ECON 230- FInal exam questions

According to Net Applications, in the summer of 2014, Google's Chrome browser exceeded a 20% market share for the first time, with a 20.37% share of the browser market (Forbes website). For a randomly selected group of 20 internet browser users, answer the following. -Compute the probability that exactly 8 of the 20 internet browser users use Chrome as their Internet browser.

.0243

If A and B are independent events with P(A) = 0.35 and P(B) = 0.20, then, P(A ∪ B) =

0.48 (addition law formula)

Eg. Probabilities of different types of vehicle-to-vehicle accidents are shown below. Accident Probability ------------ ------------ Car to Car 0.62 Car to Truck 0.10 Truck to Truck 0.28 Find the probability that an accident involves a car.

0.72

Find the t value(s) for each of the following cases. 1). Upper tail area of .01 with 30 degrees of freedom. 2) Where 95% of the area falls between these two t values with 45 degrees of freedom.

1) 2.457 2) Use .025 column, -2.014 and 2.014

A student has to take 7 more courses before he can graduate. If none of the courses are prerequisite to others, how many groups of 3 courses can he select for the next semester?

35

A simple survey consists of three multiple choice questions. The first question has 3 possible answers, the second has 4 possible answers and the third has 3 possible answers. What is the total number of different ways in which this survey could be completed?

36

Five individuals are candidates for positions of president, vice president, and treasurer of an organization. How many possibilities of selections exist?

60

How Teenagers Listen to Music. For its Music 360 survey, Nielsen Co. asked teenagers and adults how each group has listened to music in the past 12 months. Nearly two-thirds of U.S. teenagers under the age of 18 say they use Google Inc.'s video-sharing site to listen to music and 35% of the teenagers said they use Pandora Media Inc.'s custom online radio service (The Wall Street Journal). Suppose 10 teenagers are selected randomly to be interviewed about how they listen to music. A) Is randomly selecting 10 teenagers and asking whether or not they use Pandora Media Inc.'s online service a binomial experiment? B) What is the probability that none of the 10 teenagers uses Pandora Media Inc.'s online radio service? C) What is the probability that 4 of the 10 teenagers use Pandora Media Inc.'s online radio service? D) What is the probability that at least 2 of the 10 teenagers use Pandora Media Inc.'s online radio service?

A) Yes. Because the teenagers are selected randomly, p is the same from trial to trial and the trials are independent. The two outcomes per trial are use Pandora Media Inc.'s online radio service or do not use Pandora Media Inc.'s online radio service. Binomial n = 10 and p = .35 f (x) = 10! ----------- (.35)^x (1−.35)^10−x x!(10−x)! b) f(0)= .0135 c) f(4) = .2377 d) Probability (x > 2) = 1 - f(0) - f(1) From part b, f(0) = .0135 f(1)= .0725 Probability (x > 2) = 1 - f(0) - f(1) = 1 - (.0135+ .0725) = .9140

The critical value z for a 97.8% confidence interval estimation is ?

Answer: 2.29 Confidence coefficient = 0.978 𝛼 = 1 − .978 = 0.022 𝛼/2 = 0.022/2 = 0.011 Find 𝑧0.011 - This means the z-score that has an area of 0.011 to its right. (in the right tail) 𝑧0.011 = 2.29

The average hourly wage of computer programmers with 12 years of experience has been $80. Because of high demand for computer programmers, it is believed there has been a significant increase in the average hourly wage of computer programmers. To test whether or not there has been an increase, the correct hypotheses to be tested are ?

H0: μ≤ 80 Ha: μ>80

An insurance company sets up a statistical test with a null hypothesis that the average time for processing a claim is 5 days, and an alternative hypothesis that the average time for processing a claim is greater than 5 days. After completing the statistical test, it is concluded that the average time exceeds 5 days. However, it is eventually learned that the mean process time is really 5 days. What type of error occurred in the statistical test?

Type 1 error

The probability function for the number of insurance policies John will sell is given by: f(x)= 0.5- (x/6) for x= 0, 1 , 2 A) What is the expected value. of policies John will sell? B) What is the variance of the number of policies John will sell?

a) 0.667 b) 0.556

Suppose that the mean retail price per gallon of regular grade gasoline in the United States is $3.43 with a standard deviation of $.10 and that the retail price per gallon has a bell-shaped distribution. A) What percentage of regular grade gasoline sold between $3.33 and $3.53 per gallon? B) What percentage of regular grade gasoline sold between $3.33 and $3.63 per gallon? C) What percentage of regular grade gasoline sold for more than $3.63 per gallon?

