BA275 Assignment 1

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question
The Commerce Department reported receiving the following applications for the Malcolm Baldrige National Quality Award: 23 from large manufacturing firms, 18 from large service firms, and 30 from small businesses. What percentage of the applications came from small businesses (to 1 decimal)?
answer
The number of small business applied / total number of firms
question
A Bloomberg Businessweek North American subscriber study collected data from a sample of 2861 subscribers. Fifty-nine percent of the respondents indicated an annual income of $75,000 or more, and 50% reported having an American Express credit card. What is the population of interest in this study?
answer
The population of interest of this study is the set of all the poeple of North America who are subscribed to Bloomberg Businessweek.
question
A Bloomberg Businessweek North American subscriber study collected data from a sample of 2861 subscribers. Fifty-nine percent of the respondents indicated an annual income of $75,000 or more, and 50% reported having an American Express credit card. Is annual income a qualitative or quantitative variable?
answer
quantitative
question
A Bloomberg Businessweek North American subscriber study collected data from a sample of 2861 subscribers. Fifty-nine percent of the respondents indicated an annual income of $75,000 or more, and 50% reported having an American Express credit card. Is ownership of an American Express card a qualitative or quantitative variable?
answer
quantitative
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A Bloomberg Businessweek North American subscriber study collected data from a sample of 2861 subscribers. Fifty-nine percent of the respondents indicated an annual income of $75,000 or more, and 50% reported having an American Express credit card. Does this study involve cross-sectional or time series data?
answer
cross-sectional
question
A seven-year medical research study reported that women whose mothers took the drug DES during pregnancy were twice as likely to develop tissue abnormalities that might lead to cancer as were women whose mothers did not take the drug. Do you suppose the data were obtained in a survey or in an experiment?
answer
experiment
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A seven-year medical research study reported that women whose mothers took the drug DES during pregnancy were twice as likely to develop tissue abnormalities that might lead to cancer as were women whose mothers did not take the drug. For the population of women whose mothers took the drug DES during pregnancy, a sample of 3980 women showed that 63 developed tissue abnormalities that might lead to cancer. Provide a descriptive statistic (to 1 decimal) that could be used to estimate the number of women out of 1000 in this population who have tissue abnormalities. ___ out of 1,000 women developed tissue abnormalities
answer
63/3980=.0158 15.8%
question
A seven-year medical research study reported that women whose mothers took the drug DES during pregnancy were twice as likely to develop tissue abnormalities that might lead to cancer as were women whose mothers did not take the drug. For the population of women whose mothers did not take the drug DES during pregnancy, what is the estimate (to 1 decimal) of the number of women out of 1000 who would be expected to have tissue abnormalities? (Hint: remember that women whose mothers took the drug were twice as likely to develop tissue abnormalities.) ___ out of 1,000 women developed tissue abnormalities
answer
15.8/2 7.9
question
A seven-year medical research study reported that women whose mothers took the drug DES during pregnancy were twice as likely to develop tissue abnormalities that might lead to cancer as were women whose mothers did not take the drug. True or false: medical studies often use a relatively large sample (in this case, 3980) because disease occurrences can be rare and difficult to observe when only isolated populations are considered.
answer
true
question
Pew Research Center is a nonpartisan polling organization that provides information about issues, attitudes, and trends shaping America. In a recent poll, Pew researchers found that 47% of American adult respondents reported getting at least some local news on their cell phone or tablet computer (Pew Research website, May 14, 2011). Further findings showed that 42% of respondents who own cell phones or tablet computers use those devices to check local weather reports and 37% use the devices to find local restaurants or other businesses. a. One statistic reported concerned using cell phones or tablet computers for local news. What population is that finding applicable to?
answer
a) 47% of the Americcan adult respondents are applicable to this finding.
question
Pew Research Center is a nonpartisan polling organization that provides information about issues, attitudes, and trends shaping America. In a recent poll, Pew researchers found that 47% of American adult respondents reported getting at least some local news on their cell phone or tablet computer (Pew Research website, May 14, 2011). Further findings showed that 42% of respondents who own cell phones or tablet computers use those devices to check local weather reports and 37% use the devices to find local restaurants or other businesses. Another statistic concerned using cell phones or tablet computers to check local weather reports and to find local restaurants. What population is this finding applicable to?
