Question

In: Economics

Please answer question A, B, and C with the solution Suppose we take a random sample...

Please answer question A, B, and C with the solution

Suppose we take a random sample of 30 companies in an industry of 200 companies. We calculate the sample mean of the ratio of cash flow to total debt for the prior year. We find that this ratio is 23 percent. Subsequently, we learn that the population cash flow to total debt ratio (taking into account of all 200 companies) is 26 percent. What is the explanation for the discrepancy between the sample mean of 23 percent and the population mean of 26 percent?

A. Sampling Error

B. Bias

C. A lack of Consistency Answer:

Solutions

Expert Solution

Bias is the case where the same estimator gives different results different times.

Lack of consistency refers to the lack of procedures followed in estimating a sample.

Sampling error means when that sample is taken into consideration which does not represent the entire population of the data and thus the result from estimating the sample and the population becomes different.

Here, a random sample of 30 companies in an industry of 200 companies gave been taken into consideration. The sample mean of the ratio of cash flow to total debt for the prior year is 23 percent. However, the population mean of the ratio is 26 percent. This discrepancy has occurred due to sample error. The difference in the values of the sample mean and the population mean has occurred because samples have been chosen of which every sample does not follow the properties of the population. Therefore, the population mean is 26 percent and that of the sample mean is 23 percent.

Therefore, option A stands correct here. The discrepancy vetebet the sample mean and the population mean has occurred due to Sampling Error.


Related Solutions

Suppose we take a poll (random sample) of 3992 students classified as Juniors and find that...
Suppose we take a poll (random sample) of 3992 students classified as Juniors and find that 2818 of them believe that they will find a job immediately after graduation. What is the 99% confidence interval for the proportion of Juniors who believe that they will, immediately, be employed after graduation. (0.694, 0.718) (0.687, 0.724) (0.699, 0.713) (0.692, 0.72)
53% of people beleive in love at first sight. Suppose we take a random sample of...
53% of people beleive in love at first sight. Suppose we take a random sample of 200 people. What is the probability that 1 a ) Exactly 95 of them beleive in love at first sight 1 b ) 120 or more of them beleive in love at first sight 1 c) Fewer than 90 of them beleive in love at first sight
53% of people believe in love at first sight. Suppose we take a random sample of...
53% of people believe in love at first sight. Suppose we take a random sample of 550 people. What is the probability that: 1. Exactly 295 of them believe in love at first sight? 2. 300 or more of them believe in love at first sight? 3. Fewer than 250 of them believe in love at first sight?
Suppose we take a poll (random sample) of 3907 students classified as Juniors and find that...
Suppose we take a poll (random sample) of 3907 students classified as Juniors and find that 2904 of them believe that they will find a job immediately after graduation. What is the 99% confidence interval for the proportion of GSU Juniors who believe that they will, immediately, be employed after graduation. (0.725, 0.761) (0.732, 0.755) (0.73, 0.757) (0.736, 0.75)
Suppose we take a random sample X1,…,X5 from a normal population with unknown variance σ2 and...
Suppose we take a random sample X1,…,X5 from a normal population with unknown variance σ2 and unknown mean μ. We test the hypotheses H0:σ=2 vs. H1:σ<2 and we use a critical region of the form {S2<k} for some constant k. (1) - Determine k so that the type I error probability α of the test is equal to 0.05. (2) - For the value of k found in part (1), what is your conclusion in this test if you see...
Suppose we take a random sample X1,…,X5 from a normal population with an unknown variance σ2...
Suppose we take a random sample X1,…,X5 from a normal population with an unknown variance σ2 and unknown mean μ. Construct a two-sided 95% confidence interval for σ2 if the observations are given by −1.25,1.91,−0.09,2.71,2.70
Suppose we take a random sample X1,…,X7 from a normal population with an unknown variance σ21...
Suppose we take a random sample X1,…,X7 from a normal population with an unknown variance σ21 and unknown mean μ1. We also take another independent random sample Y1,…,Y6 from another normal population with an unknown variance σ22 and unknown mean μ2. Construct a two-sided 90% confidence interval for σ21/σ22 if the observations are as follows: from the first population we observe 3.50,0.64,2.68,6.32,4.04,1.58,−0.45 and from the second population we observe 1.85,1.53,1.57,2.13,0.74,2.17
A representative sample: c. is frequently a random sample d. both b and c b. reflects...
A representative sample: c. is frequently a random sample d. both b and c b. reflects the characteristics of the population a. eliminates the problem of response bias
Suppose we take a random sample X1,…,X6 from a normal population with a known variance σ21=4...
Suppose we take a random sample X1,…,X6 from a normal population with a known variance σ21=4 and unknown mean μ1. We also collect an independent random sample Y1,…,Y5 from another normal population with a known variance σ22=1 and unknown mean μ2. We use these samples to test the hypotheses H0:μ1=μ2 vs. H1:μ1>μ2 with a critical region of the form {X¯−Y¯>k} for some constant k. Here Y¯ denotes the average of Yis just like X¯ denotes the average of Xis. (1)...
We take a random sample of 16 scores from a Math Exam and the sample mean...
We take a random sample of 16 scores from a Math Exam and the sample mean is 80 (the population scores are normally distributed) . The known standard deviation = 16. What is the value of Lower Confidence Limit at 90% CI.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT