E370 Exam 3

A golfer claims that his average golf score at the course he plays regularly is less than 90. The correct hypothesis statement for this golfer to prove his claim would be

0:μ≥90; H1:μ<90

In a one-tailed hypothesis test (upper tail), the z test statistic is determined to be 1.13. The p-value for this test is

1 -NORM.DIST(B17,0,1,TRUE) =0.1292

Refer to the Exhibit Average Income. We want to test to determine if there has been a decrease in the average yearly income of dentists (at the significance level 0.04). Construct the null and the alternative hypotheses for this test choosing the fields below: H0 : μ H1: μ

1. ≥ 2. 110,000 3. < 4. 110,000

In a two-tailed hypothesis test, the z test statistic is determined to be -1.56. The p-value for this test is

2*NORM.DIST(-1.56,0,1,1) = 0.1188

In a two-tailed hypothesis test, the z test statistic is determined to be 2. The p-value for this test is

2*NORM.DIST(-2,0,1,TRUE) = 0.0455

An accountant claims to be able to complete a standard tax return in under an hour. For a random sample of 24 tax returns, the accountant averaged 63.2 minutes. Assume that the population standard deviation is equal to 7.7 minutes and the population is normally distributed. What is the test statistic for the hypothesis test to check accountant's claim?

2.04

Suppose that you set up the hypothesis test: H0:μ≤20; H1:μ>20H0:μ≤20; H1:μ>20 A sample of 23 observations provides a sample mean of 22.4. Assume that the population standard deviation is known and equals 3. The hypothesis test can lead to an erroneous conclusion because: The sample size is not large enough to apply the Central Limit Theorem The population may have a distribution different from the normal distribution The sampling distribution of x¯x¯ may be not a normal distribution

All suggested alternatives are correct:

When conducting a hypothesis test for the population mean when σ is known and the sample size is 30 or more, the distribution we use for computing the critical value and the p-value is

the standard normal distribution

Breyers is a major producer of ice cream and would like to test if the average American consumes more than 17 ounces of ice cream per month. A random sample of 25 Americans was found to consume an average of 19 ounces of ice cream last month. The standard deviation for this sample was 5 ounces.Breyers would like to set α=0.025α=0.025 for the hypothesis test. It is known that zα=1.96zα=1.96 and tα=2.06tα=2.06 for the df = 24. Also, it is established that the ice cream consumption follows the normal distribution in the population. The conclusion for this hypothesis test would be

tx¯<tαtx¯<tα. So, we do not reject the null and cannot conclude that the average amount of ice cream consumed per month is greater than 17 ounces.

Think about the hypothesis test: H0:μ≥40; H1:μ<40H0:μ≥40; H1:μ<40. A sample of 49 observations provides a sample mean of 38 and a sample standard deviation of 7. The null hypothesis will be rejected if

tx¯<−tα

In the past, the average age of employees of a large corporation has been 40 years. Recently, the company has been hiring older individuals. In order to determine whether there has been an increase in the average age of all the employees, a sample of 64 employees was selected. The average age in the sample was 45 years with a standard deviation of 16 years.The company would like to set α=0.05α=0.05 for the hypothesis test. It is known that zα=1.645zα=1.645 and tα=1.669tα=1.669 for the df = 63. The conclusion for this hypothesis test would be

tx¯>tα. So, we reject the null and can conclude that there has been an increase in the average age of the employees in the corporation.

Refer to the Exhibit Costco Customers. Provide the null and the alternative hypotheses.

H0:μ≤130; H1:μ>130

Provide the null and the alternative hypotheses.

H0:μ≤25;H1:μ>25

Refer to Exhibit 5. Compute the test statistic for this test. Round off your solutions to two decimal digits.

(24.7-20)/ (11-sqrt(32)) = 2.42

H0:μ≥40; H1:μ<40H0:μ≥40; H1:μ<40 A sample of 49 observations provides a sample mean of 38 and a sample standard deviation of 7. Compute the value of the test statistic.

-2

The supervisor of a production line wants to check if the average time to assemble an electronic component is different from 14 minutes. Assume that the population of assembly time is normally distributed with a standard deviation of 3.4 minutes. The supervisor times the assembly of 14 components, and finds that the average time for completion is 11.6 minutes. How would you calculate the p-value for the hypothesis test?