answer
Proportion of respondants who use devices to check local weather = 42/100 = 0.42 Proportion of respondants who use devices to check local restaurants = 37/100 = 0.37 Proportion of respondants who use devices to check local weather and local restaurants = 0.42 × 0.37 = 0.1554 Percentage of respondants who use devices to check local weather and local restaurants = 15.54%
question
Pew Research Center is a nonpartisan polling organization that provides information about issues, attitudes, and trends shaping America. In a recent poll, Pew researchers found that 47% of American adult respondents reported getting at least some local news on their cell phone or tablet computer (Pew Research website, May 14, 2011). Further findings showed that 42% of respondents who own cell phones or tablet computers use those devices to check local weather reports and 37% use the devices to find local restaurants or other businesses. Do you think the Pew researchers conducted a census or a sample survey to obtain their results? Why? The input in the box below will not be graded, but may be reviewed and considered by your instructor.
answer
Pew researchers conducted a sample survey to obtain the results. Pew researcher conducted a poll, in poll whole population is not covered only the interested people take part in the survey.
question
Pew Research Center is a nonpartisan polling organization that provides information about issues, attitudes, and trends shaping America. In a recent poll, Pew researchers found that 47% of American adult respondents reported getting at least some local news on their cell phone or tablet computer (Pew Research website, May 14, 2011). Further findings showed that 42% of respondents who own cell phones or tablet computers use those devices to check local weather reports and 37% use the devices to find local restaurants or other businesses. If you were a restaurant owner, would you find these results interesting? Why? How could you take advantage of this information? The input in the box below will not be graded, but may be reviewed and considered by your instructor.
answer
If we are restaurent owner these results would be interesting because 37% of the Americam adluts respondants uses the device to find local restaurants and other bussiness. We can take advantage of this information by providing updated details of our restaurants on the online search engines.
question
Consider a sample with data values of 10, 20, 12, 17, and 16. Compute the z-score for each of the five observations (to 2 decimals). Observed value z-score 10 20 12 17 16
answer
mean = (10+20+12+17+16)/5=15 standard deviation = 4 (based on the data) Z-score = (X-mean)/standard deviation Observed Value Z-score 10 (10-15)/4 =-1.25 20 (20-15)/4=1.25 12 (12-15)/4=-0.75 17 (17-15)/4=0.5 16 (16-15)/4=0.25
question
Consider a sample with data values of 10, 20, 12, 17, and 16. Compute the median Compute the mean.
answer
15, 16
question
The national average for the math portion of the College Board's Scholastic Aptitude Test (SAT) is 515 (The World Almanac, 2009). The College Board periodically rescales the test scores such that the standard deviation is approximately 100. Answer the following questions using a bell-shaped distribution and the empirical rule for the verbal test scores. a. What percentage of students have an SAT verbal score greater than 615?
answer
Gievn information: mu=515,sigma=100 (a) Here 615 is one standard deviation above mean so mu+sigma=515+100=615 So according to empirical rule percentage of scores between 515 and 615 is 68%/2 = 34%. So percentage of students have an SAT math score greater than 615 is 50% - 34% = 16%
question
The national average for the math portion of the College Board's Scholastic Aptitude Test (SAT) is 515 (The World Almanac, 2009). The College Board periodically rescales the test scores such that the standard deviation is approximately 100. Answer the following questions using a bell-shaped distribution and the empirical rule for the verbal test scores. What percentage of students have an SAT verbal score greater than 715? (to 1 decimal)
answer
(b) Here 615 is two standard deviations above mean so mu+2sigma=515+200=715 So according to empirical rule percentage of scores between 515 and 715 is 95%/2 = 47.5%. So percentage of students have an SAT math score greater than 715 is 50% - 47.5% = 2.5%
question
The national average for the math portion of the College Board's Scholastic Aptitude Test (SAT) is 515 (The World Almanac, 2009). The College Board periodically rescales the test scores such that the standard deviation is approximately 100. Answer the following questions using a bell-shaped distribution and the empirical rule for the verbal test scores. What percentage of students have an SAT verbal score between 415 and 515?
