. Use this information to…

A: Confidence Interval: In statistics, we can say a range of values of the specified population…

Q: (c) If the confidence levels were 98% rather than 80%, would the margin of error be larger or…

A: The confidence levels are 98% and 80%.

Q: Use the magnitudes (Richter scale) of the 120 earthquakes listed in the accompanying data table.…

A: Here given the data of magnitudes of 120 earthquakes. need to find the range variance and standard…

Concept explainers

Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.

Question

How large a sample should be taken if the desired margin of error is
b.$500?

c.$200?

d.$100?

e.Would you recommend trying to obtain the $100 margin of error? Explain.

Definition Definition Method in statistics by which an observation’s uncertainty can be quantified. The main use of interval estimating is for describing a range that is made by transforming a point estimate by determining the range of values, or interval within which the population parameter is likely to fall. This range helps in measuring its precision.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

Step 2

Step 3

Step 4

Step 5

Step 6

Trending now

This is a popular solution!

Step by step

Solved in 6 steps with 5 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Point Estimation, Limit Theorems, Approximations, and Bounds$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, probability and related others by exploring similar questions and additional content below.

Similar questions

Need some Help!
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of 49 business days, the mean closing price of a certain stock was $109.89 . Assume the population standard deviation is $10.27 .
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 55 home theater systems has a mean price of $120.00. Assume the population standard deviation is $18.20.
You work for a soft-drink company in the quality control division. You are interested in the standard deviation of one of your production lines as a measure of consistency. The product is intended to have a mean of 12 ounces, and your team would like the standard deviation to be as low as possible. You gather a random sample of 14 containers. Estimate the population standard deviation at a 90% level of confidence. 12.07 11.96 11.93 12.16 12.06 11.9 12.06 12.21 11.86 12.03 11.86 11.83 11.99 12.09 (Data checksum: 168.01) Note: Keep as many decimals as possible while making these calculations. If possible, keep all answers exact by storing answers as variables on your calculator or computer. a) Find the sample standard deviation: b) Find the lower and upper χ2χ2 critical values at 90% confidence:Lower: Upper: c) Report your confidence interval for σσ: ( , )
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of 38 business days, the mean closing price of a certain stock was $119.97. Assume the population standard deviation is $10.88.
On average, accountants earn approximately an average of $6500 per year in the United States. Assume that the distribution for electricians’ yearly earnings is normally distributed and that the standard deviation is $8680. Given a sample of 64 electricians, what is the standard deviation for the sampling distribution of the sample mean? a. 1205 b. 8680 c. 6000 d. 1085
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 45 home theater systems has a mean price of $114.00. Assume the population standard deviation is $15.10. Construct a 90% confidence interval for the population mean. The 90% confidence interval is (\,\ (Round to two decimal places as needed.)
Use the magnitudes (Richter scale) of the 120 earthquakes listed in the accompanying data table. Use technology to find the range, variance, and standard deviation. If another value, 8.00, is added to those listed in the data set, do the measures of variation change much? 3.30 2.77 2.80 1.96 1.69 2.54 2.43 3.44 3.92 1.58 2.87 1.63 2.56 3.95 2.50 2.44 1.82 2.19 2.44 2.99 2.92 2.35 1.98 3.06 2.78 3.86 2.95 2.07 1.86 2.36 2.37 3.46 3.45 1.54 2.54 1.49 2.22 3.06 2.27 3.20 1.96 1.91 2.40 2.95 2.61 1.49 2.16 2.33 1.91 2.70 2.92 1.80 3.66 2.69 1.43 3.60 3.13 2.56 1.54 1.41 2.83 2.85 2.20 1.68 3.18 1.38 1.73 2.37 1.15 2.35 2.46 1.81 2.01 3.00 1.94 2.42 1.85 2.24 2.35 3.22 3.97 2.10 1.48 2.31 2.31 2.62 2.55 2.20 2.77 2.46 2.69 3.57 2.84 2.75 3.26 1.74 4.72 3.23 2.36 2.00 3.83 2.41 2.88 2.67 2.32 2.80 2.80 2.40 3.41 2.34 1.49 2.41 2.45 2.50 2.73 2.44 2.78 2.65 2.68 2.43 Without the extra data value, the range is 3.573.57. (Type an integer or…
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 50 home theater systems has a mean price of $117.00. Assume the population standard deviation is $15.50. Construct a 90% confidence interval for the population mean. The 90% confidence interval is ( ). (Round to two decimal places as needed.) Construct a 95% confidence interval for the population mean. The 95% confidence interval is ( ). (Round to two decimal places as needed.) Interpret the results. Choose the correct answer below. O A. With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is narrower…
The incomes of trainees at a local mill are normally distributed with a mean of $1,100 and a standard deviation of $150. What percentage of trainees earn more than $1,300 a month? Group of answer choices 9.18% 40.82% 90.82% 35.31%