Question

Q11. A sample of 20 taken from the 500 employees of Target Insurance Company produced a sample average weekly salary of $500
0 0
Add a comment Improve this question Transcribed image text
Answer #1

n=20 s=15 sł=500 g9yce . d=0,01 dtam = 19 = 2,861 txizm-t = $0.005,1g : margin of error, E= tt e in = 2.861* 15 V20 -9,5961

Add a comment
Know the answer?
Add Answer to:
Q11. A sample of 20 taken from the 500 employees of Target Insurance Company produced a...
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. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • A sample of 49 observations is taken from a normal population with a standard deviation of 10. The sample mean is 55.

    A sample of 49 observations is taken from a normal population with a standard deviation of 10. The sample mean is 55. Determine the 99% confidence interval for the population mean. (Round your answers to 2 decimal places.) Confidence interval for the population mean is _______ and _______ .A research firm conducted a survey to determine the mean amount Americans spend on coffee during a week. They found the distribution of weekly spending followed the normal distribution with a population standard deviation...

  • 2. (20 points) A random sample of 16 senior employees at a large company yielded a...

    2. (20 points) A random sample of 16 senior employees at a large company yielded a sample mean annual salary of $109,450 with a sample standard deviation of $9,400. Determine a 95 percent confidence interval of the average salary of all senior employees at that company

  • A) Sample of size 15 is taken from among a population of egrets and used to...

    A) Sample of size 15 is taken from among a population of egrets and used to calculate a 99% confidence interval for the true average wingspan in the population. Suppose the sample standard deviation is calculated to be s =6.2. What is the margin of error for the confidence interval? Round your answer to 3 decimal places. B) Suppose you are asked to construct a 99% confidence interval for an unknown population mean μ. The population standard deviation is unknown...

  • The average time taken by a sample of 68 runners to complete a 10km race was...

    The average time taken by a sample of 68 runners to complete a 10km race was 42 minutes. Assuming a population standard deviation of 8.65 minutes, construct a 99% confidence interval for the true average time taken by all runners to complete the race. Interpret your answer. (6) Confidence Level Z - Limits 90% ± 1.645 95% ±1.96 99% ±2.58

  • QUESTION 6 A sample was taken of the salaries of 20 employees of a large company....

    QUESTION 6 A sample was taken of the salaries of 20 employees of a large company. The following s a boxplot of the salanes (ün thousands of dollars) Reference Ret 1-10 Based on this boxplot, which of the following statements is true? A. About 10 employees make over S 50,000. O B. All of the answers are correct. O c. The salary distribution is fairly symmetric. O D.Nobody makes over $80,000.

  • Kroger indicates the average salary per hour of their employees is $9 per hour. A sample...

    Kroger indicates the average salary per hour of their employees is $9 per hour. A sample of 13 Kroger employees revealed a mean of $11 per hour with a standard deviation of $3 per hour. Costco indicates the average salary per hour of their employees is $10 per hour. A sample of 15 Costco employees revealed a mean of $9.50 per hour with a standard deviation of $2 per hour. At the .05 significance level, can we conclude there is...

  • A sample was taken of the salaries of 20 employees from a large company. The following...

    A sample was taken of the salaries of 20 employees from a large company. The following are the salaries (in thousands of dollars) for this year (the data are ordered): 28, 31, 34, 35, 37, 41, 42, 42, 42, 47, 49, 51, 52, 52, 60, 61, 67, 72, 75, 77. Suppose each employee in the company receives a $3000 raise for next year (each employees salary is increased by $3000). The interquartile range of the salaries will A) increase by...

  • a) A random sample of 15 employees of airtel call Centre was taken and each employee...

    a) A random sample of 15 employees of airtel call Centre was taken and each employee took a competency test. The mean of the scores achieved by these employees was 56.3 with a standard deviation of 7.1. Results of this test have been found to be normally distributed in the past. Construct a 95% confidence interval for the mean test score of all the employees of the call Centre. [6 marks] b) Find a 95% confidence interval estimating the mean...

  • 4. In a survey conducted to determine, among other things, the cost of vacations taken by...

    4. In a survey conducted to determine, among other things, the cost of vacations taken by single adults in a particular region, 144 individuals were randomly sampled. Each person was asked to assess the total cost of her or his most recent vacation. The average cost was $2386 and the standard deviation was $400. (a) (2) Estimate with a 99% confidence interval the average cost of a trip for all single adults in the region (b) [2] How large a...

  • A random sample of 49 lunch customers was taken at a restaurant. The average amount of...

    A random sample of 49 lunch customers was taken at a restaurant. The average amount of time the customers in the sample stayed in the restaurant was 45 minutes with a standard deviation of 14 minutes. a. Compute the standard error of the mean. b. Construct a 99% confidence interval for the true average amount of time customers spent in the restaurant. c. With a .99 probability, how large of a sample would have to be taken to provide a...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT