Construct a 95% confidence interval for the true average monthly salary earned by all employees of People Plus Pty, if it was found that the average monthly salary earned by a sample of 19 employees of the company was R18 500, with a standard deviation of R1 750. Interpret your answer

defazajx 2021-02-24 Answered
Construct a 95% confidence interval for the true average monthly salary earned by all employees of People Plus Pty, if it was found that the average monthly salary earned by a sample of 19 employees of the company was R18 500, with a standard deviation of R1 750. Interpret your answer
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Ezra Herbert
Answered 2021-02-25 Author has 99 answers
Step 1 The sample size is 19, mean is R18500, standard deviation is R1750. The degrees of freedom is, df=n1
=191
=18 The degrees of freedom is 18. The confidence level is 95%, than the level of significance is 0.05. Computation of critical value: The critical value of t-distribution for 0.05 level of significance at 18 degrees of freedom can be obtained using the excel formula “=T.INV.2T(0.05,18)”. The critical value is 2.1009. Step 2 The 95% confidence interval for the true average monthly salary earned by all employees of People Plus Pty is, CI=x±tα2(sn)
=18500±2.1009(175019)
=18500±843.464
=(18500843.464,18500+843.464)
=(17656.536,19343.464)
(17657,19343) Thus, the 95% confidence interval for the true average monthly salary earned by all employees of People Plus Pty is (17657,19343). Interpretation: When repeated samples of size 19 are taken, about 95% of those confidence intervals would contain the true average monthly salary earned by all employees of People Plus Pty that lies between 17657 and 19343. That is, there is 95% confidence that the true average monthly salary earned by all employees of People Plus Pty that lies between 17657 and 19343.
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-08-04
A random sample of 100 automobile owners in the state of Virginia shows that an automobile is driven on average 23,500 kilometers per year with a standard deviation of 3900 kilometers.
Assume the distribution of measurements to be approximately normal.
a) Construct a 99% confidence interval for the average number of kilometers an automobile is driven annually in Virginia.
b) What can we assert with 99% confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23,500 kilometers per year?
asked 2021-01-10

The average zinc concentration recovered from a sample of measurements taken in 36 different locations in a river is found to be 2.6 grams per liter. Find the 95% confidence intervals for the mean zinc concentration in the river. Assume that the population standard deviation is 0.3 gram per liter. (Z0.025=1.96,Z0.005=2.575)

asked 2021-03-09

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 298 accurate orders and 51 that were not accurate. a. Construct a 90​% confidence interval estimate of the percentage of orders that are not accurate. b. Compare the results from part​ (a) to this 90​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B: 0.127<p<0.191. What do you​ conclude? a. Construct a 90​% confidence interval. Express the percentages in decimal form. ___

asked 2021-02-23
Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let j: denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting 95% confidence intervals (7-8, 9.6)
(a) Would 2 90%% confidence interval calculated from this same sample have been narrower or wider than the glven interval? Explain your reasoning.
(b) Consider the following statement: There is 9 95% chance that Is between 7.8 and 9.6. Is this statement correct? Why or why not?
(c) Consider the following statement: We can be highly confident that 95% of al bottles ofthis type of cough syrup have an alcohol content that is between 7.8 and 9.6. Is this statement correct? Why or why not?
asked 2022-03-29
I assume that this problem is trivial but I am not sure what to do: In a survey of 900 people in the US a journalist says that 60% of people support a new law. If the margin of error is 2.7% for the percentage, what is the level of confidence?
asked 2021-08-14
When σ is unknown and the sample size is n30, there are tow methods for computing confidence intervals for μμ. Method 1: Use the Student's t distribution with d.f. = n - 1. This is the method used in the text. It is widely employed in statistical studies. Also, most statistical software packages use this method. Method 2: When n30, use the sample standard deviation s as an estimate for σσ, and then use the standard normal distribution. This method is based on the fact that for large samples, s is a fairly good approximation for σσ. Also, for large n, the critical values for the Student's t distribution approach those of the standard normal distribution. Consider a random sample of size n = 31, with sample mean x¯=45.2 and sample standard deviation s = 5.3. (c) Compare intervals for the two methods. Would you say that confidence intervals using a Student's t distribution are more conservative in the sense that they tend to be longer than intervals based on the standard normal distribution?
asked 2022-05-11

For the provided sample​ mean, sample​ size, and population standard​ deviation, complete parts​ (a) through​ (c) below.

x=32​,

n=100​,

σ=3

 

Question content area bottom

Part 1

a. Find a​ 95% confidence interval for the population mean.