Concept explainers
Videos
a.
Construct box plot of the variable price.
Identify whether there are outliers or not.
Find the
Find the first
Find the third quartile value.
a.
Answer to Problem 37CE
Output of box plot for the variable price using MINITAB software is,
Yes, there are 3 outliers in the dataset.
The median price is 3,733.
The first quartile value is 1,478.
The third quartile value is 6,141.
Explanation of Solution
Calculation:
Step by step procedure to obtain boxplot using MINITAB software is given as,
- Choose Graph > Boxplot.
- In Graph variables enter the columns Price.
- Click OK.
Outliers:
In the boxplot, the outlier is represented using asterisk. In the boxplot of data set there are 3 asterisks representing outliers. Hence, there are three outliers in the dataset.
Median:
The median is the middle value of the data set. In the boxplot, the line in middle of the box represents median of the dataset. The line corresponds to value 3,733.
Hence, the median value is 3,733.
First quartile:
The border line towards the left side of the box represents the value of first quartile. In this box plot, the line of the box on left side corresponds to the value approximately 1,478.
Hence, the third quartile value is 6,141.
Third quartile:
The border line towards the right side of the box represents the value of third quartile. In this box plot, the line of the box on right side corresponds to the value approximately 6,141.
Hence, the first quartile value is 1,478.
b.
Construct box plot of the variable size.
Identify whether there are outliers or not.
Find the median price.
Find the first quartile value.
Find the third quartile value.
b.
Answer to Problem 37CE
Output of box plot for the variable size using MINITAB software is,
Yes, there are 3 outliers in the dataset.
The median price is 0.84.
The first quartile value is 0.515.
The third quartile value is 1.12.
Explanation of Solution
Calculation:
Step by step procedure to obtain boxplot using MINITAB software is given as,
- Choose Graph > Boxplot.
- In Graph variables enter the columns Size.
- Click OK.
Outliers:
In the boxplot, the outlier is represented using asterisk. In the boxplot of data set there are 3 asterisks representing outliers. Hence, there are three outliers in the dataset.
Median:
The median is the middle value of the data set. In the boxplot, the line in middle of the box represents median of the dataset. The line corresponds to value 0.84.
Hence, the median value is 0.84.
First quartile:
The border line towards the left side of the box represents the value of first quartile. In this box plot, the line of the box on left side corresponds to the value approximately 0.515.
Hence, the third quartile value is 0.515.
Third quartile:
The border line towards the right side of the box represents the value of third quartile. In this box plot, the line of the box on right side corresponds to the value approximately 1.12.
Hence, the first quartile value is 1.12.
c.
Construct
Identify whether there is association between the two variables or not.
Identify whether association is direct or indirect.
Identify whether any point seems to be different from the others.
c.
Answer to Problem 37CE
Output of scatter diagram for variables price and size using MINITAB software is,
Yes, there is association between the variables price and size.
The association is direct.
Yes, the first observation of both the price and size is large when compared to other observations.
Explanation of Solution
Calculation:
Step by step procedure to obtain scatter diagram using MINITAB software is given as,
- Choose Graph > Scatterplot > select Simple.
- In Y variable enter the column Price.
- In X variable enter the column Size.
- Click OK.
In the scatter diagram it can be observed that, the Price has increased as the Size increases indicating that the association between the variables.
Hence, there is association between the variables price and size
The relation is said to be direct if value of one variable increases due to effect of another variable. From the scatter diagram, the value of Price has increased as the Size increases indicating a direct or positive association.
Hence, the association is direct.
From the scatter diagram, it can be observed that one of the observations corresponding to the value of 5.03 carats for size and $44,312 for price is far from all the other observations. Hence, one point seems to be different from the others.
d.
Construct a
Find the most common cut grade.
Find the most common shape.
Find the most common combination of cut grade and shape.
d.
