Module 3 – Hypothesis Testing

Question 1

A manager of a factory wants to know if a new quality check protocol has decreased the number of units a worker produces in a day. Before the new protocol, a worker could produce 27 units per day. What alternative hypothesis should the manager use to test this claim?

• µ ≠ 27 units
• µ ≤ 27 units
• µ < 27 units
• µ > 27 units

Answer 1

µ ≠ 27 units

The alternative hypothesis is the claim the manager would like to substantiate. The manager does not want to test whether the average number of units a worker can produce has increased or decreased.

µ ≤ 27 units

The alternative hypothesis is the claim the manager would like to substantiate. The manager does not want to test whether the average number of units a worker can produce has remained the same or decreased.

µ < 27 units

The manager wants to know if the new quality check protocol has decreased the average number of units a worker can produce per day. For a one-sided test, the manager should use the alternative hypothesis Ha: μ<27 units. This is the claim the manger wishes to substantiate.

µ > 27 units

The alternative hypothesis is the claim the manager would like to substantiate. The manager does not want to test whether the average number of units a worker can produce has increased.

Question 2

A manager of a factory wants to know if a new quality check protocol has decreased the number of units a worker produces in a day. Before the new protocol, a worker could produce 27 units per day. What null hypothesis should the manager use to test this claim?

• µ = 27 units
• µ ≥ 27 units
• µ > 27 units
• µ < 27 units

Answer 2

µ = 27 units

The null and alternative hypotheses are always opposites. If the manager’s alternative hypothesis is that the average daily units produced has decreased, then the null hypothesis is that the average is the same or has increased. How would we represent that mathematically?

µ ≥ 27 units

This is the null hypothesis. Remember that the null and alternative hypotheses are opposites.

µ > 27 units

The null and alternative hypotheses are always opposites. If the manager’s alternative hypothesis is that the average daily units produced has decreased, then the null hypothesis is that the average is the same or has increased.

µ < 27 units

The null and alternative hypotheses are always opposites. If the manager’s alternative hypothesis is that the average daily units produced has decreased, then the null hypothesis is that the average is the same or has increased.

Question 3

If you are performing a hypothesis test based on a 90% confidence level, what are your chances of making a type II error?

• 90%
• 10%
• 5%
• It is not possible to tell without more information

Answer 3

It is not possible to tell without more information

The confidence level does not provide any information about the likelihood of making a type II error. Calculating the chances of making a type II error is quite complex and beyond the scope of this course.

Question 4

The mean score on a particular standardized test is 500, with a standard deviation of 100. To assess whether a training course has been effective in improving scores on the test, we take a random sample of 100 students from the course and find that the average score of this sample is 550. Which function would correctly calculate the 95% range of likely sample means under the null hypothesis?

• 550 ± CONFIDENCE.NORM(0.05,100,100)
• 550± CONFIDENCE.T(0.05,100,100)
• 500 ± CONFIDENCE.T(0.05,100,100)
• 500 ± CONFIDENCE.NORM(0.05,100,100)

Answer 4

550 ± CONFIDENCE.NORM(0.05,100,100)

The range of likely sample means is centered at the historical population mean, 500, not at the sample mean, 550.

550± CONFIDENCE.T(0.05,100,100)

Because our sample is larger than 30, we can assume the distribution of sample means is roughly normal, due to the central limit theorem, and use the CONFIDENCE.NORM function. In addition, the range of likely sample means is centered at the historical population mean, 500, not at the sample mean, 550.

500 ± CONFIDENCE.T(0.05,100,100)

Because our sample is larger than 30, we can assume the distribution of sample means is roughly normal, due to the central limit theorem, and use the CONFIDENCE.NORM function.

500 ± CONFIDENCE.NORM(0.05,100,100)

The range of likely sample means is centered at the historical population mean, 500. Because our sample is larger than 30, we can assume the distribution of sample means is roughly normal, due to the central limit theorem, and use the CONFIDENCE.NORM function.

Question 5

Now suppose we take a sample of 25 students, taking the same standardized test, which has a mean score of 500 and a standard deviation of 100, and find that the average score of this sample is 530. Which function would correctly calculate the 95% range of likely sample means under the null hypothesis?

• 530 ± CONFIDENCE.NORM(0.05,100,25)
• 530± CONFIDENCE.T(0.05,100,25)
• 500 ± CONFIDENCE.T(0.05,100,25)
• 500 ± CONFIDENCE.NORM(0.05,100,25)

Answer 5

530 ± CONFIDENCE.NORM(0.05,100,25)

The range of likely sample means is centered at the historical population mean, 500, not at the sample mean, 530. In addition, because of the small sample size, we cannot assume that the sample means are normally distributed, so we should not use the CONFIDENCE.NORM function.

530± CONFIDENCE.T(0.05,100,25)

The range of likely sample means is centered at the historical population mean, 500, not at the sample mean, 530.

500 ± CONFIDENCE.T(0.05,100,25)

The range of likely sample means is centered at the historical population mean, 500. Because our sample is less than 30, we cannot assume that the sample means are normally distributed, and so we should use CONFIDENCE.T rather than the CONFIDENCE.NORM function.