2×P(z<−2.64)

Exhibit 5 The screening process for detecting a rare disease is not perfect. Researchers have developed a blood test that is considered fairly reliable. It gives a positive reaction in 98% of the people who have a disease. However, it erroneously gives a positive reaction for 3% of the people who do not have the disease. Consider the null hypothesis "the individual does not have the disease" Refer to Exhibit 5. What is the probability of Type I error (α) if the new blood test is used?

3%

The screening process for detecting a rare disease is not perfect. Researchers have developed a blood test that is considered fairly reliable. It gives a positive reaction in 98% of the people who have a disease. However, it erroneously gives a positive reaction for 3% of the people who do not have the disease. Consider the null hypothesis "the individual does not have the disease". What is the probability of Type I error if the new blood test is used?

3%

Refer to Exhibit 5. What is the power (1-probability of Type II error (β)) for this test?

98%

When the following hypotheses are being tested at a level of significance αα H0:μ≥500; H1:μ<500H0:μ≥500; H1:μ<500 the null hypothesis will be rejected if the p-value is

<α

In a one-tailed hypothesis test (lower tail), the t test statistic is determined to be -2.95. The sample size is 40. The p-value for this test is

= T.DIST(-2.95, 40-1, 1) = 0.0027

Refer to Exhibit 5. Compute the p-value for this test. Round off your solutions to two decimal digits.

=2*NORM.DIST(test sat,0,1,1) = 1.98

Refer to Exhibit 5. Compute the critical value for this test. Round off your solutions to two decimal digits.

=ABS(NORM.INV(.05/2,0,1)) = 1.96

For a one-tail test at 8.35% significance level, the critical value of z is

=ABS(NORM.INV(.0835,0,1)) =1.3819

For a two-tailed test at 12.66% significance level, the critical value of z is

=ABS(NORM.INV(.1266/2,0,1)) = 1.5276

For a one-tailed test and a sample size of 20 observations at 10% significance level, the critical value of t is (Hint: you are given the sample size, not the degrees of freedom)

=ABS(T.INV(.1,20-1)) = 1.3277

In a one-tailed hypothesis test (lower tail), the z test statistic is determined to be -1.87. The p-value for this test is

=NORM.DIST(-1.13,0,1,TRUE) = 0.0307

In a two-tailed hypothesis test, the t test statistic is determined to be -1.7. The degrees of freedom for the test is equal 15. The p-value for this test is

=T.DIST.2T(ABS(-1.7),15) = 0.1098

In a two-tailed hypothesis test, the t test statistic is determined to be 2.15. The sample size is 35. The p-value for this test is

=T.DIST.2T(ABS(2.15),35-1) = 0.0388

In a one-tailed hypothesis test (upper tail), the t test statistic is determined to be 1.35. The sample size is 35. The p-value for this test is

=T.DIST.RT(1.35,35-1) = 0.093

For a two-tailed test and a sample of 15 observations at 20% significance level, the critical value of t is (Hint: you are given the sample size, not the degrees of freedom)

=T.INV(.2/2,15-1) =1.345

Which of the following statements are valid null and alternative hypotheses?

H0:μ≤10; H1:μ>10

Exhibit 2 Over the past several years, the proportion of one-person households has been increasing. The Census Bureau would like to test the hypothesis that the proportion of one-person households differs from 0.27. A random sample of 125 households found that 43 consisted of one person. Refer to Exhibit 2. The correct null hypothesis statement and alternative

H0 : p = 0.27, H1 : p =/= 0.27

Exhibit 5 Pizza X would like to test the hypothesis that the average online order value equals $20. A random sample of 32 Pizza X customers provided an average order value of $24.70. It is believed that the population standard deviation for the order value is $11.00. Pizza X would like to set α = 0.05. Q26 Refer to Exhibit 5. The correct hypothesis statement for this test would

H0 : µ = 20 vs H1 : µ =/= 20

The Department of Labor would like to test the hypothesis that the average hourly wage for recent college graduates is greater than $20. A random sample of 32 recent college graduates averaged $22.30 per hour with a standard deviation of $3.20 per hour.

