stat2160

To design a new advertising campaign, Ford Motor Company would like to estimate the proportion of drivers of the new Ford Fusion that are women. In a random sample of 97 Fusion owners, 50 of them were women. What is the 99% confidence interval estimating the proportion of all drivers who are women?

( 0.38476 , 0.64617 )

Professors in the Economics Department at Western want to determine how challenging the program was for students. Out of a random sample of 20 students, 17 indicated that the program was either "challenging" or "very challenging". The 99% confidence interval estimating the proportion of all students in the department who thought the program was challenging is given by which of the following?

( 0.64434 , 1.05566 )*** closest answer i got .6440035194, 1.055996481 same as above but take p hat and add and subtract the moe (find like above)

The owner of a local golf course wants to estimate the difference between the average ages of males and females that play on the golf course. He randomly samples 12 men and 11 women that play on his course. He finds the average age of the men to be 34.65 with a standard deviation of 7.75. The average age of the women was 33.83 with a standard deviation of 11.47. Construct a 99% confidence interval to estimate the difference of (average age of men - average age of women). Assume the population standard deviations are the same.

(-10.65, 12.29)

A restaurant wants to test a new in-store marketing scheme in a small number of stores before rolling it out nationwide. The new ad promotes a premium drink that they want to increase the sales of. 5 locations are chosen at random and the number of drinks sold are recorded for 2 months before the new ad campaign and 2 months after. The average difference in the sales quantity (after - before) is -28.376 with a standard deviation of 42.6429. Calculate a 99% confidence interval to estimate the true average difference in nationwide sales quantity before the ad campaign and after.

(-116.1783, 59.4263) * calc in radian mode? mod 7.3.2 w7 xbard (mean) +- t*(sd (std dev/err)/sqr rt n) -28.376 +- t* (42.6429/sqrt 5) t= invT (.995, n-1) -28.376+- 87.8023204

**A new drug to treat high cholesterol is being tested by pharmaceutical company. The cholesterol levels for 8 patients were recorded before administering the drug and after. The mean difference in total cholesterol levels (after - before) was 15.539 mg/dL with a standard deviation of 19.2612 mg/dL. Create a 90% confidence interval for the true average difference in cholesterol levels by the drug.

(-2.6372, 28.4408) WRONG 2.6372, 28.4408 RIGHT chap 7.3.2?

The owner of a local phone store wanted to determine how much customers are willing to spend on the purchase of a new phone. In a random sample of 10 phones purchased that day, the sample mean was $340.833 and the standard deviation was $23.555. Calculate a 90% confidence interval to estimate the average price customers are willing to pay per phone.

(327.179, 354.487)

The owner of a local phone store wanted to determine how much customers are willing to spend on the purchase of a new phone. In a random sample of 13 phones purchased that day, the sample mean was $393.593 and the standard deviation was $22.6101. Calculate a 99% confidence interval to estimate the average price customers are willing to pay per phone.

(374.438, 412.748)

Suppose that the middle 68% of average scores in the NBA per player per game fall between 9.95 and 23.55. Give an approximate estimate of the standard deviation of the number of the points scored. Assume the points scored has a normal distribution.

(μ-σ,μ+σ) = (9.95, 23.55) => μ - σ = 9.95 .............(1) => μ + σ = 23.55 .............(2) Subtracting equation (1) from equation (2), we get: (μ + σ) - (μ - σ) = 23.55 - 9.95 => 2σ = 13.6 => σ = 6.8 [ANSWER]

When rolling a die 195 times, what is the probability of rolling a 6 between 30 and 33 (inclusively) times?

**.2970 (binomcdf n, p= 1/6= 1.6667, k=33) - (binomcdf n, 1.6667, j-1= 29) = answer

Suppose that the distribution of income in a certain tax bracket is approximately normal with a mean of $50,226.6 and a standard deviation of $4,548.574. Approximately 4.59% of households had an income less than what dollar amount? 57,895.4 34,932,320 42,557.8 34,831,867 not enough info

**57,895.4 WRONG?? 100-4.59= 95.41 put in area .9541 2, vars, invNorm= 57895.39656 if i put in.459 for area, my asnwer is 49758.30914

Ten students took a statistics final and their scores were 84.9; 80.1; 78.7; 77; 75.7; 79.1; 80.6; 79.4; 73.8; 79.2. Calculate the coefficient of variation of the dataset.

.038

Pinterest claims that 0.3563 of their app users are men. In a sample of 71 randomly chosen app users, what is the probability that exactly 21 of them will be men?

.0576 (l5) 2, vars, binompdf, 71, .3563, 21=

Ten students took a statistics final and their scores were 71.9; 85; 72.1; 71; 80.5; 77.7; 88.8; 80.8; 73.9; 82.5. Calculate the coefficient of variation of the dataset.