500 ± CONFIDENCE.NORM(0.05,100,25)

Because of the small sample size, we cannot assume that the sample means are normally distributed, so we should not use the CONFIDENCE.NORM function.

Question 6

If you are performing a hypothesis test based on a 90% confidence level, what are your chances of making a type I error?

• 90%
• 10%
• 5%
• It is not possible to tell without more information

Answer 6

90%

The probability of a type I error is equal to the significance level, which is 1–confidence level.

10%

The probability of a type I error is equal to the significance level, which is 1–confidence level. A 90% confidence level indicates that the significance level is 10%. Therefore there is a 10% chance of making a type I error.

5%

The probability of a type I error is equal to the significance level, which is 1–confidence level.

It is not possible to tell without more information

The confidence level provides the necessary information. The probability of a type I error is equal to the significance level, which is 1–confidence level.

Question 7

If the one-sided p-value of a given sample mean is 0.0150, what is the two-sided p-value for that sample mean?

• 0.0075
• 0.0150
• 0.0300
• The answer cannot be determined without further information

Answer 7

0.0075

The two-sided p-value is double the one-sided p-value.

0.0150

The two-sided p-value is double the one-sided p-value.

0.0300

The two-sided p-value is double the one-sided p-value. Since the one-sided p-value is 0.0150, the two-sided p-value is 0.0150*2=0.0300.

The answer cannot be determined without further information

Since we know the one-sided p-value, we can calculate the two-sided p-value of the sample mean.

Question 8

A manager of a factory wants to know if the average number of workplace accidents is different for workers who attended an equipment safety training compared to those who did not attend. What alternative hypothesis should the manager use to test this claim?

• µattended ≠ µdid not attend
• µattended > µdid not attend
• µattended < µdid not attend
• µattended = µdid not attend

Answer 8

µattended ≠ µdid not attend

The manager has reason to believe that the training has changed the average number of workplace accidents between the two groups of workers. For a two-sided test, the manager should use the alternative hypothesis Ha: µattended ≠ µdid not attend. This is the claim the manger wishes to substantiate.

µattended > µdid not attend

The alternative hypothesis is the claim the manager would like to substantiate. The manager does not want to test whether the average number of workplace accidents has increased for those that attended the training.

µattended < µdid not attend

The alternative hypothesis is the claim the manager would like to substantiate. The manager does not want to test whether the average number of workplace accidents has decreased for those that attended the training. While we may guess that accidents would decrease after training, the manager wishes to test for a change

µattended = µdid not attend

The alternative hypothesis is the claim the manager would like to substantiate. The manager does not want to test whether the average number of workplace accidents has remained the same.

Question 9

A manager of a factory wants to know if the average number of workplace accidents is different for workers who attended an equipment safety training compared to those who did not attend. What null hypothesis should the manager use to test this claim?

• µattended > µdid not attend
• µattended ≥ µdid not attend
• µattended ≤ µdid not attend
• µattended = µdid not attend

Answer 9

µattended > µdid not attend

The null and alternative hypotheses are always opposites. If the manager’s alternative hypothesis is that the average number of workplace accidents has changed, then the null hypothesis is that average number of workplace accidents has remained the same. How would we represent that mathematically?

µattended ≥ µdid not attend

The null and alternative hypotheses are always opposites. If the manager’s alternative hypothesis is that the average number of workplace accidents has changed, then the null hypothesis is that average number of workplace accidents has remained the same. How would we represent that mathematically?

µattended ≤ µdid not attend

The null and alternative hypotheses are always opposites. If the manager’s alternative hypothesis is that the average number of workplace accidents has changed, then the null hypothesis is that average number of workplace accidents has remained the same. How would we represent that mathematically?

µattended = µdid not attend

If the manager’s alternative hypothesis is that the average number of workplace accidents has changed between the two groups of workers, then the null hypothesis is that the average number of accidents has remained the same.

Question 10

If you are performing a hypothesis test based on a 20% significance level, what are your chances of making a type I error?

• 80%
• 10%
• 20%
• It is not possible to tell without more information

Answer 10

80%

The probability of a type I error is equal to the significance level, which is 1–confidence level.

10%

The probability of a type I error is equal to the significance level, which is 1–confidence level.

20%

The probability of a type I error is equal to the significance level, which is 1–confidence level.

It is not possible to tell without more information

The significance level provides the necessary information. The probability of a type I error is equal to the significance level, which is 1–confidence level.

Question 11

If the two-sided p-value of a given sample mean is 0.0040, what is the one-sided p-value for that sample mean?

• 0.0020
• 0.0040
• 0.0080
• The answer cannot be determined without further information

Answer 11

0.0020

The one-sided p-value is half of the two-sided p-value. Since the two-sided p-value is 0.0040, the one-sided p-value is 0.0040/2=0.0020.

0.0040

The one-sided p-value is half of the two-sided p-value.

0.0080

The one-sided p-value is half of the two-sided p-value.