answer
Here 415 is one standard deviation below mean so mu-sigma=515-100=415 So according to empirical rule percentage of scores between 415 and 515 is 68%/2 = 34%.
question
The national average for the math portion of the College Board's Scholastic Aptitude Test (SAT) is 515 (The World Almanac, 2009). The College Board periodically rescales the test scores such that the standard deviation is approximately 100. Answer the following questions using a bell-shaped distribution and the empirical rule for the verbal test scores. What percentage of students have an SAT verbal score between 315 and 615? (to 1 decimal)
answer
Here 615 is one standard deviation above mean so mu+sigma=515+100=615 So according to empirical rule percentage of scores between 515 and 615 is 68%/2 = 34%. Here 315 is two standard deviations below mean so mu-2sigma=515-200=315 So according to empirical rule percentage of scores between 315 and 515 is 95%/2 = 47.5%. So percentage of students have an SAT math score between 315 and 615 is 34% +47.5% = 81.5%
question
Given that z is a standard normal random variable, compute the following probabilities (to 4 decimals). P(-1.98<=z<=.49) P(.52<=z<=1.22) P(-1.75<=z<=-1.04)
answer
.6879-.0239=0.664 .6985-.8888=0.1903 .1492-.0401=0.1091
question
Given that z is a standard normal random variable, find z for each situation (to 2 decimals). The area to the left of z is .2119.
answer
We get the z score from the given left tailed area. As Left tailed area = 0.2119 Then, using table or technology, z = -0.799846048 [ANSWER] ***************
question
Given that z is a standard normal random variable, find z for each situation (to 2 decimals). The area between -z and z is .9030
answer
b) Getting the outward areas of the two extreme tails, alpha/2 = (1-middle area)/2 = 0.0485 Thus, the z values bounding these tails, using table/technology, -z = lower z value = -1.659574906 z = upper z value = 1.659574906 [ANSWER] ***************
question
Given that z is a standard normal random variable, find z for each situation (to 2 decimals). The area between -z and z is .2052.
answer
Getting the outward areas of the two extreme tails, alpha/2 = (1-middle area)/2 = 0.3974 Thus, the z values bounding these tails, using table/technology, -z = lower z value = -0.260082677 z = upper z value = 0.260082677 [ANSWER] ****************
question
Given that z is a standard normal random variable, find z for each situation (to 2 decimals). The area to the left of z is .9948.
answer
d) We get the z score from the given left tailed area. As Left tailed area = 0.9948 Then, using table or technology, z = 2.562238076 [ANSWER] ******************
question
Given that z is a standard normal random variable, find z for each situation (to 2 decimals). The area to the right of z is .6915.
answer
e) We get the z score from the given left tailed area. As Left tailed area = 0.3085 Then, using table or technology, z = -0.500106627 [ANSWER] *********************
question
The mean cost of domestic airfares in the United States rose to an all-time high of $385 per ticket (Bureau of Transportation Statistics website, November 2, 2012). Airfares were based on the total ticket value, which consisted of the price charged by the airlines plus any additional taxes and fees. Assume domestic airfares are normally distributed with a standard deviation of $110. a. What is the probability that a domestic airfare is $550 or more (to 4 decimals)?
answer
a) P(X => 550) = (550-385)/110 = 165/110 = 1.5 = P ( Z >1.5) From Standard Normal Table = 0.0668
question
The mean cost of domestic airfares in the United States rose to an all-time high of $385 per ticket (Bureau of Transportation Statistics website, November 2, 2012). Airfares were based on the total ticket value, which consisted of the price charged by the airlines plus any additional taxes and fees. Assume domestic airfares are normally distributed with a standard deviation of $110. What is the probability than a domestic airfare is $250 or less (to 4 decimals)?
answer
b) P(X <= 250) = (250-385)/110 = -135/110= -1.2273 = P ( Z <-1.2273) From Standard Normal Table = 0.1099
question
The mean cost of domestic airfares in the United States rose to an all-time high of $385 per ticket (Bureau of Transportation Statistics website, November 2, 2012). Airfares were based on the total ticket value, which consisted of the price charged by the airlines plus any additional taxes and fees. Assume domestic airfares are normally distributed with a standard deviation of $110. What if the probability that a domestic airfare is between $300 and $500 (to 4 decimals)?