.078

The Student Recreation Center wanted to determine what sort of physical activity was preferred by students. In a survey of 84 students, 59 indicated that they preferred outdoor exercise over exercising in a gym. When estimating the proportion of all students at the university who prefer outdoor exercise with 90% confidence, what is the margin of error

.0818*** closest answer to what i got i got .0820621667? mod 7.2, w8, lect #1 std error= sq rt p hat * (1-phat)/n p hat= 59/84= .7023809584 se= sqrt of .00249964806 moe= se*z score bc talking in props 90%= z score of 1.645 se* 1.645= .0820621667

Pinterest claims that 0.3382 of their app users are men. In a sample of 66 randomly chosen app users, what is the probability that between 27 and 30 (inclusively) of them will be men?

.1206

Suppose that the probability of a baseball player getting a hit in an at-bat is 0.3153. If the player has 28 at-bats during a week, what's the probability that he gets exactly 7 hits?

.1288 binompdf

The stock price for International Business Machines (IBM) historically has followed an approximately normal distribution (when adjusting for inflation) with a mean of $189.218 and standard deviation of $4.868. What is the probability that on a selected day the stock price is below $184.14?

.1484 2, vars, normcdf

The stock price for International Business Machines (IBM) historically has followed an approximately normal distribution (when adjusting for inflation) with a mean of $185.288 and standard deviation of $3.5141. What is the probability that on a selected day the stock price is between $185.3 and 186.81?

.1662

Suppose that the probability of a baseball player getting a hit in an at-bat is 0.3089. If the player has 25 at-bats during a week, what's the probability that he gets greater than 9 hits?

.2176 2, vars, binomcdf, 25, .3089, 9= .7823639051 1-above= answer

Suppose that the mean and standard deviation of the scores on a statistics exam are 75.3 and 6.2, respectively, and are approximately normally distributed. Calculate the proportion of scores above 77.

.3920 2, vars, normcdf, 77, e99, mean, sd=

Suppose that the number of gallons of milk sold per day at a local supermarket are normally distributed with mean and standard deviation of 486.9 and 24.01, respectively. What is the probability that on a given day the supermarket will sell between 477 and 525 gallons of milk?

.6037

If the scores per round of golfers on the PGA tour are approximately normally distributed with mean 61.8 and standard deviation 2.17, what is the probability that a randomly chosen golfer's score is between 59 and 63 strokes?

.6114

hw3 attempt 2- If the scores per round of golfers on the PGA tour are approximately normally distributed with mean 69.7 and standard deviation 1.8, what is the probability that a randomly chosen golfer's score is between 68 and 72 strokes?

.7269 2, vars, normcdf

Red Bull is the most popular energy drink in sales in the United States. Red Bull GmbH (the parent company) has observed that daily sales are normally distributed with an average of 7,437,952 drinks sold with a standard deviation of 8,512.06. What is the probability that on a given day above 7,431,820 drinks are sold?

.7643

Pandora Radio has determined that the probability of a user liking a new song enough to give a "thumbs up" is 0.7729. Given that Pandora plays about 587 new songs per user per month on average, what is the probability that at least 445 of those plays get a "thumbs up"?

.8179

The five-number summary for a dataset on Inches of Rain Per Year is given in the table below. Using this information, calculate the percent of values below 1.31. min 1.31, q1 5.47, med 9.72, q3 12.06, max 19.71

0%

hw3- (l4 and 5) The stock price for International Business Machines (IBM) historically has followed an approximately normal distribution (when adjusting for inflation) with a mean of $154.463 and standard deviation of $3.1171. What is the probability that on a selected day the stock price is below $152.08?

0.2223 2 vars, normal cdf, -e99, 152.08, mean, sd=

A professor at a university wants to estimate the average number of hours of sleep students get during exam week. The professor wants to find a sample mean that is within 1.362 hours of the true average for all college students at the university with 99% confidence. If the professor knows the standard deviation of the sleep times for all college students is 5.368, what sample size will need to be taken?

104

Suppose that the average and standard deviation of the number of points scored in an NBA game per player are 18.51 and 6.71, respectively. Calculate an interval that is symmetric around the mean such that it contains approximately 68% of players scores. Assume that the points scored has a normal distribution.

11.8, 25.22

Fill in the blanks. Ivy League colleges are known for their high level of denials to prospective students, especially for graduate school. In fact, the raw probability of admission to a graduate program is around 0.07. Assuming each student is independent and 18,762 students apply, about __________ students, give or take __________, will be accepted.