The answer cannot be determined without further information

Since we know the two-sided p-value, we can calculate the one-sided p-value of the sample mean.

Question 12

One-sided or Two-sided hypothesis test?

TEST WHETHER INCOMING STUDENTS AT A BUSINESS SCHOOL RECEIVE BETTER GRADES IN THEIR CLASSES IF THEY’VE TAKEN AN ON-LINE PROGRAM COVERING BASIC MATERIAL

Answer 12

One

Question 13

One-sided or Two-sided hypothesis test?

TEST WHETHER USERS OF A COMMERCIAL WEBSITE ARE LESS LIKELY TO MAKE A PURCHASE IF THEY ARE REQUIRED TO SET UP A USER ACCOUNT ON THE SITE

Answer 13

One

Question 14

One-sided or Two-sided hypothesis test?

TEST WHETHER THERE IS A DIFFERENCE BETWEEN MEN’S AND WOMEN’S USAGE OF A MOBILE FITNESS APP

Answer 14

Two

Question 15

One-sided or Two-sided hypothesis test?

TEST WHETHER THE NUMBER OF LISTENERS OF A STREAMING MUSIC SERVICE HAS CHANGED AFTER THEY CHANGED THE USER INTERFACE

Answer 15

Two

Question 16

A movie theater manager wants to determine whether popcorn sales have increased since the theater switched from using “butter-flavored topping” to real butter. Historically the average popcorn revenue per weekend day was approximately $3,500. After the theater started using real butter, the manager randomly sampled 12 weekend days and calculated the sample’s summary statistics. The average revenue per weekend day in the sample was approximately $4,200 with a standard deviation of $140. Select the function that would correctly calculate the 90% range of likely sample means.

• 3,500±CONFIDENCE.T(0.10,140,12)
• 4,200±CONFIDENCE.T(0.10,140,12)
• 3,500±CONFIDENCE.NORM(0.10,140,12)
• 4,200±CONFIDENCE.NORM(0.10,140,12)

Answer 16

3,500±CONFIDENCE.T(0.10,140,12)

Correct. The range of likely sample means is centered at the historical population mean, in this case $3,500. Because the sample contains fewer than 30 data points, we use CONFIDENCE.T. Excel’s CONFIDENCE.T function syntax is CONFIDENCE.T(alpha, standard_dev, size). Because we wish to construct a 90% range of likely sample means, alpha equals 0.10.

4,200±CONFIDENCE.T(0.10,140,12)

The range of likely sample means is centered at the historical population mean, in this case $3,500, rather than the sample mean, in this case $4,200. Because the sample contains fewer than 30 data points, we use CONFIDENCE.T. Excel’s CONFIDENCE.T function syntax is CONFIDENCE.T(alpha, standard_dev, size). Because we wish to construct a 90% range of likely sample means, alpha equals 0.10.

3,500±CONFIDENCE.NORM(0.10,140,12)

The range of likely sample means is centered at the historical population mean, in this case $3,500. Because the sample contains fewer than 30 data points, we use CONFIDENCE.T, rather than CONFIDENCE. NORM. Excel’s CONFIDENCE.T function syntax is CONFIDENCE.T(alpha, standard_dev, size). Because this is a 90% range of likely sample means, alpha equals 0.10. Excel’s CONFIDENCE.T function syntax is CONFIDENCE.T(alpha, standard_dev, size).

4,200±CONFIDENCE.NORM(0.10,140,12)

The range of likely sample means is centered at the historical population mean, in this case $3,500, rather than the sample mean, in this case $4,200. Furthermore, because the sample contains fewer than 30 cases, we use CONFIDENCE.T rather than CONFIDENCE.NORM. Because this is a 90% range of likely sample means, alpha equals 0.10. Excel’s CONFIDENCE.T function syntax is CONFIDENCE.T(alpha, standard_dev, size).

Question 17

The owner of a local health food store recently started a new ad campaign to attract more business and wants to test whether average daily sales have increased. Historically average daily sales were approximately $2,700. After the ad campaign, the owner took a random sample of forty-five days and found that daily average sales had increased to $2,984. What is store owner’s null hypothesis?

• μ≥$2,984
• μ≤$2,984
• μ=$2,984
• μ≥$2,700
• μ≤$2,700
• μ=$2,700

Answer 17

μ≥$2,984

The null hypothesis is the opposite of the hypothesis you are trying to substantiate. The owner wants to test for an increase. In addition, the null hypothesis is always based on historical information.

μ≤$2,984

The null hypothesis is always based on historical information.

μ=$2,984

The null hypothesis is the opposite of the hypothesis you are trying to substantiate. The owner wants to test for an increase. In addition, the null hypothesis is always based on historical information.

μ≥$2,700

The null hypothesis is the opposite of the hypothesis you are trying to substantiate. The owner wants to test for an increase. This answer would be correct if the owner wanted to test for a decrease.

μ≤$2,700

The null hypothesis is the opposite of the hypothesis you are trying to substantiate. Since the owner wants to test for an increase, the null hypothesis is μ≤$2,700. Remember that the null hypothesis is always based on historical information.

