(a)
To graph: A stem-and-leaf plot for the data of the age distribution.
(a)

Explanation of Solution
The data shows the ages of 50 drivers arrested while driver under the influence of alcohol.
Graph: To construct stem-and-leaf plot by using the Minitab, the steps are as follows:
Step 1: Enter the data in C1.
Step 2: Go to Graph > Stem-and-Leaf plot and select ‘C1’ in Graph variable.
Step 3: Click on OK.
The Stem-and-Leaf plot for these data is obtained as:
Stem-and-leaf of the age distribution N = 50 | ||
1 | 6 8 | |
2 | 0 1 1 2 2 2 3 4 4 5 6 6 6 7 7 7 9 | |
3 | 0 0 1 1 2 3 4 4 5 5 6 7 8 9 | |
4 | 0 0 1 3 5 6 7 7 9 9 | |
5 | 1 3 5 6 8 | |
6 | 3 4 |
(b)
To find: The frequency table for the data..
(b)

Answer to Problem 8CR
Solution: The complete frequency table is as:
Class Limits | Class boundaries | Midpoints | Frequency | Relative Frequency | Cumulative Frequency |
16-22 | 15.5-22.5 | 19 | 8 | 0.16 | 8 |
23-29 | 22.5-29.5 | 26 | 11 | 0.22 | 19 |
30-36 | 29.5-36.5 | 33 | 11 | 0.22 | 30 |
37-43 | 36.5-43.5 | 40 | 7 | 0.14 | 37 |
44-50 | 43.5-50.5 | 47 | 6 | 0.12 | 43 |
51-57 | 50.5-57.5 | 54 | 4 | 0.08 | 47 |
58-64 | 57.5-64.5 | 61 | 3 | 0.06 | 50 |
Explanation of Solution
Calculation: To find the class width for the whole data of 50 values, it is observed that largest value of the data set is 64 and the smallest value is 16 in the data. Using 7 classes, the class width calculated in the following way:
The value is round up to the nearest whole number. Hence, the class width of the data set is 7. The class width for the data is 7 and the lowest data value (16) will be the lower class limit of the first class. Because the class width is 7, it must add 7 to the lowest class limit in the first class to find the lowest class limit in the second class. There are 7 desired classes. Hence, the class limits are 16-22, 23-29, 30-36, 37-43, 44-50, 51-57, and 58-64. Now, to find the class boundaries subtract 0.5 from lower limit of every class and add 0.5 to the upper limit of the every class interval. Hence, the class boundaries are 15.5-22.5, 22.5-29.5, 29.5-36.5, 36.5-43.5, 43.5-50.5, 50.5-57.5, and 57.5-64.5.
Next to find the midpoint of the class is calculated by using formula,
Midpoint of first class is calculated as:
The frequencies for respective classes are 8, 11, 11, 7, 6, 4 and 3.
Relative frequency is calculated by using the formula,
The frequency for 1st class is 8 and total frequencies are 50 so the relative frequency is
The calculated frequency table is as follows:
Class limits | Class boundaries | midpoints | Frequency | Relative Frequency | Cumulative Frequency |
16-22 | 15.5-22.5 | 19 | 8 | 0.16 | 8 |
23-29 | 22.5-29.5 | 26 | 11 | 0.22 | |
30-36 | 29.5-36.5 | 33 | 11 | 0.22 | |
37-43 | 36.5-43.5 | 40 | 7 | 0.14 | |
44-50 | 43.5-50.5 | 47 | 6 | 0.12 | |
51-57 | 50.5-57.5 | 54 | 4 | 0.08 | |
58-64 | 57.5-64.5 | 61 | 3 | 0.06 |
Interpretation: Hence, the complete frequency table is as:
Class limits | Class boundaries | Midpoints | Frequency | Relative Frequency | Cumulative Frequency |
16-22 | 15.5-22.5 | 19 | 8 | 0.16 | 8 |
23-29 | 22.5-29.5 | 26 | 11 | 0.22 | |
30-36 | 29.5-36.5 | 33 | 11 | 0.22 | |
37-43 | 36.5-43.5 | 40 | 7 | 0.14 | |
44-50 | 43.5-50.5 | 47 | 6 | 0.12 | |
51-57 | 50.5-57.5 | 54 | 4 | 0.08 | |
58-64 | 57.5-64.5 | 61 | 3 | 0.06 |
(c)
To graph: A histogram for the data of the age distribution.