H0 : µ ≤ 20, H1 : µ > 20

Exhibit 3 YouTube would like to test the hypothesis that the average length of an online video watched by a user is more than 8 minutes. A random sample of 37 people watched online videos that averaged 8.7 minutes in length. It is believed that the population standard deviation for the length of online videos is 2.5 minutes. YouTube would like to set α = 0.1. 3 Refer to Exhibit 3. The correct hypothesis statement for this hypothesis test would be

H0 : µ ≤ 8.0 H1 : µ > 8.0

Representatives of a large national union announced that the fraction of women in the union was equal to one-half in the previous year. You are interested in testing whether there has been a change in the fraction of women this year. How would you set up the hypothesis test?

H0:p=0.5; H1:p≠0.5

Refer to the Exhibit Soft Drinks. Provide the null and the alternative hypotheses.

H0:p≤0.21;H1:p>0.21

The National Association of Realtors reported that 26% of home buyers in the state of Florida were foreigners in 2015. In 2017, a group of students from Indiana University conducted a study and, based on a sample of 100 people, concluded that 28% of home buyers in the state of Florida are foreigners. So, the members of the group think that the proportion has increased since then. The correct hypothesis statement for the student's group to test is

H0:p≤0.26;H1:p>0.26

Refer to the Exhibit Female Employees. Provide the null and the alternative hypotheses.

H0:p≥0.5;H1:p<0.5

Exhibit: Average Debt Load. The Department of Education would like to check if the average debt load of graduating students with a bachelor's degree is different from $17,000. A random sample of 34 students had an average debt load of $18,200. It is believed that the population standard deviation for student debt load is $4,200. The significance level α is set to 0.02 for the hypothesis test.

H0:μ=$17,000;H1:μ≠$17,000

Travelocity would like to test if the average roundtrip airfare between New York and London differs from $1,400. The correct hypothesis statement for this test would be

H0:μ=1,400; H1:μ≠1,400

A production line operation is designed to fill cartons with laundry detergent to a mean weight of 32 ounces. A sample of cartons is periodically selected and weighted to determine whether overfilling or underfilling is occurring. If the sample data led to a conclusion of underfilling or overfilling, the production line would be shut down and adjusted to obtain a proper filling. The correct hypothesis statement for this test would be

H0:μ=32; H1:μ≠32

Provide the null and the alternative hypotheses.

H0:μ=3;H1:μ≠3H0:μ=3;H1:μ≠3

Refer to the Exhibit Female Employees. Suppose that, instead of the critical value approach, you decided to use the p-value approach and found that the p-value for the test in this problem is equal to 0.0228. This p-value is best interpreted as the following:

If the true proportion of female employees in the company was 50%, then there would be a 2.28% chance of observing a proportion of females equal to or below 45% in the sample of 400 individuals.

Refer to Exhibit 2. The test statistic follows

Normal distribution because np ≥ 5 and nq ≥ 5.

Refer to Exhibit 3. Assume that the test statistic is a where a > 0. How would you compute the p-value?

P(z > a)

A professor of statistics wants to test that the average amount of money a typical college student spends per day during spring break is over $70. Based upon previous research, the population standard deviation is estimated to be $17.32. The professor surveys 35 students and finds that the mean spending is $72.43. How would you calculate the p-value for this test?

P(z>0.83)

Refer to the Exhibit Average Debt Load. Which of the following statements is the most accurate given your p-value? Note: the choices refer to significance levels different from 0.02. So, the conclusion might be different from what you stated earlier.

Reject the null hypothesis at α = 0.10, but not at α = 0.05

Customers at Costco spend an average of $130 per trip (The Wall Street Journal, October 6, 2010). One of Costco's rivals would like to determine whether Costco's customers spend more per trip. A survey of the receipts of 25 customers found that the sample mean was $135.25. Assume that the population standard deviation of spending is $10.50 and the spending follows a normal distribution (use the significance level 0.07).

SPACER

Exhibit: Check Out Times. The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes with a standard deviation of 0.5 minutes. We want to test to determine whether or not the mean waiting time of all customers is any different from 3 minutes (use the significance level 0.02 for the test).

SPACER

Exhibit: Female Employees. Last year, 50% of MNM, Inc. employees were female. It is believed that there has been a reduction in the percentage of females in the company. This year, in a random sample of 400 employees, 180 were female. Based on this information, you want to determine if there has been a decrease in the percentage of females in the company (at the significance level of 5%).