μ=$2,700

The null hypothesis is the opposite of the hypothesis you are trying to substantiate. The owner wants to test for an increase. This answer would be correct if the owner wanted to test for a change in either direction.

Question 18

The owner of a local health food store recently started a new ad campaign to attract more business and wants to test whether average daily sales have increased. Historically average daily sales were approximately $2,700. After the ad campaign, the owner took another random sample of forty-five days and found that average daily sales were $2,984 with a standard deviation of approximately $585. Calculate the upper bound of the 95% range of likely sample means for this one-sided hypothesis test using the CONFIDENCE.NORM function.

Answer 18

To construct a 95% range of likely sample means, calculate the margin of error using the function CONFIDENCE.NORM(alpha, standard_dev, size). However, CONFIDENCE.NORM finds the margin of error for a two-sided hypothesis test and this question asks for the upper bound of a one-sided test. To find the upper bound for the one-sided test you must first determine what two-sided test would have a 5% rejection region on the right side. Since the distribution of sample means is symmetric, a two-sided test with a 10% significance level would have a 5% rejection region on the left side of the normal distribution and a 5% rejection region on the right side. Thus, the upper bound for a two-sided test with alpha=0.1 will be the same as the upper bound on a one-sided test with alpha=0.05. The margin of error is CONFIDENCE.NORM(alpha, standard_dev, size)= CONFIDENCE.NORM(0.1,C3,C4)=CONFIDENCE.NORM(0.1,585,45)=$143.44. The upper bound of the 95% range of likely sample means for this one-sided hypothesis test is the population mean plus the margin of error, which is approximately $2,700+$143.44=$2,843.44.

Question 19

Recall that the owner of a local health food store recently started a new ad campaign to attract more business and wants to know if average daily sales have increased. Historically average daily sales were approximately $2,700. The upper bound of the 95% range of likely sample means for this one-sided test is approximately $2,843.44. If the owner took a random sample of forty-five days and found that daily average sales were now $2,984, what can she conclude at the 95% confidence level?

• Average daily sales have increased
• Average daily sales have decreased
• Average daily sales have remained the same
• Average daily sales have not increased
• The answer cannot be determined without further information

Answer 19

Average daily sales have increased

Since the sample mean, $2,984, falls outside the range of likely sample means (which has an upper bound=$2,843.44), the store owner can reject the null hypothesis that μ≤$2,700μ≤$2,700 at a 95% confidence level. Since she can reject the null hypothesis, she can essentially accept the alternative hypothesis and conclude the average daily sales have increased.

Average daily sales have decreased

Since the sample mean falls outside the range of likely sample means (which has an upper bound=$2,843.44), the store owner can reject the null hypothesis that μ≤$2,700μ≤$2,700 at a 95% confidence level. Therefore, she cannot conclude that average daily sales have decreased.

Average daily sales have remained the same

Since the sample mean falls outside the range of likely sample means (which has an upper bound=$2,843.44), the store owner can reject the null hypothesis that μ≤$2,700μ≤$2,700 at a 95% confidence level. Therefore, she cannot conclude that average daily sales have remained the same.

Average daily sales have not increased

Since the sample mean falls outside the range of likely sample means (which has an upper bound=$2,843.44), the store owner can reject the null hypothesis that μ≤$2,700μ≤$2,700 at a 95% confidence level. If the sample mean had fallen in the range of likely sample means, the owner would have failed to reject the null hypothesis and hence would have insufficient evidence to conclude that average daily sales had increased. Either way, she cannot conclude that average daily sales have not increased, as that would be equivalent to accepting the null hypothesis.

The answer cannot be determined without further information

Since you know the upper bound of the 95% range of likely sample means for this one-sided test, and you know the sample mean, there is enough information to answer this question.

Question 20

A business school professor is interested to know if watching a video about the Central Limit Theorem helps students understand it. To assess this, the professor tests students’ knowledge both immediately before they watch the video and immediately after. The professor takes a sample of students, and for each one compares their test score after the video to their score before the video. Using the data below, calculate the p-value for the following hypothesis test:

Answer 20

The p-value of the one-sided hypothesis test is T.TEST(array1, array2, tails, type)=T.TEST(B2:B31,C2:C31,1,1), which is approximately 0.0128. You must designate this test as a one-sided test (that is, assign the value 1 to the tails argument) and as a type 1 (a paired test) because you are testing the same students on the same knowledge at two points in time. You must link directly to values in order to obtain the correct answer.

Question 21

A college student is interested in testing whether business majors or liberal arts majors are better at trivia. The student gives a trivia quiz to a random sample of 30 business majors and finds the sample’s average score is 86. He gives the same quiz to 30 randomly selected liberal arts majors and finds the sample’s average score is 89. What is the alternative hypothesis of this test?

• μBusiness>μLiberal Arts
• μBusiness=μLiberal Arts
• μBusiness≠μLiberal Arts
• μBusiness<μLiberal Arts

Answer 21

μ Business > μLiberal Arts

This would be the alternative hypothesis if the student wanted to test whether business school majors are better at trivia than liberal arts majors.

