
Concept explainers
a.
To construct: The
a.

Answer to Problem 20CQ
Output using the MINITAB software is given below:
Explanation of Solution
Given info:
The data shows the age of a child (x) and the number of cavities (y) values.
Calculation:
Step by step procedure to obtain scatterplot using the MINITAB software:
- Choose Graph > Scatterplot.
- Choose Simple and then click OK.
- Under Y variables, enter a column ofNo.of cavities.
- Under X variables, enter a column ofAge of child.
- Click OK.
b.
To compute: The value of the
b.

Answer to Problem 20CQ
The value of the
Explanation of Solution
Calculation:
Correlation coefficient r:
Software Procedure:
Step-by-step procedure to obtain the ‘correlation coefficient’ using the MINITAB software:
- Select Stat >Basic Statistics > Correlation.
- In Variables, select x and y from the box on the left.
- Click OK.
Output using the MINITAB software is given below:
From the MINITAB output, the value of the correlation is 0.842.
c.
To test: The significance of the correlation coefficient at
c.

Answer to Problem 20CQ
The conclusion is that,there is a linear relation between the age of a child and the number of cavities.
Explanation of Solution
Given info:
The level of significance is
Calculation:
The hypotheses are given below:
Null hypothesis:
That is, there is no linear relation betweenthe age of a child and the number of cavities.
Alternative hypothesis:
That is, there is a linear relation between the age of a child and the number of cavities.
The sample size is 6.
The formula to find the degrees of the freedom is
That is,
From the “TABLE –I: Critical Values for the PPMC”, the critical value for 4 degrees of freedom and
Rejection Rule:
If the absolute value of r is greater than the critical value then reject the null hypothesis.
Conclusion:
From part (b), the value of r is0.842that is the absolute value of r is 0.842.
Here, the absolute value of r is greater than the critical value
That is,
By the rejection rule,reject the null hypothesis.
There is asufficient evidence to support the claim that “there is alinear relation betweenthe age of a child and the number of cavities”.
d.
To find: The regression equation for the given data.
d.

Answer to Problem 20CQ
The regression equation for the given datais
Explanation of Solution
Calculation:
Regression:
Software procedure:
Step by step procedure to obtain the regression equation using the MINITAB software:
- Choose Stat > Regression > Regression.
- In Responses, enter the column ofNo.of cavities.
- In Predictors, enter the column ofAge of child.
- Click OK.
Output using the MINITAB software is given below:
Thus, regression equation for the given datais
e.
To construct: The scatterplot for the variablesthe age of a child and the number of cavitieswith regression line.
e.

Answer to Problem 20CQ
Output using the MINITAB software is given below:
Explanation of Solution
Calculation:
Step by step procedure to obtain scatterplot using the MINITAB software:
- Choose Graph > Scatterplot.
- Choose with line and then click OK.
- Under Y variables, enter a column of No.of cavities.
- Under X variables, enter a column ofAge of child.
- Click OK.
f.
To obtain: The predicted value of the number of cavities for a child of 11.
f.

