**DSME2011B Business Statistics**

**Homework 3**

**Reminders:**

- Due date: Nov-12-2020 (Thursday) before midnight.
- Please show the names, ID numbers and group number of all your group members on the cover page.
- Please submit your word file to VeriGuide and submit
__both__word__and__R scripts file to Blackboard.

- In basketball, a player who is fouled in the act of shooting gets to shoot two free throws. Suppose we hear that one player is an “85% free throw shooter.”
- If this player is fouled 25 times in the act of shooting (maybe over a period of several games), find the distribution of occasions where he makes both free throws. That is, if X is the number of times he makes both free throws, find P(X = k) for each k from 0 to 25.
- How likely is it that he will make both free throws on at least 20 of the 25 occasions?

- A family is considering a move from a midwestern city to a city in California. The distribution of housing costs where the family currently lives is normal, with mean $105,000 and standard deviation $18,200. The distribution of housing costs in the California city is normal with mean $235,000 and standard deviation $30,400. The family’s current house is valued at $110,000.
- What percentage of houses in the family’s current city cost less than theirs?
- If the family buys a $200,000 house in the new city, what percentage of houses there will cost less than theirs?
- What price house will the family need to buy to be in the same percentile (of housing costs) in the new city as they are in the current city?

- Suppose that if a presidential election were held today, 53% of all voters would vote for candidate Smith over candidate Jones. (You can substitute the names of the most recent presidential candidates.) If 2500 voters are sampled randomly, what is the probability that the sample will indicate (correctly) that Smith is preferred to Jones?
- Calculate the probability with binomial probability distribution.
- Calculate the probability with normal probability distribution (with
*m*=*np*and*s*^{2}=*np*(1 –*p*)). - Compare the answers obtained in (a) and (b) above. Are they close?

- It is known that the distribution of purchase amounts by customers entering a popular retail store is approximately normal with mean $25 and standard deviation $10.
- What is the probability that a randomly selected customer spends between $12 and $28 at this store?
- What is the probability that the average amount spent by 16 randomly selected customers is between $21 and $28?

- A randomly collected sample consists of the following data: 158, 233, 121, 261, 294, 149, 224, 156, 293, 273, 182, 223, 244, 130, 265, 244, 150. Construct a 95% confidence interval for
*µ.* - In a sample of 1200 randomly selected voters, 575 of them said that they would vote for Candidate A while 625 of them indicated that they would vote for Candidate B. There are only two candidates in this election. Construct a 95% confidence interval to estimate the percentage of votes that Candidate B will get. Do you have 95% confidence to predict that Candidate B will win the election?
- Suppose that you want to estimate an unknown population mean
*m*and the population standard deviation is known to be*s*= 3500.- If you want the error of margin to be 500 with 95% confidence, how big a sample size should you take?
- If you want the error of margin to be 100 with 99% confidence, how big a sample size should you take?

- In the coming US presidential election, if a pollster wants to predict the percentage of votes for President Trump to be within ±1% (margin of error) with 99% confidence, how big a sample size should he take? How about if he wants the margin of error to be 1.5% with 90% confidence?

For Questions 9 and 10, please refer to the file “budget_mart.csv”, which contains information of 1000 customers randomly selected from the database of Budget Mart – a large retailing firm. The data has the following variables:

** Age:** 1 if 30 or younger, 2 if 31 to 55, 3 if 56 or older

** Gender:** 1 if male, 2 if female

** Married: **1 if married, 2 otherwise

** Religion:** 1 if Buddhist, 2 if Muslim, 3 if Christian, and 4 Otherwise

** Education:** 1 if Below High School, 2 if High School, 3 if College, and 4 if Graduate

** Salary: **Annual salary

** Children:** Number of children

** AmountSpent: **Total amount spent on purchases this year

- Consider only those customers who are Christian. What proportion of them spend more than (≥) $3,000? Construct a 95% confidence interval to estimate the proportion of Christians who spend more than (≥) $3,000. Do the same for each of the other three religious groups of customers. Discuss the results.
- Let
*m*_{1}and*m*_{2}be the mean amount spent by*married*and*unmarried*customers, respectively.- Construct and interpret a 99% confidence interval for
*m*_{1}. - Construct and interpret a 99% confidence interval for
*m*_{2}. - Based on the results of (a) and (b) above, can you conclude with 99% confidence whether married or unmarried customers spend more on average?

- Construct and interpret a 99% confidence interval for