μ Business = μLiberal Arts

This is the null hypothesis for testing whether business school and liberal arts school majors’ trivia knowledge is different.

μBusiness ≠ μLiberal Arts

The alternative hypothesis is the claim that is being tested. Since the student wants to test whether there is a difference between business school majors’ and liberal arts majors’ trivia scores, the alternative hypothesis is that the mean scores are not equal.

μBusiness < μLiberal Arts

This would be the alternative hypothesis if the student wanted to test whether liberal arts majors do better at trivia than business school majors.

Question 22

A college student is interested in testing whether business majors or liberal arts majors are better at trivia. The student gives a trivia quiz to a random sample of 30 business school majors and finds the sample’s average score is 86. He gives the same quiz to 30 randomly selected liberal arts majors and finds the sample’s average score is 89. Using the data provided below, calculate the p-value for the following hypothesis test:

• H0: μBusiness = μLiberal Arts
• Ha: μBusiness ≠ μLiberal Arts

Answer 22

The p-value of the two-sided hypothesis test is T.TEST(array1, array2, tails, type)=T.TEST(A2:A31,B2:B31,2,3), which is approximately 0.0524. You must designate this test as a two-sided test (that is, assign the value 2 to the tails argument) and as a type 3 test (an unpaired test with unequal variances) because you are testing two different samples. You must link directly to values in order to obtain the correct answer.

Question 23

A college student is interested in testing whether business majors or liberal arts majors are better at trivia. The student gives a trivia quiz to a random sample of 30 business school majors and finds the sample’s average test score is 86. He gives the same quiz to 30 randomly selected liberal arts majors and finds the sample’s average quiz score is 89. The student finds that the p-value for the hypothesis test equals approximately 0.0524. What can be concluded at α=5%?

• The student should reject the null hypothesis and conclude that there is insufficient evidence of difference between business and liberal arts majors’ knowledge of trivia.
• The student should reject the null hypothesis and conclude that there is a significant difference between business and liberal arts majors’ knowledge of trivia.
• The student should fail to reject the null hypothesis and conclude that there is insufficient evidence of difference between business and liberal arts majors’ knowledge of trivia.
• The student should fail to reject the null hypothesis and conclude that there is a significant difference between business and liberal arts majors’ knowledge of trivia.

Answer 23

The student should reject the null hypothesis and conclude that there is insufficient evidence of difference between business and liberal arts majors’ knowledge of trivia.

Since the p-value, 0.0524, is greater than the significance level, 0.05, the student should fail to reject the null hypothesis.

The student should reject the null hypothesis and conclude that there is a significant difference between business and liberal arts majors’ knowledge of trivia.

Since the p-value, 0.0524, is greater than the significance level, 0.05, the student should fail to reject the null hypothesis.

The student should fail to reject the null hypothesis and conclude that there is insufficient evidence of difference between business and liberal arts majors’ knowledge of trivia.

Since the p-value, 0.0524, is greater than the significance level, 0.05, the student should fail to reject the null hypothesis and conclude that there is insufficient evidence of difference between business and liberal arts majors’ knowledge of trivia. Because the null hypothesis is that there is no difference between the two types of majors, this answer is correct.

The student should fail to reject the null hypothesis and conclude that there is a significant difference between business and liberal arts majors’ knowledge of trivia.

Since the p-value, 0.0524, is greater than the significance level, 0.05, the student should fail to reject the null hypothesis. However, because the null hypothesis is that there is no difference between the two types of majors, the student should conclude that there is insufficient evidence of difference between business and liberal arts majors’ knowledge of trivia. This answer is incorrect.

Question 24

Select the p-value(s) at which you would reject the null hypothesis for a two-sided test at the 90% confidence level. SELECT ALL THAT APPLY.

• 0.9500
• 0.9000
• 0.8900
• 0.1100
• 0.0900
• 0.0500
• 0.0250

Answer 24

0.9500

To reject the null hypothesis at the 90% confidence level, the p-value must be less than 1–0.90=0.10. 0.9500 is not less than 0.10 so we cannot reject the null hypothesis.

0.9000

To reject the null hypothesis at the 90% confidence level, the p-value must be less than 1–0.90=0.10. 0.9000 is not less than 0.10 so we cannot reject the null hypothesis.

0.8900

To reject the null hypothesis at the 90% confidence level, the p-value must be less than 1–0.90=0.10. 0.8900 is not less than 0.10 so we cannot reject the null hypothesis.

0.1100

To reject the null hypothesis at the 90% confidence level, the p-value must be less than 1–0.90=0.10. 0.1100 is not less than 0.10 so we cannot reject the null hypothesis.

0.0900

To reject the null hypothesis at the 90% confidence level, the p-value must be less than 1–0.90=0.10. 0.0900 is less than 0.10 so we can reject the null hypothesis.

0.0500

To reject the null hypothesis at the 90% confidence level, the p-value must be less than 1–0.90=0.10. 0.0500 is less than 0.10 so we can reject the null hypothesis.