Answer to Problem 20CQ
Thepredicted value of the number of cavities is 4.143.
Explanation of Solution
Calculation:
Thus, regression equation for the given datais
Substitute x as 11 in the regression equation
Thus, the predicted value of the number of cavities is 4.143.
Want to see more full solutions like this?
Chapter 10 Solutions
ELEMENTARY STATISTICS W/CONNECT >IP<
- A normal distribution has a mean of 50 and a standard deviation of 4. Solve the following three parts? 1. Compute the probability of a value between 44.0 and 55.0. (The question requires finding probability value between 44 and 55. Solve it in 3 steps. In the first step, use the above formula and x = 44, calculate probability value. In the second step repeat the first step with the only difference that x=55. In the third step, subtract the answer of the first part from the answer of the second part.) 2. Compute the probability of a value greater than 55.0. Use the same formula, x=55 and subtract the answer from 1. 3. Compute the probability of a value between 52.0 and 55.0. (The question requires finding probability value between 52 and 55. Solve it in 3 steps. In the first step, use the above formula and x = 52, calculate probability value. In the second step repeat the first step with the only difference that x=55. In the third step, subtract the answer of the first part from the…arrow_forwardIf a uniform distribution is defined over the interval from 6 to 10, then answer the followings: What is the mean of this uniform distribution? Show that the probability of any value between 6 and 10 is equal to 1.0 Find the probability of a value more than 7. Find the probability of a value between 7 and 9. The closing price of Schnur Sporting Goods Inc. common stock is uniformly distributed between $20 and $30 per share. What is the probability that the stock price will be: More than $27? Less than or equal to $24? The April rainfall in Flagstaff, Arizona, follows a uniform distribution between 0.5 and 3.00 inches. What is the mean amount of rainfall for the month? What is the probability of less than an inch of rain for the month? What is the probability of exactly 1.00 inch of rain? What is the probability of more than 1.50 inches of rain for the month? The best way to solve this problem is begin by a step by step creating a chart. Clearly mark the range, identifying the…arrow_forwardClient 1 Weight before diet (pounds) Weight after diet (pounds) 128 120 2 131 123 3 140 141 4 178 170 5 121 118 6 136 136 7 118 121 8 136 127arrow_forward
- Client 1 Weight before diet (pounds) Weight after diet (pounds) 128 120 2 131 123 3 140 141 4 178 170 5 121 118 6 136 136 7 118 121 8 136 127 a) Determine the mean change in patient weight from before to after the diet (after – before). What is the 95% confidence interval of this mean difference?arrow_forwardIn order to find probability, you can use this formula in Microsoft Excel: The best way to understand and solve these problems is by first drawing a bell curve and marking key points such as x, the mean, and the areas of interest. Once marked on the bell curve, figure out what calculations are needed to find the area of interest. =NORM.DIST(x, Mean, Standard Dev., TRUE). When the question mentions “greater than” you may have to subtract your answer from 1. When the question mentions “between (two values)”, you need to do separate calculation for both values and then subtract their results to get the answer. 1. Compute the probability of a value between 44.0 and 55.0. (The question requires finding probability value between 44 and 55. Solve it in 3 steps. In the first step, use the above formula and x = 44, calculate probability value. In the second step repeat the first step with the only difference that x=55. In the third step, subtract the answer of the first part from the…arrow_forwardIf a uniform distribution is defined over the interval from 6 to 10, then answer the followings: What is the mean of this uniform distribution? Show that the probability of any value between 6 and 10 is equal to 1.0 Find the probability of a value more than 7. Find the probability of a value between 7 and 9. The closing price of Schnur Sporting Goods Inc. common stock is uniformly distributed between $20 and $30 per share. What is the probability that the stock price will be: More than $27? Less than or equal to $24? The April rainfall in Flagstaff, Arizona, follows a uniform distribution between 0.5 and 3.00 inches. What is the mean amount of rainfall for the month? What is the probability of less than an inch of rain for the month? What is the probability of exactly 1.00 inch of rain? What is the probability of more than 1.50 inches of rain for the month? The best way to solve this problem is begin by creating a chart. Clearly mark the range, identifying the lower and upper…arrow_forward
- Problem 1: The mean hourly pay of an American Airlines flight attendant is normally distributed with a mean of 40 per hour and a standard deviation of 3.00 per hour. What is the probability that the hourly pay of a randomly selected flight attendant is: Between the mean and $45 per hour? More than $45 per hour? Less than $32 per hour? Problem 2: The mean of a normal probability distribution is 400 pounds. The standard deviation is 10 pounds. What is the area between 415 pounds and the mean of 400 pounds? What is the area between the mean and 395 pounds? What is the probability of randomly selecting a value less than 395 pounds? Problem 3: In New York State, the mean salary for high school teachers in 2022 was 81,410 with a standard deviation of 9,500. Only Alaska’s mean salary was higher. Assume New York’s state salaries follow a normal distribution. What percent of New York State high school teachers earn between 70,000 and 75,000? What percent of New York State high school…arrow_forwardPls help asaparrow_forwardSolve the following LP problem using the Extreme Point Theorem: Subject to: Maximize Z-6+4y 2+y≤8 2x + y ≤10 2,y20 Solve it using the graphical method. Guidelines for preparation for the teacher's questions: Understand the basics of Linear Programming (LP) 1. Know how to formulate an LP model. 2. Be able to identify decision variables, objective functions, and constraints. Be comfortable with graphical solutions 3. Know how to plot feasible regions and find extreme points. 4. Understand how constraints affect the solution space. Understand the Extreme Point Theorem 5. Know why solutions always occur at extreme points. 6. Be able to explain how optimization changes with different constraints. Think about real-world implications 7. Consider how removing or modifying constraints affects the solution. 8. Be prepared to explain why LP problems are used in business, economics, and operations research.arrow_forward
- ged the variance for group 1) Different groups of male stalk-eyed flies were raised on different diets: a high nutrient corn diet vs. a low nutrient cotton wool diet. Investigators wanted to see if diet quality influenced eye-stalk length. They obtained the following data: d Diet Sample Mean Eye-stalk Length Variance in Eye-stalk d size, n (mm) Length (mm²) Corn (group 1) 21 2.05 0.0558 Cotton (group 2) 24 1.54 0.0812 =205-1.54-05T a) Construct a 95% confidence interval for the difference in mean eye-stalk length between the two diets (e.g., use group 1 - group 2).arrow_forwardAn article in Business Week discussed the large spread between the federal funds rate and the average credit card rate. The table below is a frequency distribution of the credit card rate charged by the top 100 issuers. Credit Card Rates Credit Card Rate Frequency 18% -23% 19 17% -17.9% 16 16% -16.9% 31 15% -15.9% 26 14% -14.9% Copy Data 8 Step 1 of 2: Calculate the average credit card rate charged by the top 100 issuers based on the frequency distribution. Round your answer to two decimal places.arrow_forwardPlease could you check my answersarrow_forward
- Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw HillFunctions and Change: A Modeling Approach to Coll...AlgebraISBN:9781337111348Author:Bruce Crauder, Benny Evans, Alan NoellPublisher:Cengage LearningBig Ideas Math A Bridge To Success Algebra 1: Stu...AlgebraISBN:9781680331141Author:HOUGHTON MIFFLIN HARCOURTPublisher:Houghton Mifflin Harcourt
- College AlgebraAlgebraISBN:9781305115545Author:James Stewart, Lothar Redlin, Saleem WatsonPublisher:Cengage LearningAlgebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:CengageAlgebra and Trigonometry (MindTap Course List)AlgebraISBN:9781305071742Author:James Stewart, Lothar Redlin, Saleem WatsonPublisher:Cengage Learning