1313.34, 34.949 n*p= mean q= 1-p sd= sq root of (n*p*q)

A factory has added a new incentive program to encourage employees to use payroll deductions to contribute to their 401K retirement account. The company will now contribute funds based on the employee deduction amount. Before the program, employees contributed on average $108.503 with standard deviation of $7.975. To calculate the new contribution amount (employee and company combined), every observation in the dataset is multiplied by 1.264. What will the new mean be?

137.15

The daily stock price for International Business Machines (IBM) historically has followed an approximately normal distribution (when adjusting for inflation) with a mean of $138.918 and standard deviation of $2.8408 Approximately 35.53% of days IBM had a stock price less than what dollar amount?

137.86

The site traffic for a business's website was monitored for 10 days, showing daily traffic of 14,782; 14,801; 15,264; 15,907; 14,794; 15,079; 15,439; 15,464; 15,580; 14,774. Calculate the mean of the dataset.

15188.4

A certain list of movies were chosen from lists of recent Academy Award Best Picture winners, highest grossing movies, series movies (e.g. the Harry Potter series, the Spiderman series), and from the Sundance Film Festival and are being analyzed. The mean box office gross was $147.653 million with a standard deviation of $11.2004 million. Given this information, 93.41% of movies grossed less than how much money (in millions)? Assume the distribution is approximately normal.

164.53 2, vars, invNorm, .9341, mean, sd=

Suppose that 9 people lost an average of 14.84 pounds over the course of the first month of being on a new diet. If another person begins the diet and the average weight lost after one month for all 10 people is 15.31, how much weight did the last person lose?

19.5

Suppose that the middle 95% of a certain kind of battery have a shelf life between 6.09 and 14.53 years. Give an approximate estimate of the standard deviation of the shelf life. Assume shelf life has a normal distribution.

2.11 ***interval is 4 std devs wide 14.53-6.09=8.44 8.44/4=2.11

Fill in the blanks. According to estimates by the office of the Treasury Inspector General of IRS, approximately 0.07 of the tax returns filed are fraudulent or will contain errors that are purposely made to cheat the IRS. In a random sample of 296 independent returns from this year, around __________ returns, give or take __________, will be fraudulent or will contain errors that are purposely made to cheat the IRS.

20.72, 4.39

Fill in the blanks. According to estimates by the office of the Treasury Inspector General of IRS, approximately 0.07 of the tax returns filed are fraudulent or will contain errors that are purposely made to cheat the IRS. In a random sample of 332 independent returns from this year, around __________ returns, give or take __________, will be fraudulent or will contain errors that are purposely made to cheat the IRS.

23.24, 4.649 mean= n*p= 332*.07 =23.24 sd= n*p*q q= 1-p= 1-.07= .93 .93*23.24= 21.6132 !!!!! SQUARE ROOT of 21.6132= answer

hw5- An analysis of 43 Wall Street traders showed that 29 of their stock picks beat the market average. What is the estimate of the population proportion? What is the standard error of this estimate? n=43

29/43= .6744186047 std error= sr root of (p*(1-p)/n) .6744186047*(1- .6744186047)= .6744186047*.3255813953= .2195781503 .2195781503/n= .0051064686 sqrt (.0051064686)= .0714595592 est prop .674 std error .0715

The average domestic economy car gets around 28.4 miles per gallon (MPG) with a standard deviation of 2.65. Suppose domestic manufacturers all vow to increase fuel economy over the next 2 years. If successesful, 3.4 is added to every observation in the dataset. What is the new mean?

31.8

The owner of a golf course wants to determine if his golf course is more difficult than the one his friend owns. He has 21 golfers play a round of 18 holes on his golf course and records their scores. Later that week, he has the same 21 golfers play a round of golf on his friend's course and records their scores again. The average difference in the scores (treated as the scores on his course - the scores on his friend's course) is 4.987 and the standard deviation of the differences is 6.828. If he calculates a 99% confidence interval to estimate the average difference in scores for all golfers, what is the margin of error of the interval?

4.2395

**hw6- (w& out of order see notes) You own a small storefront retail business and are interested in determining the average amount of money a typical customer spends per visit to your store. You take a random sample over the course of a month for 24 customers and find that the average dollar amount spent per transaction per customer is $92.963 with a standard deviation of $12.5982. When creating a 90% confidence interval for the true average dollar amount spend per customer, what is the margin of error?

4.4074

Suppose that the distribution of income in a certain tax bracket is approximately normal with a mean of $44,831.41 and a standard deviation of $3,139.062. Approximately 56.82% of households had an income GREATER than what dollar amount?

45,370.68*** INSTEAD of 2nd VARS invNorm .5682, 44831.41, 3139.062 YES 1-.5682= .4318 .4318, 44831.41, 3139.062

The five-number summary for a dataset on Temperature is given in the table below. Using this information, calculate the percent of values below 56.1. min 10.6, q1 23.7, med 56.1, q3 83.4, max 103.1

50%

Suppose that the distribution of income in a certain tax bracket is approximately normal with a mean of $53,114.38 and a standard deviation of $1,350.89. Approximately 2.87% of households had an income greater than what dollar amount?