0.0250

To reject the null hypothesis at the 90% confidence level, the p-value must be less than 1–0.90=0.10. 0.0250 is less than 0.10 so we can reject the null hypothesis.

Question 25

An engineer designing a new type of bridge wants to test the stress and load bearing capabilities of a prototype before beginning construction. Her null hypothesis is that the bridge’s stress and load capabilities are safe. Select which type of error would be worse. Make sure that the type of error is matched with the correct definition. * Type I; the engineer does not deem the bridge safe even though it is * Type I; the engineer deems the bridge safe and moves onto construction even though it is not actually safe * Type II; the engineer does not deem the bridge safe even though it is * Type II; the engineer deems the bridge safe and moves onto construction even though it is not actually safe

Answer 25

Type I; the engineer does not deem the bridge safe even though it is This is the correct pairing of the error and its definition but it would be worse to proceed with construction of a bridge that is not safe than to presume that a safe bridge is unsafe. Type I; the engineer deems the bridge safe and moves onto construction even though it is not actually safe It would be worse to proceed with construction of a bridge that is not safe but this is not the type I error. Type II; the engineer does not deem the bridge safe even though it is This is not the correct pairing of the error and its definition. Moreover, it would be worse to proceed with construction of a bridge that is not safe than to presume that a safe bridge is unsafe. **Type II; the engineer deems the bridge safe and moves onto construction even though it is not actually safe** **The type II error is that the engineer deems the bridge safe and moves onto construction even though it is not actually safe. This would be worse than presuming that a safe bridge is unsafe.**

Question 26

A streaming music site changed its format to focus on previously unreleased music from rising artists. The site manager now wants to determine whether the number of unique listeners per day has changed. Before the change in format, the site averaged 131,520 unique listeners per day. Now, beginning three months after the format change, the site manager takes a random sample of 30 days and finds that the site has an average of 124,247 unique listeners per day. SELECT THE TWO ANSWERS below that represent the correct null and alternative hypotheses. * μ=124,247 * μ≠124,247 * μ≥124,247 * μ≤124,247 * μ=131,520 * μ≠131,520 * μ≥131,520 * μ≤131,520

Answer 26

μ=124,247 The null and alternative hypotheses are always based on historical information. μ≠124,247 The null and alternative hypotheses are always based on historical information. μ≥124,247 The null and alternative hypotheses are always based on historical information. μ≤124,247 The null and alternative hypotheses are always based on historical information. **μ = 131,520** **The null hypothesis is that number of unique listeners per day has not changed. Thus, μ=131,520μ=131,520 is the null hypothesis.** **μ ≠ 131,520** **The alternative hypothesis is that the number of unique listener per day has changed. Thus, μ≠131,520μ≠131,520 is the alternative hypothesis.** μ≥131,520 The owner wants to test for a change in either direction. Thus, this option can be neither the null nor the alternative hypothesis. μ≤131,520 The owner wants to test for a change in either direction. Thus, this option can be neither the null nor the alternative hypothesis.

Question 27

A streaming music site changed its format to focus on previously unreleased music from rising artists. The site manager now wants to determine whether the number of unique listeners per day has changed. Before the change in format, the site averaged 131,520 unique listeners per day. Now, beginning three months after the format change, the site manager takes a random sample of 30 days and finds that the site now has an average of 124,247 unique listeners per day. Using the data provided below, calculate the p-value for the following hypothesis test: H0:μ=131,520 Ha:μ≠131,520

Answer 27

To use Excel’s T.TEST function for a hypothesis test with one sample, you must create a second column of data that will act as a second sample. So, first enter the historical average unique listeners into each cell in column B associated with a day in the sample; that is, enter 131,520 into cells B2 to B31. Then, the p-value of the two-sided hypothesis test is T.TEST(array1, array2, tails, type)=T.TEST(A2:A31,B2:B31,2,3), which is approximately 0.0743. You must link directly to values in order to obtain the correct answer.

Question 28

A streaming music site changed its format to focus on previously unreleased music from rising artists. The site manager now wants to determine whether the number of unique listeners per day has changed. Before the change in format, the site averaged 131,520 unique listeners per day. Now, beginning three months after the format change, the site manager takes a random sample of 30 days and finds that the site now has an average of 124,247 unique listeners per day. The manager finds that the p-value for the hypothesis test is approximately 0.0743. What can be concluded at the 95% confidence level? * The manager should reject the null hypothesis; there is sufficient evidence that the number of unique daily listeners has likely changed. * The manager should reject the null hypothesis; there is not enough evidence to conclude that the number of unique daily listeners has changed. * The manager should fail to reject the null hypothesis; there is sufficient evidence that the number of unique daily listeners has likely changed. * The manager should fail to reject the null hypothesis; there is not enough evidence to conclude that the number of unique daily listeners has changed.