a) 68% b) 81.5% c) 2.5%

The mean hourly wage for employees in goods-producing industries is currently $24.57. Suppose we take a sample of employees from the manufacturing industry to see if the mean hourly wage differs from the reported mean of $24.57 for the goods-producing industries. a) State the null and alternative hypotheses we should use to test whether the population mean hourly wage in the manufacturing industry differs from the population mean hourly wage in the goods-producing industries. b) Suppose a sample of 30 employees from the manufacturing industry showed a sample mean of $23.89 per hour. Assume a population standard deviation of $2.40 per hour and compute the p-value. c) With α = .05 as the level of significance, what is your conclusion? d) Repeat the preceding hypothesis test using the critical value approach.

a) H0: u = 24.57 Ha: u ≠ 24.57 b) test statistic: z= -1.55 because p value is 2 x the lower tail are 2 (.0606) = .1212 c) do not reject d) do not reject

The average starting salary of students who graduated from colleges of Business in 2009 was $48,400. A sample of 100 graduates of 2010 showed an average starting salary of $50,000. Assume the standard deviation of the population is known to be $8000. We want to determine whether or not there has been a significant increase in the starting salaries. a) State the null and alternative hypotheses to be tested. b) Compute the test statistic. c) The null hypothesis is to be tested at the 5% level of significance. Determine the critical value for this test. d. What do you conclude? e. Compute the p-value.

a) H0: μ ≤ $48,400 Ha: μ > $48,400 b) Test statistic z = 2.0 (refer to test statistic formula on formula sheet) c) z = 1.645 d) Reject H0 and conclude that there has been a significant increase in the starting salaries. ** reject because z value 2.0> critical value of 1.645 e) p-value = .0228 ** look up 2.0 in z table and get the area of .9772, 1-.9772 = .0228

The average price of homes sold in the U.S. in the past year was $220,000. A random sample of 81 homes sold this year showed an average price of $210,000. It is known that the standard deviation of the population is $36,000. At a 5% level of significance, test to determine if there has been a significant decrease in the average price of homes. a) State the null and alternative hypotheses to be tested. b) Compute the test statistic. c) Determine the critical value for this test. d) What do you conclude? e) Compute the p-value.

a) H0: μ ≥ $220,000 Ha: μ < $220,000 b) Test statistic z = -2.5 c) Critical z = -1.645 (refer to confidence level chart, where 𝜎/2 = .05 the z that corresponds = 1.645 and it is negative since it is a left tailed test) d) Reject H0 and conclude that there has been a significant decrease in the average price of homes. (***If z- value > critical value, then you reject the null hypothesis) e) p-value = .0062 (look up -2.5 z value are in table) ** if you were using the p-value approach, then you reject the null hypothesis if p-value ≤ level of significance: .0062 < .05 so you would still reject using this approach.

The Los Angeles Times regularly reports the air quality index for various areas of Southern California. A sample of air quality index values for Pomona provided the following data: 28, 42, 58, 48, 45, 55, 60, 49, and 50. a. Compute the range and interquartile range. b. Compute the sample variance and sample standard deviation. c. A sample of air quality index readings for Anaheim provided a sample mean of 48.5, a sample variance of 136, and a sample standard deviation of 11.66. What comparisons can you make between the air quality in Pomona and that in Anaheim on the basis of these descriptive statistics?

a) Range = 60 - 28 = 32 IQR = Q3 - Q1 = 56.5 - 43.5 = 13 (Refer to study guide for steps on how to find IQR) b) sample variance= 92.75 sample standard deviation= 9.65 (refer to study guide on how to find solution)

Given that z is a standard normal random variable, find z for each situation. a) The area to the left of z is .2119. b) The area between −z and z is .9030. c) The area between −z and z is .2052. d) The area to the left of z is .9948. e) The area to the right of z is .6915.

a) The z-value corresponding to a cumulative probability of .2119 is z = -.80. b) Compute .9030/2 = .4515; z corresponds to a cumulative probability of .5000 + .4515 = .9515. So z = 1.66. c) Compute .2052/2 = .1026; z corresponds to a cumulative probability of .5000 + .1026 = .6026. So z = .26. d) The z-value corresponding to a cumulative probability of .9948 is z = 2.56. e) The area to the left of z is 1 - .6915 = .3085. So z = -.50.