Answer to Problem 37CE
The contingency table for the variables shape and cut grade is,
| Shape | Cut Grade | |||||
| Average | Good | Ideal | Premium | Ultra Ideal | Total | |
| Emerald | 0 | 0 | 1 | 0 | 0 | 1 |
| Marquise | 0 | 2 | 0 | 1 | 0 | 3 |
| Oval | 0 | 0 | 0 | 1 | 0 | 1 |
| Princess | 1 | 0 | 2 | 2 | 0 | 5 |
| Round | 1 | 3 | 3 | 13 | 3 | 23 |
| Total | 2 | 5 | 6 | 17 | 3 | 33 |
The most common cut grade is premium.
The most common shape is round.
The most common combination of cut grade and shape is premium and round.
Explanation of Solution
Calculation:
Contingency table:
A table that is used for classifying observations based on the two identifiable characteristics is termed as contingency table. It is used for summarizing two variables.
The variable cut grade is classified into 5 different categories ‘average, good, ideal, premium, ultra ideal’. The variable shape is classified into 5 different categories ‘emerald, marquise, oval, princess, and round’.
Count the number of cut grades are average with shape of emerald. From the data, there is no combination of average cut grades with shape of emerald. Hence, the frequency is 0.
Similarly, count the frequency for each of the possible combination of cut grade and shape. Then calculate the totals for each column and row. The contingency table is obtained as below,
| Shape | Cut Grade | |||||
| Average | Good | Ideal | Premium | Ultra Ideal | Total | |
| Emerald | 0 | 0 | 1 | 0 | 0 | 1 |
| Marquise | 0 | 2 | 0 | 1 | 0 | 3 |
| Oval | 0 | 0 | 0 | 1 | 0 | 1 |
| Princess | 1 | 0 | 2 | 2 | 0 | 5 |
| Round | 1 | 3 | 3 | 13 | 3 | 23 |
| Total | 2 | 5 | 6 | 17 | 3 | 33 |
The cut grade ‘Premium’ has a total of 17, which is large when compared to other cut grades. This shows that, the most common cut grade of diamonds is ‘Premium.
Hence, the most common cut grade is premium.
The shape ‘Round’ has a total of 23, which is large when compared to other shapes. This shows that, the most common shape of diamonds is ‘Round’.
Hence, the most common shape is round.
The combination of cut grade ‘Premium’ and shape ‘Round’ has a total of 13, which is large when compared to other combinations. This shows that, the most common combination of diamonds is cut grade ‘Premium’ and shape ‘Round’.
Hence, the most common combination of cut grade and shape is premium and round.
Want to see more full solutions like this?
Chapter 4 Solutions
Loose Leaf for Statistical Techniques in Business and Economics
- Suppose you are gambling on a roulette wheel. Each time the wheel is spun, the result is one of the outcomes 0, 1, and so on through 36. Of these outcomes, 18 are red, 18 are black, and 1 is green. On each spin you bet $5 that a red outcome will occur and $1 that the green outcome will occur. If red occurs, you win a net $4. (You win $10 from red and nothing from green.) If green occurs, you win a net $24. (You win $30 from green and nothing from red.) If black occurs, you lose everything you bet for a loss of $6. a. Use simulation to generate 1,000 plays from this strategy. Each play should indicate the net amount won or lost. Then, based on these outcomes, calculate a 95% confidence interval for the total net amount won or lost from 1,000 plays of the game. (Round your answers to two decimal places and if your answer is negative value, enter "minus" sign.) I worked out the Upper Limit, but I can't seem to arrive at the correct answer for the Lower Limit. What is the Lower Limit?…arrow_forwardLet us suppose we have some article reported on a study of potential sources of injury to equine veterinarians conducted at a university veterinary hospital. Forces on the hand were measured for several common activities that veterinarians engage in when examining or treating horses. We will consider the forces on the hands for two tasks, lifting and using ultrasound. Assume that both sample sizes are 6, the sample mean force for lifting was 6.2 pounds with standard deviation 1.5 pounds, and the sample mean force for using ultrasound was 6.4 pounds with standard deviation 0.3 pounds. Assume that the standard deviations are known. Suppose that you wanted to detect a true difference in mean force of 0.25 pounds on the hands for these two activities. Under the null hypothesis, 40 0. What level of type