55, 681.42 100-2.87, put in area 2, vars, invNorm

Assume that the monthly food expenditures of families of four in one demographic are approximately normally distributed with mean $637.46 and standard deviation $82.23. Calculate an interval that is symmetric around the mean such that it contains approximately 95% of expenditures.

555.23, 719.69 **WRONG take + and - (82.23*2) from the mean

Suppose that the distribution of income in a certain tax bracket is approximately normal with a mean of $59,194.86 and a standard deviation of $4,454.165. Approximately 22.1% of households had an income greater than what dollar amount? 15,312,273 55,770.4 not enough info 15,193,883 62619.31

55770.4 isWRONG* 62619.31247 COrr :) stat vars invNorm (100-22.1= 77.9) .779, mean, std dev = corr

The average height of an adult male is 60.45" with a standard deviation of 1.02". Calculate an interval that is symmetric around the mean such that it contains approximately 95% of heights. Assume that height has a normal distribution.

58.41, 62.49

You are in the market for a new car. You want to check whether there is a significant difference between the fuel economy of mid-size domestic cars and mid-size import cars. You sample 17 domestic car makes and find an average fuel economy of 31.535 MPG with a standard deviation of 3.079 MPG. For imports, you sample 10 cars and find an average MPG of 31.123 MPG with a standard deviation of 8.224. If a 99% confidence interval is calculated to estimate the difference between the average fuel economy of domestic and import mid-size cars, what is the MARGIN OF ERROR? Assume both population standard deviations are equal.

6.126

The site traffic for a business's website was monitored for 10 days, showing daily traffic of 15,101; 15,265; 15,362; 15,206; 14,669; 15,129; 14,015; 16,341; 16,259; 15,919. Calculate the standard deviation of the dataset.

708.02

The five-number summary for a dataset on Exam Scores is given in the table below. Using this information, calculate the percent of values below 75.1. min 37.6, q1 57.3, med 61.1, q3 75.1, max 96.5

75%

A sample proportion is calculated from a sample size of 68. How large of a sample would we need in order to decrease the standard error by a factor of 7?

7^2= 49 68*49= 3332

hw9 **- week 11 on chap (based on w10 later lectures of chap 8) It is reported in USA Today that the average flight cost nationwide is $350.11. You have never paid close to that amount and you want to perform a hypothesis test that the true average is actually greater than $350.11. What are the appropriate hypotheses for this test?

HO: μ ≤ 350.11 HA: μ > 350.11

Does the average internet speed of an ISP depend on continent? Specifically, you would like to test whether customers in North America have an average internet download speed that is greater than the average download speed of customers in Europe. If North American customers are in group 1 and European customers are in group 2, what are the hypotheses for your test of interest?

HO: μ1 ≤ μ2 HA: μ1 > μ2

Suppose you work for a political pollster during an election year. You are tasked with determining the projected winner of the November election. That is, you wish to determine if the number of votes for Candidate 1 is greater than the votes for Candidate 2. What are the hypotheses for this test?

HO: μ1 ≤ μ2 HA: μ1 > μ2 *hint: look at hA what you WANT to happen. h0 is opp of that, if nothing happens

A USA Today article claims that the proportion of people who believe global warming is a serious issue is 0.58, but given the number of people you've talked to about this same issue, you believe it is greater than 0.58. The hypotheses for this test are Null Hypothesis: p ≤ 0.58, Alternative Hypothesis: p > 0.58. If you randomly sample 27 people and 18 of them believe that global warming is a serious issue, what is your test statistic and p-value?

Test stat: .912, p val: .181 proportion= 2nd link, w9, 8.5 find p hat and test stat z= (phat-p0)/sqrt of (p0*(1-p0))/n) phat= 18/27= .66666 p0= .58 n= 27 (.666-.58)/sqrt (.58*(1-.58)/27 .086666/sqrt of .00902222= test stat= .9124204273 find p val (if greater than sign in hypA, shade to right so # all the way to e99, if less than -E99 then #,, IF EQUAL pick one and times by 2)**** normcdf(.9124204273, E99, 0, 1)= .180773691

In a packing plant, one of the machines packs jars into a box. A sales rep for a packing machine manufacturer comes into the plant saying that a new machine he is selling will pack the jars faster than the old machine. To test this claim, each machine is timed for how long it takes to pack 10 cartons of jars at randomly chosen times. Given a 95% confidence interval of (-2.48, 2.54) for the true difference in average times to pack the jars (old machine - new machine), what can you conclude from this interval?