The following data are from a simple random sample. 5 8 10 7 10 14 A) What is the point estimate of the population mean? B) What is the point estimate of the population standard deviation?

a) x=Exi /n = 54 / 6 = 9 b) s= 3.1 (refer to study guide for worked out solution)

(Chebyshev's Theorem) In a statistics class, the average grade on the final examination was 75 with a standard deviation of 5. If nothing is known about the shape of the distribution, answer the following questions. a. At least what percentage of the students received grades between 50 and 100? b. Determine an interval for the grades that will be true for at least 70% of the students. (Hint: First, compute the Z-score.)

a) 𝑧50 = (50−75)/ 5 = −5 𝑧100 = (100−75) /5 = 5 Applying Chebyshev's theorem with z=5 1 − (1/ z^2) = 1 − (1 /5^2 )= 0.96 At least 96% b. 1 − (1/z^2) = 70% 100%-70%=30% .30=1/z^2 square root of (1/.30^2)= 1.826 so z = 1.826 75 ± (1.826)(5) = 65.87 to 84.13

The following probability distributions of job satisfaction scores for a sample of information systems (IS) senior executives and middle managers range from a low of 1 (very dissatisfied) to a high of 5 (very satisfied). Probability Satisfaction Score Executives Managers 1 .05 .04 2 .09 .10 3 .03 .12 4 .42 .46 5 .41 .28 a. What is the expected value of the job satisfaction score for senior executives? b. What is the expected value of the job satisfaction score for middle managers? c. Compute the variance of job satisfaction scores for executives and middle managers. d. Compute the standard deviation of job satisfaction scores for both probability distributions.

a. E(x) = Σx f(x) = 0.05(1) + 0.09(2) + 0.03(3) + 0.42(4) + 0.41(5) = 4.05 b. E(x) = Σx f(x) = 0.04(1) + 0.10(2) + 0.12(3) + 0.46(4) + 0.28(5) = 3.84 c. Executives: o^2 = Σ(x - μ)^2 f(x) = 1.25 Middle Managers: o^2 = Σ(x - μ)2 f(x) =1.13 d. Executives: o = 1.12 Middle Managers: o= 1.07 (refer to study guide for guided solution)

A population has a mean of 200 and a standard deviation of 50. A sample of size 100 will be taken and the sample mean will be used to estimate the population mean. a. What is the expected value of 𝑥̅? b. What is the standard deviation of 𝑥̅?

a. E(𝑥̅)=u=200 b. o/ square root of n= 50/ 100 = 50/10 = 5

Given that z is a standard normal random variable, compute the following probabilities. P(0 ≤ z ≤ .83) P(−1.57 ≤ z ≤0) P(z > .44) P(z ≥ −.23) P(z < 1.20) P(z ≤ −.71)

a. P(0 ≤ z ≤ .83) = .7967 - .5000 = .2967 P(-1.57 ≤ z ≤ 0) = .5000 - .0582 = .4418 P(z>.44)=1-.6700=.3300 P(z ≥ -.23) = 1 - .4090 = .5910 P(z<1.20)=.8849 P(z ≤ -.71) = .2389

Classify the random variables below according to whether they are discrete or continuous: a. The number of phone calls made in a day b. The number of guests at a wedding c. The number of turtles in a pond d. The area of a postage stamp e. The distance a javelin is thrown

a. The number of phone calls made in a day is discrete because the value of the random variable can be counted. b. The number of guests at a wedding is discrete because the value of the random variable can be counted c. The number of turtles in a pond is discrete because the value of the random variable can be counted. d. The area of a postage stamp is continuous because area is measured on an interval. e. The distance a javelin is thrown is continuous because distance is measured on an interval.

Zip Code consist of numeric values. Zip Code is an example of

a. a quantitative variable. b. either a quantitative or a categorical variable. c. an exchange variable. d. a categorical variable Answer: D Categorical, because they cannot be meaningfully added, subtracted, multiplied, or divided

Data were collected on the amount spent by 64 customers for lunch at a major Houston restaurant resulted in a sample mean of 21.52. Based upon past studies the population standard deviation is known with σ = $6. a) What is the standard error of the mean, 𝜎𝑥̅ ? b) At 99% confidence, what is the margin of error? c) Develop a 99% confidence interval estimate of the mean amount spent for lunch.

a. 𝜎𝑥̅= 𝜎/√n = 6 / √64 =0.75 b. With 99% confidence z 𝜎/ 2 = z.005 = 2.576 Margin of error = 2.576(𝜎/ √ n) = 2.576(6 / √64 ) = 1.93 *** just multiply .75 x 2.576 c. Confidence Interval: 21.52 + or - 1.93 or 19.59 to 23.45

Which of the following is a quantitative variable?

a.Favorite food b. Types of birds seen at a feeder c. Weight loss d. Albums in a collection Answer: C

Categorical data

are measurements that cannot be measured on a natural numerical scale EXAMPLE: favorite food

A discrete random variable

can assume a countable number of values. That is, if we can list the values of a random variable x, even though the list is never ending, the list is called countable and the corresponding random variable is a discrete random variable.