Answer 28

The manager should reject the null hypothesis; there is sufficient evidence that the number of unique daily listeners has likely changed. Since the p-value, 0.0743, is greater than the significance level, 0.05, the manager should fail to reject the null hypothesis. The manager should reject the null hypothesis; there is not enough evidence to conclude that the number of unique daily listeners has changed. Since the p-value, 0.0743, is greater than the significance level, 0.05, the manager should fail to reject the null hypothesis. The manager should fail to reject the null hypothesis; there is sufficient evidence that the number of unique daily listeners has likely changed. Since the p-value, 0.0743, is greater than the significance level, 0.05, the manager should fail to reject the null hypothesis. By failing to reject the null hypothesis, the manager cannot accept the null hypothesis. Thus, the manager cannot conclude that the number of unique daily listeners has changed. **The manager should fail to reject the null hypothesis; there is not enough evidence to conclude that the number of unique daily listeners has changed.** **Since the p-value, 0.0743, is greater than the significance level, 0.05, the manager should fail to reject the null hypothesis.**

Question 29

A streaming music site changed its format to focus on previously unreleased music from rising artists. The site manager now wants to determine whether the number of unique listeners per day has changed. Before the change in format, the site averaged 131,520 unique listeners per day. Now, beginning three months after the format change, the site manager takes a random sample of 30 days and finds that the site has an average of 124,247 unique listeners per day. The manager finds that the p-value for the hypothesis test is approximately 0.0743. How would you interpret the p-value? * The likelihood that the average number of unique daily listeners per day is no longer 131,520 is 7.43%. * The likelihood that the manager should reject the null hypothesis is 7.43%. * If the average number of unique daily listeners per day is still 131,520, the likelihood of obtaining a sample with a mean at least as extreme as 124,247 is 7.43%. * If the average number of unique daily listeners per day is no longer 131,520, the likelihood of obtaining a sample with a mean at least as extreme as 124,247 is 7.43%.

Answer 29

The likelihood that the average number of unique daily listeners per day is no longer 131,520 is 7.43%. 7.43% is the likelihood of selecting a sample with a mean at least as extreme as 124,247 if the average number of unique daily listeners per day has not changed, that is, if it is still 131,520. There is no information on the likelihood that the mean changed. The likelihood that the manager should reject the null hypothesis is 7.43%. The null hypothesis is rejected if the p-value is less than the significance level. There is no information on the likelihood that the manager should reject the null hypothesis. **If the average number of unique daily listeners per day is still 131,520, the likelihood of obtaining a sample with a mean at least as extreme as 124,247 is 7.43%.** **The null hypothesis is that the average number of unique daily listeners per day has not changed, that is, it is still 131,520. Therefore, the p-value of 0.0743 indicates that if the average number of unique daily listeners is still 131,520, the likelihood of obtaining a sample with a mean at least as extreme as 124,247 is 7.43%%.** If the average number of unique daily listeners per day is no longer 131,520, the likelihood of obtaining a sample with a mean at least as extreme as 124,247 is 7.43%. The p-value is the likelihood of obtaining a sample with a mean at least as extreme as 124,247, given the null hypothesis is true. This statement assumes the null hypothesis is false.

Question 30

If the two-sided p-value of a given sample is 0.0020, what is the one-sided p-value for that sample mean? * 0.0010 * 0.0020 * 0.0040 * The answer cannot be determined without further information

Answer 30

**0.0010** **The one-sided p-value is half of the two-sided p-value. Thus, the one-sided p-value is 0.00202=0.00100.00202=0.0010.** 0.0020 The one-sided p-value is half of the two-sided p-value. 0.0040 The one-sided p-value is half of the two-sided p-value. The answer cannot be determined without further information Since you know the two-sided p-value, you can calculate the one-sided p-value of the sample mean.

Question 31

A food truck operator has traditionally sold 75 bowls of noodle soup each day. He moves to a new location and after a week sees that he has averaged 85 bowls of noodle soup sales each day. He runs a one-sided hypothesis test to determine if his daily sales at the new location have increased. The p-value of the test is 0.031. How should he interpret the p-value? * There is a 3.1% chance that the true mean of soup sales at the new location is 85 bowls a day. * There is a 96.9% chance that the true mean of soup sales at the new location is greater than 75 bowls a day. * There is a 96.9% chance that the sample mean of soup sales at the new location is 85 bowls a day. * There is a 3.1% chance of obtaining a sample with a mean of 85 or higher assuming that the true mean sales at the new location is still equal to or less than 75 bowls a day. * There is a 96.9% chance that the true mean of soup sales at the new location is within 3.1 bowls of 85 bowls a day.

Answer 31

There is a 3.1% chance that the true mean of soup sales at the new location is 85 bowls a day. P-values do not provide information about the likelihood that the mean’s true value is equal to any specific number, such as the point forecast of 85 bowls per day. There is a 96.9% chance that the true mean of soup sales at the new location is greater than 75 bowls a day. The p-value does not provide us with the likelihood of the alternative hypothesis being true; it provides us with the likelihood of the sample being observed if the null hypothesis were true. There is a 96.9% chance that the sample mean of soup sales at the new location is 85 bowls a day. We have already determined that the sample mean at the new location is 85 bowls a day; therefore there is a 100% chance that the sample mean is 85. **There is a 3.1% chance of obtaining a sample with a mean of 85 or higher assuming that the true mean sales at the new location is still equal to or less than 75 bowls a day.** **The p-value provides us with the likelihood of the sample value equal to or more extreme than the observed sample value if the null hypothesis is true. In this case the p-value of 0.031 tells us that there would be a 3.1% chance of the sample value of 85 or above being observed if the null hypothesis were true.** There is a 96.9% chance that the true mean of soup sales at the new location is within 3.1 bowls of 85 bowls a day. The p-value is a probability, not a measure of variability and therefore cannot be used to construct a confidence interval.