There is no significant difference between the average packing times of the two machines. The sales rep does not appear to be telling the truth. (can only tell if both are pos or both neg)

***hw7- (mod8) You are tracking Chipotle Mexican Grill's stock price, attempting to figure out a good price point for entry into the market. You look at 35 random days over the course of a few months and see that the closing price has an average of $358.71 with a standard deviation of $21.883. You construct a 90% confidence interval for the average stock price to be (352.46, 364.96). Of the following choices, what is the appropriate interpretation of this interval?

We are 90% confident the AVERAGE daily closing price for ANY day is between $352.46 and $364.96

A new drug to treat high cholesterol is being tested by pharmaceutical company. The cholesterol levels for 20 patients were recorded before administering the drug and after. The average difference in cholesterol levels (after - before) was -0.52 mg/dL with a standard deviation of 6.16 mg/dL. Using this information, the calculated 95% confidence paired-t interval is (-3.403, 2.363). Which of the following is the best interpretation of this interval?

We are 95% confident that the AVG DIFF in cholesterol levels between those who would take the drug and those who would not is between -3.403 and 2.363. (NOT just the ppl in the survey)

You are interested in getting an investment portfolio started with any extra money you make from your part time job while also going to school. While flipping through the latest edition of Money magazine, you read an article that of a survey of magazine subscribers, 194 were randomly selected and analyzed. A 95% confidence interval was constructed for the proportion of all subscribers who made money in the previous year in their investments, which was ( 0.7255 , 0.8415 ). What is the correct interpretation of this confidence interval?

We are 95% confident that the proportion of ALL Money magazine subscribers that made money in the previous year from their investments is between 0.7255 and 0.8415.

The Student Recreation Center wanted to determine what sort of physical activity was preferred by students. In a survey of 50 students, 33 indicated that they preferred outdoor exercise over exercising in a gym. They estimated the proportion of students at the university who prefer outdoor exercise as ( 0.5287 , 0.7913 ), with 95% confidence. Which of the following is an appropriate interpretation of this confidence interval?

We are 95% confident that the proportion of ALL students AT the university who prefer outdoor exercise is between 0.5287 and 0.7913.

Suppose you work for a pharmaceutical company that is testing a new drug to treat ulcerative colitis. You conduct a random sample of 25 patients on the new drug and 9 of them indicated a decrease in symptoms since starting the drug regimen. You construct a 99% confidence interval for this proportion to be ( 0.1123 , 0.6077 ). What is the correct interpretation of this confidence interval?

We are 99% confident that the proportion of all patients on the drug regimen that have experienced a reduction in symptoms is between 0.1123 and 0.6077. i chose cant determine proper interp bc 9/25= .36 which is in the range. FEEDBACK= we can always interp conf interval

You work for a pharmaceutical company that is testing a new cholesterol drug. The proportion of patients on the previous drug who had a positive treatment was 0.341. You want to see if the new drug is more effective than the previous drug. You conduct a sample of 86 patients on the new drug and find that 51 have had a positive treatment effect. The 95% confidence interval is ( 0.4892 , 0.6969 ). What is the best conclusion you can make of those listed below?

We can claim that the proportion of patients who see a positive treatment effect with the new drug is larger than 0.341. bc .341 isnt in the range, and both are positive

Based on past data, the producers of Ice Mountain bottled water knew that the proportion of people who preferred Ice Mountain to tap water was 0.706. To see how consumer perception of their product changed, they decided to conduct a survey. Of the 139 respondents, 122 indicated that they preferred Ice Mountain to the tap water in their homes. The 90% confidence interval for this proportion is ( 0.832 , 0.9234 ). What is the best conclusion of those listed below?

We can conclude that the proportion of consumers who prefer Ice Mountain to their tap water is larger than 0.706. bc both positive and .706 not in the range

A medical researcher wants to determine if the average hospital stay of patients that undergo a certain procedure is less than 7.6 days. The hypotheses for this scenario are as follows: Null Hypothesis: μ ≥ 7.6, Alternative Hypothesis: μ < 7.6. If the researcher takes a random sample of patients and calculates a p-value of 0.5014 based on the data, what is the appropriate conclusion? Conclude at the 5% level of significance.