A continuous random variable

can assume values corresponding to any of the points contained in an interval. That is, when the values of a random variable are not countable, but instead correspond to the points on some interval, where the set of all possible values for the variable cannot be listed (not countable), the variable is a continuous random variable.

The Central Limit Theorem is important in statistics because _.

for a large n, it says the sampling distribution of the sample mean is approximately normal, regardless of the population

Stem. Leaf ----- ----- 1 2 3 2 0 1 5 5 8 3 4 6 9 5 3 4 4 4 6 6 2 5 6 6 7 8 8 0 3 5 9 8 What are the mean, median, and mode for the stem and leaf plot?

mean= 50.56 (The total number of values recorded in the stem and leaf plot is the number of leaves = 25 The values written out in full are: 12, 13, 20, 21, 25, 25, 28, 34, 36, 39, 53, 54, 54, 54, 56, 62, 65, 66, 66, 67, 68, 80, 83, 85, 98 So the mean = (12 + 13 + 20 + 21 + 25 + 25 + 28 + 34 + 36 + 39 + 53 + 54 + 54 + 54 + 56 + 62 + 65 + 66 + 66 + 67 + 68 + 80 + 83 + 85 + 98) ÷ 25 = 1,264 ÷ 25 = 50.56 (Note that this is not the same as the mean of the stems plus the mean of the leaves) median= 54 mode= 54

quantitative data

measurements that are recorded on a naturally occurring numerical scale.

How large a sample should be selected to provide a 95% confidence interval with a margin of error of 3? Assume that the population standard deviation is 9.

n= 34.57 ---->use n=35 (1.96^2) (9^2) / 3^2 (use sample size for an interval estimate of a population mean formula on formula sheet)

A safety inspector conducts air quality tests on a randomly selected group of 7 classrooms at a high school. Identify the population and sample in this setting.

population: all classrooms at the highschool sample: the 7 classrooms selected

The American Automobile Association (AAA) reported that families planning to travel over the Labor Day weekend spend an average of $749. Assume that the amount spent is normally distributed with a standard deviation of $225. a) What is the probability of family expenses for the weekend being less that $400? b) What is the probability of family expenses for the weekend being $800 or more? c) What is the probability that family expenses for the weekend will be between $500 and $1000? d) What would the Labor Day weekend expenses have to be for the 5% of the families with the most expensive travel plans?

u = 749 and o = 225 ---------------------- a) z = (x − u) / o = (400 − 749) / 225 = −1.55 P(z < −1.55) = .0606 The probability that expenses will be less than $400 is .0606. ------------ b) z= 0.23 1-P (z > .23) =1 - .5910 = .4090 ------------ c) P(500 £ x £ 1000) = P ( z £ 1.12) - P ( z £ -1.11) = .8686 - .1335 = .7351 ------------ d) The upper 5%, or area = 1- .05 = .95 occurs for z = 1.645 x = u + zo = 749 + 1.645(225) = $1119 The 5% most expensive travel plans will be $1,119 or more.

The mean preparation fee H&R Block charged retail customers last year was $183. Use this price as the population mean and assume the population standard deviation of preparation fees is $50. A) What is the probability that the mean price for a sample of 30 H&R Block retail customers is within $8 of the population mean? B) What is the probability that the mean price for a sample of 50 H&R Block retail customers is within $8 of the population mean? C) What is the probability that the mean price for a sample of 100 H&R Block retail customers is within $8 of the population mean?

u= 183 o= 50 within $8 means 183 + & - 8 so 175 < 𝑥̅ < 191 𝑥̅ - u / (o/√n) ---------------- a) n= 30 8 / (50 / √30) = .88 p (-.88 < z < .88) = .8106 - .1894 = .6212 _____________________ b) .7416 c) .8904

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 65 weekly reports showed a sample mean of 19.5 customer contacts per week. The sample standard deviation was 5.2. If we want to determine a 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel. a) the degrees of freedom for reading the t value is: b) the value of t is c) The value of the margin of error at 95% confidence is d) The 95% confidence interval for the average hourly wage (in $) of all information system managers is

𝑥̅ ± t 𝜎 /2 (s/ √ n) a. df=64 b. 95% confidence: df = 64 t.025 = 1.998 c. 1.998(5.2/ √65)=1.29 d. 19.5 ± 1.29 or 18.21 to 20.79