Module 3 – Hypothesis Testing Flashcards
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
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.
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
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.
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
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.
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)
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.
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)
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.
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
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.
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
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.
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
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.
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
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.
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
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.
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
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.
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
One
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
One
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
Two
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
Two
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.
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.
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:
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.
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
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.
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
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.