We did not find enough evidence to say the true average hospital stay of patients is shorter than 7.6 days.

hw4- For each of the following variables, identify the type of variable (categorical vs. numeric).(I) Local name for soft drinks (e.g. Soda, Pop, Coke, etc)(II) Satisfaction level on a survey (e.g. very dissatisfied, satisfied, very satisfied, etc.) num, num, cat, num num, cat no corr match cat, cat

cat and cat

(hw10) You own a small storefront retail business and are interested in determining the average amount of money a typical customer spends per visit to your store. You take a random sample over the course of a month for 15 customers and find that the average dollar amount spent per transaction per customer is $96.914 with a standard deviation of $19.5505. Create a 90% confidence interval for the true average spent for all customers per transaction.

chap 7.1? inVT(.95, n-1= 14)= 1.761310111= t* xbar +- (t* * s/sqrtn) 96.914 +- (1.761310111 * 19.5505/sqrt15) 96.914 +- (1.761310111 * 5.047917394) 96.914-8.890947946= 88.02305205 96.914+8.890947946= 105.8049479 (88.023, 105.805)

(hw10) A student at a university wants to determine if the proportion of students that use iPhones is greater than 0.46. The hypotheses for this scenario are as follows. Null Hypothesis: p ≤ 0.46, Alternative Hypothesis: p > 0.46. If the student randomly samples 22 other students and finds that 11 of them use iPhones, what is the test statistic and p-value?

chap 8.5 see card below Test Statistic: 0.376, P-Value: 0.353

In the year 2000, the average car had a fuel economy of 24.1 MPG. You are curious as to whether the average in the present day is DIFFERENT from the historical value. What are the appropriate hypotheses for this test?

h0: mu = 24.1 ha: mu NO = 24.1 different means ANY change not just greater or less

Consumers Energy states that the average electric bill across the state is $60.07. You want to test the claim that the average bill amount is actually greater than $60.07. What are the appropriate hypotheses for this test?

ho: mu is less than or equal to 60.07 ha: mu is greater than 60.07

A student at a university wants to determine if the proportion of students that use iPhones is different from 0.37. If the student conducts a hypothesis test, what will the null and alternative hypotheses be?

ho= p = .37 ha= p does NOT equal .37

**hw8- week 9 c8 intro and 8.5 A USA Today article claims that the proportion of people who believe global warming is a serious issue is 0.8, but given the number of people you've talked to about this same issue, you believe it is less than 0.8. If you conduct a hypothesis test, what will the null and alternative hypotheses be?

ho= p greater than or equal to .8 ha= p is less than .8

choose descrip which best describes shape of boxplot (boxes on right, dots on left)

left skewed

he closing price of the Standard and Poor 500 Stock Covered Call Index is being recorded and put in a dataset each day. Suppose that the average index price is $2,421.4. If the distribution of the index price is right-skewed (e.g. in the plot below), will the median be larger, smaller, or approximately equal to $2,421.4? most of data is on L side

median under 2421.4

Suppose that one-way commute times in a particular city are normally distributed with a mean of 26.33 minutes and a standard deviation of 1.951 minutes. Would it be unusual for a commute time to be above 22 minutes?

not unusual s, vars, normcdf, 22, e99, mean, sd=

the revenue of 200 companies is plotted and found to follow a bell curve. The mean is $554.01 million with a standard deviation of $33.3188 million. Would it be unusual for a randomly selected company to have a revenue between $387.63 and 576.16 million?

not unusual s, vars, normcdf, 387.63, 576.16, mean, sd=

A medical researcher wants to determine if the average hospital stay after a certain procedure is greater than 10.7 days. The hypotheses for this scenario are as follows: Null Hypothesis: μ ≤ 10.7, Alternative Hypothesis: μ > 10.7. If the researcher randomly samples 20 patients that underwent the procedure and determines their average hospital stay was 12.13 days with a standard deviation of 5.427 days, what is the test statistic and p-value of this test?

same as above ts: 1.178, p val: .1266 12.13-10.7/5.427/sqrt20 1.43/1.213514091= 1.178395876 hypa greater tcdf( 1.17...above, e99, 19)= .1265935596

157 employees of your firm were asked about their job satisfaction. Out of the 157, 78 said they were unsatisfied. What is the estimate of the population proportion? What is the standard error of this estimate?

same process as above est pop= .497 std error= .0399

??Historically, the average score of PGA golfers for one round is 65.9 with a standard deviation of 3.56. A random sample of 104 golfers is taken. What is the probability that the sample mean is greater than 65.6?

same steps as above 3.56/sqrt(104)= 3.56/10.19803903= std error= .3490867204 normcdf(65.5, e99, 65.9, .3490867204)= .8740707329 closest answer .8049??

**??Approximately 40.39% of all businesses are owned by women. If you take a sample of 115 businesses in Michigan, what is the probability that greater than 42.2% of them would be owned by women?

same steps as above std err= .047559493 .3517590109 closest answer .3462?