(c)

Explanation of Solution
The data shows the ages of 50 drivers arrested while driver under the influence of alcohol.
Graph: To construct the histogram by using the MINITAB, the steps are as follows:
Step 1: Enter the class boundaries in C1 and frequency in C2.
Step 2: Go to Graph > Histogram > Simple.
Step 3: Enter C1 in Graph variable then go to Data options > Frequency > C2.
Step 4: Click on OK.
The obtained histogram is
(d)
The shape of the histogram of age distribution..
(d)

Answer to Problem 8CR
Solution: The shape of histogram of age distribution is skewed to the right.
Explanation of Solution
A right-skewed distribution has a long right tail. Right-skewed distributions are also called positive-skew distributions. That’s because there is a long tail in the positive direction on the number line.
From above histogram, there are two class boundaries (22.5-29.5 and 29.5-36) have higher frequencies 11 on left side and most of the data values fall on the left side of the graph. The data to lean towards right side of the graph and also there is tail on right side.
Hence, the shape of histogram of age distribution is skewed to the right.
Want to see more full solutions like this?
Chapter 2 Solutions
EBK UNDERSTANDING BASIC STATISTICS
- ian income of $50,000. erty rate of 13. Using data from 50 workers, a researcher estimates Wage = Bo+B,Education + B₂Experience + B3Age+e, where Wage is the hourly wage rate and Education, Experience, and Age are the years of higher education, the years of experience, and the age of the worker, respectively. A portion of the regression results is shown in the following table. ni ogolloo bash 1 Standard Coefficients error t stat p-value Intercept 7.87 4.09 1.93 0.0603 Education 1.44 0.34 4.24 0.0001 Experience 0.45 0.14 3.16 0.0028 Age -0.01 0.08 -0.14 0.8920 a. Interpret the estimated coefficients for Education and Experience. b. Predict the hourly wage rate for a 30-year-old worker with four years of higher education and three years of experience.arrow_forward1. If a firm spends more on advertising, is it likely to increase sales? Data on annual sales (in $100,000s) and advertising expenditures (in $10,000s) were collected for 20 firms in order to estimate the model Sales = Po + B₁Advertising + ε. A portion of the regression results is shown in the accompanying table. Intercept Advertising Standard Coefficients Error t Stat p-value -7.42 1.46 -5.09 7.66E-05 0.42 0.05 8.70 7.26E-08 a. Interpret the estimated slope coefficient. b. What is the sample regression equation? C. Predict the sales for a firm that spends $500,000 annually on advertising.arrow_forwardCan you help me solve problem 38 with steps im stuck.arrow_forward
- How do the samples hold up to the efficiency test? What percentages of the samples pass or fail the test? What would be the likelihood of having the following specific number of efficiency test failures in the next 300 processors tested? 1 failures, 5 failures, 10 failures and 20 failures.arrow_forwardThe battery temperatures are a major concern for us. Can you analyze and describe the sample data? What are the average and median temperatures? How much variability is there in the temperatures? Is there anything that stands out? Our engineers’ assumption is that the temperature data is normally distributed. If that is the case, what would be the likelihood that the Safety Zone temperature will exceed 5.15 degrees? What is the probability that the Safety Zone temperature will be less than 4.65 degrees? What is the actual percentage of samples that exceed 5.25 degrees or are less than 4.75 degrees? Is the manufacturing process producing units with stable Safety Zone temperatures? Can you check if there are any apparent changes in the temperature pattern? Are there any outliers? A closer look at the Z-scores should help you in this regard.arrow_forwardNeed help pleasearrow_forward
