Answered

Sampling Distribution (5 pts) A bank in a small town has 100,000 customers. A national survey on the banking habits of people in U.S. shows that 80% of the people with income higher than 60,000 dollars have both savings and checking accounts and also shows that the average number of banking operations that a person aged 18 and over performs per week is 10. The manager of the bank decides to do a survey among the customers of his bank and takes a simple random sample of 350 customers aged 18 and over. In the sample, the average number of banking transactions per week is 13 with standard deviation equal to 5 a) The average number of times a customer carries out banking transactions per week is give or take or so. Show how you computed your answers. b) Give a 90% and a 99% confidence interval for the average number of banking operations per week for the town residents aged 18 and over. What is the difference in Margin of error, how does it affect your confidence interval. Show your working and interpret in plain English c) Is the apparent difference in banking habits between the nation and the customers of the bank real or just due to chance? Explain for both 90% and 99% confidence levels. d) A 95% confidence interval gives a range of values for the which are plausible according to the observed data. Fill in the blanks. (Possible answer: (A) Population average, (B) Sample average) e) The sample standard deviation measures how far is from sample average. The standard deviation for the sample average measures how far average for typicalISfrom the To fill in the blanks, choose among: (A) number of bank operations, (B) average number of operations, (C) samples, (D) customers aged 18 and over, (E) bank, (F) person with high income.

Answer :

fortuneonyemuwa

Answer:

Step-by-step explanation:

(a) The average number of times a customer carries out banking transactions per week is 13 give or take 5 or so.

(b) The 90% confidence interval for the average number of banking operations per week is given by:

(Assuming a normal distribution of the number of banking operations)

CI=\overline{X}\pm 1.645\times \sigma/\sqrt{n}

CI=13\pm 1.645\times 5/\sqrt{350}

CI=13\pm 0.439645

CI=(12.560355, 13.439645)

The 99% confidence interval for the average number of banking operations per week is given by:

(Assuming a normal distribution of the number of banking operations)

CI=\overline{X}\pm 2.576\times \sigma/\sqrt{n}

CI=13\pm 2.576\times 5/\sqrt{350}

CI=13\pm 0.688465

CI=(12.311535, 13.688465)

The difference in the margin of error is 0.688465-0.439645=0.24882

The difference in the margin of error will make the confidence interval wider in the second case(99% confidence interval) as compared to the first case(90% confidence interval).

(c) Since, both our confident interval contains the value 10 hence we have sufficient evidence to conclude that the apparent difference in banking habits between the nation (It is given that the national survey suggest that on an average 10 transactions are performed per week by a single person) and the customer of the bank is not significant or is not real and is just due to the random errors/variations in sampling.

(d) A 95% confidence interval gives a range of values for the (A) Population Average which are plausible according to the observed data.

(The confidence interval always gives the range of possible values for the population values)

(e) The sample standard deviation measures how far (A) number of bank operations is from sample average.

The standard deviation for the sample average measures how far (B) Average number of bank operations is from the population average for typical (D) samples.

Other Questions