You are contacted by a phone-in technical support business that is interested in some information about the amount of time their customers spend on hold. You find out that on average, each caller spends 11.72 minutes on hold with a standard deviation of 2.03 minutes. If you were to take a random sample of 57 callers, you would expect 55% of the time the average hold time would be less than how many minutes?

same steps as above std error= .2688800685 11.75378783 11.75

Fill in the blank. In a drive thru performance study, the average service time for McDonald's is 203.21 seconds with a standard deviation of 5.67 seconds. A random sample of 90 times is taken. There is a 51% chance that the average drive-thru service time is less than ________ seconds.

std dev/sqrt(n) std error= .5976704778 invNorm(.51, 203.21, .5976704778)= 203.2249829 203.22

(similar to hw10)Approximately 43.65% of all businesses are owned by women. If you take a sample of 180 businesses in Michigan, what is the probability that greater than 46.62% of them would be owned by women? p=.4365 n=180

std error= sqrt ((p*(1-p))/n) std error= sqrt (.0013664875)= normcdf(.4662, e99, .4365, .036966037) .2109

Suppose you are interested in measuring the amount of time, on average, it takes you to make your commute to school. You've estimated that the average time is 38.23 minutes with a standard deviation of 6.642 minutes. Assuming that your estimated parameters are correct and the commute time is normally distributed, what is the probability that the average commute time of 11 random days is less than 39.67 minutes?

std error= std dev/sqrt of n 6.642/sq rt(11) 6.642/3.31662479= 2.002638351 normcdf (-E99, 39.67, 38.23, 2.002638351)= .7639454632 .7639

For each of the following variables, identify an appropriate graph or chart that could be used.(I) Duration (in minutes) of a flight between two locations(II) Year of high school graduation bar graph, bar g pie, stem and leaf stem and leaf, pie pie, bar boxplot, histogram

stem and leaf, pie*** WRONG box, histo CORR

Suppose the national average dollar amount for an automobile insurance claim is $501.294. You work for an agency in Michigan and you are interested in whether or not the state AVERAGE is greater than the national average. The hypotheses for this scenario are as follows: Null Hypothesis: μ ≤ 501.294, Alternative Hypothesis: μ > 501.294. A random sample of 79 claims shows an average amount of $486.94 with a standard deviation of $96.1184. What is the test statistic and p-value for this test?

test stat: -1.327 pval: .9059 avg= 1 mean 3rd lect link, chap 8.2 test stat (t)= (xBAR - mu0)/s/sqrtn xbar= mean= 486.94 n= 79 mu)= hyp val= 501.294 s= std dev= 96.1184 s/swrt n= std error (486.94-501.294)/(96.1184/sqrt79) -14.354/10.81416489= -1.327333192= test stat p val= (hypA sign is greater, so shade r, so # then e99) **tcdf(pval/e99,, n-1) tcdf(-1.327333192, e99, 78)= .9058647275

The rainfall in Aberdeen Reservoir, Washington (the wettest place in the contiguous United States) averages 132.81 inches per year with a standard deviation of 5.735 inches. According to the Central Limit Theorem, what happens to the histogram of averages as n increases?

the histogram will begin to look more like the normal curve

Suppose that healthy human body temperatures have an average 99.03 and standard deviation of 2.983. According to the Central Limit Theorem, what happens to the histogram of averages as n increases?

the histogram will begin to look more like the normal curve (std dev decreases as sample size increases)

Consumers Energy states that the average electric bill across the state is $141.89. You want to test the claim that the average bill amount is actually different from $141.89. The hypotheses for this situation are as follows: Null Hypothesis: μ = 141.89, Alternative Hypothesis: μ ≠ 141.89. You complete a randomized survey throughout the state and perform a one-sample hypothesis test for the mean, which results in a p-value of 0.0106. What is the appropriate conclusion? Conclude at the 5% level of significance.

the true average electric bill is significantly different from $141.89.

Your friend tells you that the proportion of active Major League Baseball players who have a batting average greater than .300 is less than 0.58, a claim you would like to test. The hypotheses for this test are Null Hypothesis: p ≥ 0.58, Alternative Hypothesis: p < 0.58. If you randomly sample 30 players and determine that 16 of them have a batting average higher than .300, what is the test statistic and p-value?

ts: -.518, pval .302 same as above, 16/30= .53333 phat p0= .58 z= .5333-.58/sqrt(.58*(1-.58)/30) -.046666/sqrt(.00812)= -.5178795553 less than so shade to L, -e99 then # normcdf(-e99, -.5178795553, 0, 1)= .3022711376

The speeds of the 100 m dash for females between 11 and 17 years old is normally distributed with an average of 11.712 seconds and standard deviation of 1.5779 seconds. Would it be unusual to see a time between 6.1 and 9.62 seconds?

value is borderline unusual 2, vars, normcdf, 6.1, 9.62, mean, sd= 0.09226 which is less than .095