SPACER

Exhibit: Soft Drinks. Last year, a soft drink manufacturer had 21% of the market. In order to increase their portion of the market, the manufacturer has introduced a new flavor in their soft drinks. A sample of 400 individuals participated in the taste test and 100 indicated that they like the taste. We are interested in determining if more than 21% of the population will like the new soft drink at the significance level 0.05.

SPACER

Exhibit: Tires Life Expectancy. A tire manufacturer has been producing tires with an average life expectancy of 25,000 miles. Now the company is advertising that its new tires' life expectancy has increased. In order to test the legitimacy of the advertising campaign, an independent testing agency tested a sample of 6 of their tires and has provided the following data Life Expectancy (In Thousands of Miles) 28 27 25 28 29 25

SPACER

The Bureau of Labor Statistics reported that the average yearly income of dentists in the year 2009 was $110,000. A sample of 34 dentists, which was taken in 2010, showed an average yearly income of $104,000. Assume the standard deviation of the population of dentists in 2010 is $21,000.

SPACER

The manager of an automobile dealership is trying to understand if a new bonus plan has increased sales. Previously, the mean sales rate per salesperson was five automobiles per month. Suppose that, the manager conducts a hypothesis test and, based on the hypothesis test, she determines that the p-value equals 0.0062. Using the significance level α=0.05α=0.05, how would you state the conclusion for the test?

Since the p-value = 0.0062 < 0.05, reject H0. Therefore, there is an evidence that the new bonus plan is more efficient than the old one.

Medicare would like to test if the average monthly rate for one-bedroom assisted-living facility is different from $3,300. Suppose that you collect sample information and, based on the hypothesis test, determine that the z test statistic is equal to 1.98 while the critical value of z is 2.05. How would you state the conclusion?

Since −zα/2=−2.05<zx¯=1.98<zα/2=2.05−zα/2=−2.05<zx¯=1.98<zα/2=2.05, do not reject H0. Therefore, there is not enough evidence to conclude that the monthly rate for one-bedroom assisted-living facility is different from $3,300.

The p-value is best interpreted as the following:

The p-value indicates the probability of observing the proportion of people who like new flavor equal to 25% or greater in the sample of 400 individuals, if actual proportion of people who like the new flavor is 21%.

Refer to the Exhibit Average Income. Suppose that, instead of the critical value approach, you decided to use the p-value approach and found that the p-value for the test in this problem is equal to 0.0479. This p-value is best interpreted as the following:

There is a 4.79% probability of observing the average yearly income in the sample of 34 dentists of $104,000 or below if the actual average yearly income of dentists is $110,000.

Refer to the Exhibit Average Income. State your conclusion for the test in words relying on the economic context of the problem.

There is not enough evidence to conclude that the yearly income of dentists has decreased since 2009.

Which of the following statements is correct? (this question is a little tricky) We cannot establish a claim if the null hypothesis is not rejected Since the sample evidence cannot be supported by the null hypothesis, we can reject the null We can accept the null hypothesis if the sample evidence is not inconsistent with the null hypothesis We can establish a claim if the sample evidence is consistent with the null hypothesis

We cannot establish a claim if the null hypothesis is not rejected

Do we need any additional assumptions about the life expectancy of the tires in the population to make sure that the conclusion stated in the previous question is reliable?

Yes. The life expectancy of the tires should follow the normal distribution in the population which would guarantee that the conclusions derived in the test are reliable.

Exhibit 4 Ben & Jerry's is a producer of ice cream and would like to test the hypothesis that the average American consumes more than 17 ounces of ice cream per month. A random sample of 15 Americans was found to consume an average of 18.2 ounces of ice cream last month. The standard deviation for this sample was 3.9 ounces. Ben & Jerry's would like to set α = 0.1. Refer to Exhibit 4. Do we need any additional assumptions to make sure that the conclusion from this test reliable?

Yes. The population should be normally distributed.

Refer to Exhibit 3. What is the test statistic for this hypothesis test?