Question 32

An automotive manufacturer has developed a new type of tire that the research team believes to increase fuel efficiency. The manufacturer wants to test if there is an increase in the mean gas mileage of mid-sized sedans that use the new type of tire, compared to 32 miles per gallon, the historic mean gas mileage of mid-sized sedans not using the new tires. The automotive manufacturer should perform a _____________ hypothesis test to \_\_\_\_\_\_\_\_\_\_\_\_\_. * one-sided, analyze a change in a single population * two-sided, analyze a change in a single population * one-sided, compare two populations * two-sided, compare two populations

Answer 32

**one-sided, analyze a change in a single population** **The manufacturer believes that the new tires change fuel efficiency in a single direction (i.e., that efficiency increases) and thus should use a one-sided hypothesis test. The automotive manufacturer is analyzing the change of a single population mean compared to the known historic population mean of gas mileage in mid-sized sedans.** two-sided, analyze a change in a single population Two-sided tests are used when there are no a priori predictions or strongly held beliefs about the direction of change. The manufacturer believes that the new tires change fuel efficiency in a single direction and thus should use a one-sided hypothesis test. The automotive manufacturer is analyzing the change of a single population mean compared to the historic population mean of gas mileage in mid-sized sedans. The historic population mean is a stated value and is not considered a separate population. (If we were comparing the fuel efficiency of a sample of cars without the new tires to a sample of cars using the new tires we would use a two population test.) one-sided, compare two populations The manufacturer believes that the new tires change fuel efficiency in a single direction and thus should use a one-sided hypothesis test. The automotive manufacturer is analyzing the change of a single population mean compared to the historic population mean of gas mileage in mid-sized sedans. The historic population mean is not considered a separate population. (If we were comparing the fuel efficiency of a sample of cars without the new tire type to a sample of cars using the new tires we would use a two population test.) two-sided, compare two populations Two-sided tests are used when there are no a priori predictions or strongly held beliefs about the direction of change. The manufacturer believes that the new tires change fuel efficiency in a single direction and thus should use a one-sided hypothesis test. The automotive manufacturer is analyzing the change of a single population mean compared to the historic population mean of gas mileage in mid-sized sedans. The historic population mean is a stated value and is not considered a separate population. (If we were comparing the fuel efficiency of a sample of cars without the new tires to a sample of cars using the new tires we would use a two population test.)

Question 33

Before beginning a hypothesis test, an analyst specified a significance level of 0.10. Which of the following is true? * There is a 90% chance that the alternative hypothesis is true. * There is a 90% chance that the confidence interval will include the true mean of the population. * There is a 10% chance of rejecting the null hypothesis when it is actually true. * There is a 90% chance of rejecting the null hypothesis when it is actually false.

Answer 33

There is a 90% chance that the alternative hypothesis is true. The alternative hypothesis is either true or not true. The significance level specifies how different the observed sample mean has to be from the mean expected under the null hypothesis before we reject the null hypothesis. It doesn’t give us the probability of the null hypothesis being true. There is a 90% chance that the confidence interval will include the true mean of the population. The confidence interval either contains the true mean or it does not. The significance level specifies how different the observed sample mean has to be from the mean expected under the null hypothesis before we reject the null hypothesis. It doesn’t give us a definitive answer about how likely we are to construct a confidence interval that contains the true mean. **There is a 10% chance of rejecting the null hypothesis when it is actually true.** **Correct. The significance level specifies how different the observed sample mean has to be from the mean expected under the null hypothesis before we reject the null hypothesis. A significance level of 0.10 means that the observed sample mean is so different from the mean expected under the null hypothesis that it would only occur 10% of the time if the null hypothesis were true.** There is a 90% chance of rejecting the null hypothesis when it is actually false. The significance level specifies how different the observed sample mean has to be from the mean expected under the null hypothesis before we reject the null hypothesis. A significance level of 0.10 means that the observed sample mean is so different from the mean expected sample that it would only occur 10% of the time if the null hypothesis were true. However, the significance level does not specify anything about what happens if the null hypothesis is false.

Question 34

When performing a hypothesis test based on a 95% confidence level, what are the chances of making a type II error? * 95% * 2.5% * 5% * It is not possible to tell without more information.

Answer 34

**It is not possible to tell without more information.** A type II error occurs when we fail to reject the null hypothesis when the null hypothesis is actually false. The confidence level does not provide any information about the likelihood of making a type II error. Calculating the chances of making a type II error is quite complex and beyond the scope of this course.