- Please conduct a step by step of these statistical tests on separate sheets of Microsoft Excel. If the calculations in Microsoft Excel are incorrect, the null and alternative hypotheses, as well as the conclusions drawn from them, will be meaningless and will not receive any points. 4. One-Way ANOVA: Analyze the customer satisfaction scores across four different product categories to determine if there is a significant difference in means. (Hints: The null can be about maintaining status-quo or no difference among groups) H0 = H1=arrow_forwardPlease conduct a step by step of these statistical tests on separate sheets of Microsoft Excel. If the calculations in Microsoft Excel are incorrect, the null and alternative hypotheses, as well as the conclusions drawn from them, will be meaningless and will not receive any points 2. Two-Sample T-Test: Compare the average sales revenue of two different regions to determine if there is a significant difference. (Hints: The null can be about maintaining status-quo or no difference among groups; if alternative hypothesis is non-directional use the two-tailed p-value from excel file to make a decision about rejecting or not rejecting null) H0 = H1=arrow_forwardPlease conduct a step by step of these statistical tests on separate sheets of Microsoft Excel. If the calculations in Microsoft Excel are incorrect, the null and alternative hypotheses, as well as the conclusions drawn from them, will be meaningless and will not receive any points 3. Paired T-Test: A company implemented a training program to improve employee performance. To evaluate the effectiveness of the program, the company recorded the test scores of 25 employees before and after the training. Determine if the training program is effective in terms of scores of participants before and after the training. (Hints: The null can be about maintaining status-quo or no difference among groups; if alternative hypothesis is non-directional, use the two-tailed p-value from excel file to make a decision about rejecting or not rejecting the null) H0 = H1= Conclusion:arrow_forward
- Please conduct a step by step of these statistical tests on separate sheets of Microsoft Excel. If the calculations in Microsoft Excel are incorrect, the null and alternative hypotheses, as well as the conclusions drawn from them, will be meaningless and will not receive any points. The data for the following questions is provided in Microsoft Excel file on 4 separate sheets. Please conduct these statistical tests on separate sheets of Microsoft Excel. If the calculations in Microsoft Excel are incorrect, the null and alternative hypotheses, as well as the conclusions drawn from them, will be meaningless and will not receive any points. 1. One Sample T-Test: Determine whether the average satisfaction rating of customers for a product is significantly different from a hypothetical mean of 75. (Hints: The null can be about maintaining status-quo or no difference; If your alternative hypothesis is non-directional (e.g., μ≠75), you should use the two-tailed p-value from excel file to…arrow_forwardPlease conduct a step by step of these statistical tests on separate sheets of Microsoft Excel. If the calculations in Microsoft Excel are incorrect, the null and alternative hypotheses, as well as the conclusions drawn from them, will be meaningless and will not receive any points. 1. One Sample T-Test: Determine whether the average satisfaction rating of customers for a product is significantly different from a hypothetical mean of 75. (Hints: The null can be about maintaining status-quo or no difference; If your alternative hypothesis is non-directional (e.g., μ≠75), you should use the two-tailed p-value from excel file to make a decision about rejecting or not rejecting null. If alternative is directional (e.g., μ < 75), you should use the lower-tailed p-value. For alternative hypothesis μ > 75, you should use the upper-tailed p-value.) H0 = H1= Conclusion: The p value from one sample t-test is _______. Since the two-tailed p-value is _______ 2. Two-Sample T-Test:…arrow_forwardPlease conduct a step by step of these statistical tests on separate sheets of Microsoft Excel. If the calculations in Microsoft Excel are incorrect, the null and alternative hypotheses, as well as the conclusions drawn from them, will be meaningless and will not receive any points. What is one sample T-test? Give an example of business application of this test? What is Two-Sample T-Test. Give an example of business application of this test? .What is paired T-test. Give an example of business application of this test? What is one way ANOVA test. Give an example of business application of this test? 1. One Sample T-Test: Determine whether the average satisfaction rating of customers for a product is significantly different from a hypothetical mean of 75. (Hints: The null can be about maintaining status-quo or no difference; If your alternative hypothesis is non-directional (e.g., μ≠75), you should use the two-tailed p-value from excel file to make a decision about rejecting or not…arrow_forward
- Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw Hill