II error would you recommend here? = Round your answer to four decimal places (e.g. 98.7654). Use α = 0.05. β = 0.0594 What sample size would be required? Assume the sample sizes are to be…arrow_forwardConsider the hypothesis test Ho: 0 s² = = 4.5; s² = 2.3. Use a = 0.01. = σ against H₁: 6 > σ2. Suppose that the sample sizes are n₁ = 20 and 2 = 8, and that (a) Test the hypothesis. Round your answers to two decimal places (e.g. 98.76). The test statistic is fo = 1.96 The critical value is f = 6.18 Conclusion: fail to reject the null hypothesis at a = 0.01. (b) Construct the confidence interval on 02/2/622 which can be used to test the hypothesis: (Round your answer to two decimal places (e.g. 98.76).) 035arrow_forward
- Using the method of sections need help solving this please explain im stuckarrow_forwardPlease solve 6.31 by using the method of sections im stuck and need explanationarrow_forwarda) When two variables are correlated, can the researcher be sure that one variable causes the other? If YES , why? If NO , why? b) What is meant by the statement that two variables are related? Discuss.arrow_forward
- SCIE 211 Lab 3: Graphing and DataWorksheetPre-lab Questions:1. When should you use each of the following types of graphs? Fill answers in the table below.Type of Graph Used to showLine graphScatter plotBar graphHistogramPie Chart2. Several ways in which we can be fooled or misled by a graph were identified in the Lab 3Introduction. Find two examples of misleading graphs on the Internet and paste them below. Besure to identify why each graph is misleading. Data Charts:Circumference vs. Diameter for circular objectsDiameter Can 1 (cm) Can 2 (cm) Can 3 (cm)Trial 1Trial 2Trial 3MeanCircumference Can 1 (cm) Can 2 (cm) Can 3 (cm)Trial 1Trial 2Trial 3MeanScatter Plot Graph – Circumference Vs. DiameterIdentify 2 points of the Trendline.Y1 = ________ Y2 = _________X1 = ________ X2 = _________Calculate the Slope of the Trendline = Post-lab Questions:1. Answer the questions below. You will need to use the following equation to answer…arrow_forwardThe U.S. Bureau of Labor Statistics reports that 11.3% of U.S. workers belong to unions (BLS website, January 2014). Suppose a sample of 400 U.S. workers is collected in 2014 to determine whether union efforts to organize have increased union membership. a. Formulate the hypotheses that can be used to determine whether union membership increased in 2014.H 0: p H a: p b. If the sample results show that 52 of the workers belonged to unions, what is the p-value for your hypothesis test (to 4 decimals)?arrow_forwardA company manages an electronic equipment store and has ordered 200200 LCD TVs for a special sale. The list price for each TV is $200200 with a trade discount series of 6 divided by 10 divided by 2.6/10/2. Find the net price of the order by using the net decimal equivalent.arrow_forward
- According to flightstats.com, American Airlines flights from Dallas to Chicago are on time 80% of the time. Suppose 10 flights are randomly selected, and the number of on-time flights is recorded. (a) Explain why this is a binomial experiment. (b) Determine the values of n and p. (c) Find and interpret the probability that exactly 6 flights are on time. (d) Find and interpret the probability that fewer than 6 flights are on time. (e) Find and interpret the probability that at least 6 flights are on time. (f) Find and interpret the probability that between 4 and 6 flights, inclusive, are on time.arrow_forwardShow how you get critical values of 1.65, -1.65, and $1.96 for a right-tailed, left- tailed, and two-tailed hypothesis test (use a = 0.05 and assume a large sample size).arrow_forwardSuppose that a sports reporter claims the average football game lasts 3 hours, and you believe it's more than that. Your random sample of 35 games has an average time of 3.25 hours. Assume that the population standard deviation is 1 hour. Use a = 0.05. What do you conclude?arrow_forward
Big Ideas Math A Bridge To Success Algebra 1: Stu...AlgebraISBN:9781680331141Author:HOUGHTON MIFFLIN HARCOURTPublisher:Houghton Mifflin Harcourt
Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw Hill
Holt Mcdougal Larson Pre-algebra: Student Edition...AlgebraISBN:9780547587776Author:HOLT MCDOUGALPublisher:HOLT MCDOUGAL