answer
c) To find P(a < = Z < = b) = F(b) - F(a) P(X < 300) = (300-385)/110 = -85/110 = -0.7727 = P ( Z <-0.7727) From Standard Normal Table = 0.21984 P(X < 500) = (500-385)/110 = 115/110 = 1.0455 = P ( Z <1.0455) From Standard Normal Table = 0.85209 P(300 < X < 500) = 0.85209-0.21984 = 0.6323
question
The mean cost of domestic airfares in the United States rose to an all-time high of $385 per ticket (Bureau of Transportation Statistics website, November 2, 2012). Airfares were based on the total ticket value, which consisted of the price charged by the airlines plus any additional taxes and fees. Assume domestic airfares are normally distributed with a standard deviation of $110. What is the cost for the 3% highest domestic airfares (to a whole number)?
answer
d) P ( Z > x ) = 0.03 Value of z to the cumulative probability of 0.03 from normal table is 1.88 P( x-u/ (s.d) > x - 385/110) = 0.03 That is, ( x - 385/110) = 1.88 --> x = 1.88 * 110+385 = 591.91
question
The average return for large-cap domestic stock funds over the three years 2009-2011 was 14.4% (AAII Journal, February, 2012). Assume the three-year returns were normally distributed across funds with a standard deviation of 4.4%. a. What is the probability an individual large-cap domestic stock fund had a three-year return of at least 20% (to 4 decimals)?
answer
Mean = u = 14.4 SD = 4.4 a) Atleast 20 : z = (x - u) / SD z = (20 - 14.4) / 4.4 z = 1.27272727272 P(z > 1.2727272727272) = 0.1016
question
The average return for large-cap domestic stock funds over the three years 2009-2011 was 14.4% (AAII Journal, February, 2012). Assume the three-year returns were normally distributed across funds with a standard deviation of 4.4%. b. What is the probability an individual large-cap domestic stock fund had a three-year return of 10% or less (to 4 decimals)?
answer
b) Less than or equal to 10 : x = 10 z = (x - u) / SD z = (10 - 14.4) / 4.4 z = -1 P(z <= -1) = 0.1587
question
The average return for large-cap domestic stock funds over the three years 2009-2011 was 14.4% (AAII Journal, February, 2012). Assume the three-year returns were normally distributed across funds with a standard deviation of 4.4%. c. How big does the return have to be to put a domestic stock fund in the top 10% for the three-year period (to 2 decimals)?
answer
Top 10% : P = 0.90 From the table, for P = 0.90, z = 1.281551 z = (x - u) / SD 1.281551 = (x - 14.4) / 4.4 Crossmultiplying : 5.6388244 = x - 14.4 x = 5.6388244 + 14.4 x = 20.0388244 x = 20.04 So, it has to be atleast 20.04%
question
New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $204 per night (USA Today, April 30, 2012). Assume that room rates are normally distributed with a standard deviation of $55. a. What is the probability that a hotel room costs $225 or more per night (to 4 decimals)?
answer
P(xx>225)=P(z>225-204/55)=P(z>0.38)=0.352
question
New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $204 per night (USA Today, April 30, 2012). Assume that room rates are normally distributed with a standard deviation of $55. b. What is the probability that a hotel room costs less than $140 per night (to 4 decimals)?
answer
P(x<140)=P(z<140-204/55)=P(z<-1.16)=0.123
question
New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $204 per night (USA Today, April 30, 2012). Assume that room rates are normally distributed with a standard deviation of $55. c. What is the probability that a hotel room costs between $200 and $300 per night (to 4 decimals)?
answer
P(200<x<300)=P(-0.07<z<1.75)=0.4878
question
New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $204 per night (USA Today, April 30, 2012). Assume that room rates are normally distributed with a standard deviation of $55. d. What is the cost of the 20% most expensive hotel rooms in New York City? Round up to the next dollar.
answer
P(Z>z)=0.20 z=0.841 x-204/55=0.841 x=250.26
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