The revenue of 200 companies is plotted and found to follow a bell curve. The mean is $185.153 million with a standard deviation of $27.4955 million. Would it be unusual for a randomly selected company to have a revenue below $133.08 million?

value is unusual 2, vars, normcdf, -e99 133.08, mean, sd= 0.0291 which is less than .045

Researchers in the corporate office of an airline wonder if there is a significant difference between the cost of a flight on Priceline.com vs. the airline's own website. A random sample of 22 flights were tracked on Priceline and the airline's website and the 90% confidence interval for the mean difference in price (Priceline - Airline Site) was (31.43, 137.51). Which of the following is the appropriate conclusion?

we are 90% confident that the AVG DiFFERence in price is positive w the higher price coming from priceline

***The owner of a local supermarket believes the average number of gallons of milk the store sells per day is 239.7. In a random sample of 23 days, the owner finds that the average number of gallons sold was 237.2 with a standard deviation of 31.6. Using this information, the owner calculated the confidence interval of (223.5, 250.9) with a confidence level of 95%. Which of the following statements is the best conclusion?

we are 95% confident that the avg # of gallons sold per day is less than 239.7= WRONG CORRECT-The average number of gallons sold per day is not significantly different from 239.7. BECAUSE the test value falls inside the interval.

A new drug to treat high cholesterol is being tested by pharmaceutical company. The cholesterol levels for 24 patients were recorded before administering the drug and after. The average difference in cholesterol levels (after - before) was 0.59 mg/dL with a standard deviation of 9.466 mg/dL. Using this information, the calculated 95% confidence paired-t interval is (-3.407, 4.587). Which of the following is the best interpretation of this interval?

we are 95% confident that the avg difference in cholesterol levels between those who would take the drug and those who would not is between -3.407 and 4.587

You are in the market for a new car. You want to check whether there is a significant difference between the fuel economy of mid-size domestic cars and mid-size import cars. You sample 22 domestic car makes and find an average fuel economy of 33.905 MPG with a standard deviation of 3.644 MPG. For imports, you sample 17 cars and find an average MPG of 33.149 MPG with a standard deviation of 7.688. You use this information to calculate a 99% confidence interval for the difference in mean fuel economy of (-4.288, 5.8). Of the following statements, what is the best interpretation of this interval?

we are 99% confident that the DIFFERENCE between the avg fuel economy of all domestic mid-size cars and all import mid-sized cars surveyed is between -4.288 and 5.8

The owner of a local golf course wanted to determine the average age (in years) of the golfers that played on the course. In a random sample of 27 golfers that visited his course, the sample mean was 37.3 years old and the standard deviation was 6.95 years. Using this information, the owner calculated the confidence interval of (33.6, 41) with a confidence level of 99% for the average age. Which of the following is an appropriate interpretation of this confidence interval?

we are 99% confident that the avg age of ALL golfers that play on the golf course is btwn 33.6 and 41yrs old

***A pharmaceutical company is testing a new drug to increase memorization ability. It takes a sample of individuals and splits them randomly into two groups. One group is administered the drug, and the other is given a placebo. After the drug regimen is completed, all members of the study are given a test for memorization ability with higher scores representing a better ability to memorize. You are presented a 99% confidence interval for the difference in population mean scores (with drug - without drug) of (-0.95, 11.75). What can you conclude from this interval?

we are 99% confident that the avg memorization ability of those NOT on the drug is higher than those who are on the drug= WRONG There is no significant difference between the average memorization abilities for those on the drug compared to those not on the drug.= CORRECT BC ZERO IS CONTAINED W/I THE INTERVAL

You hear on the local news that for the city of Kalamazoo, the proportion of people who support President Trump is 0.38. However, you think it is different from 0.38. The hypotheses you want to test are Null Hypothesis: p = 0.38, Alternative Hypothesis: p ≠ 0.38. You take a random sample around town and calculate a p-value for your hypothesis test of 0.0902. What is the appropriate conclusion? Conclude at the 5% level of significance.

we did not find enough evidence to say a sig diff exists between the prop of ppl who supp pres trump and .38

You hear on the local news that for the city of Kalamazoo, the proportion of people who support President Trump is 0.43. However, you think it is less than 0.43. The hypotheses you want to test are Null Hypothesis: p ≥ 0.43, Alternative Hypothesis: p < 0.43. You take a random sample around town and calculate a p-value for your hypothesis test of 0.6840. What is the appropriate conclusion? Conclude at the 5% level of significance.

we did not find enough evidence to say the prop of ppl who support pres trump is less than .43 fig 8.2 flow chart** if p val under or equal to alpha we reject H0 and there IS evidence to show if p val over a, we DONT reject H0 and theres NOT ev so a= .05 (5% and move the decimal place) .684 is larger than .05 no reject, no evidence