Zxbar = 1.70

The error of rejecting a true null hypothesis is

a Type I error

Suppose that you are working on an lower tail test. As the test statistic becomes larger, the p-value

becomes larger

Exhibit 1^: The p-value can be interpreted as if the population average of wage was $20, the probability of observing sample mean of wage in the sample of 32 college graduates.

c is equal or greater than $22.30

Refer to Exhibit 4. Suppose that you want to change α from 0.1 to 0.05. Then, _____ will be changed accordingly.

critical value

Suppose that you are working on an upper tail test. As the test statistic becomes larger, the p-value

gets smaller

Suppose that you set up the hypothesis test: H0:μ≤800; H1:μ>800H0:μ≤800; H1:μ>800 Given the population standard deviation, the sample size of 59 and the sample mean of 879, you calculate the value of the test statistic zx¯zx¯ and determine the p-value = 0.017 for the test. This p-value means that

if the population mean was equal to 800, then there is a 1.7% probability of obtaining the sample mean of 879 or greater in the sample of 59 observations.

If a hypothesis is rejected at 5% level of significance, it (Hint: draw a picture for this question)

may be rejected or not rejected at the 1% level

State your conclusion for the test using the p-value and the significance level of 3%.

p-value < 0.03, so we reject Ho. Therefore, there is enough evidence to conclude that the mean life expectancy of the new tires has increased.

State your conclusion for the test using the p-value.

p-value < 0.05, so we reject Ho. Therefore, there is enough evidence to conclude that the proportion of people who like new favor is above 21%.

Refer to the Exhibit Costco Customers. State your conclusion for the test using the p-value.

p-value < 0.07, so we reject Ho. Therefore, there is enough evidence to conclude that Costco's customers spend more than $130 per trip.

The level of significance is the

probability of Type I error (alpha)

When conducting a hypothesis test for the population mean when σ is unknown and the sample size is 30 or more, the distribution we use for computing the critical value and the p-value is

the Student's t-distribution

The Department of Economic and Community Development (DECD) reported that in 2009 the average number of new jobs created per county was 450 (the only information that you know from DECD). Doing a project in your international economics class, you want to determine whether there has been a decrease in the average number of jobs created, and you collect information about 10 countries. To conduct the hypothesis test, what distribution would you use to calculate the critical value and the p-value?

the Student's t-distribution with 9 degrees of freedom

In hypothesis testing, if the null hypothesis is rejected,

the alternative hypothesis is true

In hypothesis testing, the tentative assumption about the population parameter is

the null hypothesis

Over the past several years, the proportion of one-person households has been increasing. The Census Bureau would like to test the hypothesis that the proportion of one-person households exceeds 0.27. A random sample of 125 households found that 43 consisted of one person.To conduct the hypothesis test, what distribution would you use to calculate the critical value and the p-value?

the standard normal distribution

If a hypothesis is rejected at 2% level of significance, it

will always be rejected at the 5% level.

Refer to Exhibit 3. What is the critical value for this hypothesis test?

z0.1 = 1.28

Refer to the Exhibit Female Employees. State your conclusion for the test using the critical value.

zp<−zα⇒zp<−zα⇒ so, we reject Ho. Therefore, there is enough evidence to conclude that there has been a reduction in the percentage of females in the company.

Suppose that you set up the hypothesis test: H0:μ≤800; H1:μ>800H0:μ≤800; H1:μ>800 Assuming that the population standard deviation is known and αα is the level of significance, the null hypothesis will be rejected if

zx¯>zα

Refer to the Exhibit Average Income. State your conclusion for the test indicating whether you reject the null or not.

zx¯>−zα⇒zx¯>−zα⇒ we cannot reject H0.

State your conclusion for the test.

−tα/2 < tx¯ < tα/2⇒ we do not reject Ho. Therefore, there is not enough evidence to conclude that the mean waiting time of all customers is any different from 3 minutes.

Suppose that you want to test: H0:μ≥0.54; H1:μ<0.54H0:μ≥0.54; H1:μ<0.54 based on a sample of n = 25 and known population standard deviation of 13.2. What is the appropriate boarder of the rejection region for the test at 0.03 significance level?

−z0.03

Refer to the Exhibit Average Debt Load. State your conclusion for the test.

−zα/2<zx¯<zα/2⇒−zα/2<zx¯<zα/2⇒ we do not reject Ho. Therefore, there is not enough evidence to conclude that the average debt load of graduating students with a bachelor's degree is different from $17,000.