Introduction to Business Statistics
7th Edition
ISBN: 9781111792374
Author: WEIERS
Publisher: Cengage Learning
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Chapter 7.3, Problem 7.37E
To determine
The new average service life to design a new and more durable pump.
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Consider the time series X₁ and Y₁ = (I – B)² (I – B³)Xt. What transformations were
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Calculate the 90% confidence interval for the population mean difference using the data in the attached image. I need to see where I went wrong.
Microsoft Excel snapshot for random sampling: Also note the formula used for the last
column
02
x✓ fx =INDEX(5852:58551, RANK(C2, $C$2:$C$51))
A
B
1
No.
States
2
1
ALABAMA
Rand No.
0.925957526
3
2
ALASKA
0.372999976
4
3
ARIZONA
0.941323044
5
4 ARKANSAS
0.071266381
Random Sample
CALIFORNIA
NORTH CAROLINA
ARKANSAS
WASHINGTON
G7
Microsoft Excel snapshot for systematic sampling:
xfx INDEX(SD52:50551, F7)
A
B
E
F
G
1
No.
States
Rand No. Random Sample
population
50
2
1 ALABAMA
0.5296685 NEW HAMPSHIRE
sample
10
3
2 ALASKA
0.4493186 OKLAHOMA
k
5
4
3 ARIZONA
0.707914 KANSAS
5
4 ARKANSAS 0.4831379 NORTH DAKOTA
6
5 CALIFORNIA 0.7277162 INDIANA
Random Sample
Sample Name
7
6 COLORADO 0.5865002 MISSISSIPPI
8
7:ONNECTICU 0.7640596 ILLINOIS
9
8 DELAWARE 0.5783029 MISSOURI
525
10
15
INDIANA
MARYLAND
COLORADO
Chapter 7 Solutions
Introduction to Business Statistics
Ch. 7.1 - What is the difference between a continuous...Ch. 7.1 - What is a probability density function and how is...Ch. 7.1 - Why is the total area beneath a probability...Ch. 7.1 - Prob. 7.4ECh. 7.2 - It has been stated that the nomal distribution is...Ch. 7.2 - Prob. 7.6ECh. 7.2 - Prob. 7.7ECh. 7.2 - Prob. 7.8ECh. 7.2 - Prob. 7.9ECh. 7.2 - The Canada Urban Transit Association has reported...
Ch. 7.2 - Prob. 7.11ECh. 7.2 - In 2009, the average charge for tax preparation by...Ch. 7.2 - It has been reported that the average hotel...Ch. 7.2 - The average American family of four spends $5000...Ch. 7.2 - Prob. 7.15ECh. 7.3 - The normal distribution is really a family of...Ch. 7.3 - In the standard normal distribution, approximately...Ch. 7.3 - A Continuous random variable, x, is normally...Ch. 7.3 - Prob. 7.19ECh. 7.3 - Prob. 7.20ECh. 7.3 - Prob. 7.21ECh. 7.3 - Prob. 7.22ECh. 7.3 - Prob. 7.23ECh. 7.3 - Prob. 7.24ECh. 7.3 - Prob. 7.25ECh. 7.3 - Prob. 7.26ECh. 7.3 - For the normal distribution described in Exercise...Ch. 7.3 - Prob. 7.28ECh. 7.3 - For the normal distribution described in Exercise...Ch. 7.3 - For the normal distribution described in Exercise...Ch. 7.3 - For the normal distribution described in Exercise...Ch. 7.3 - Prob. 7.32ECh. 7.3 - It has been reported that households in the West...Ch. 7.3 - Andre is a fearless circus performer who gets shot...Ch. 7.3 - Drying times for newly painted microwave oven...Ch. 7.3 - KleerCo supplies an under-hood, emissionscontrol...Ch. 7.3 - Prob. 7.37ECh. 7.3 - Prob. 7.38ECh. 7.4 - What is the correction for continuity and why is...Ch. 7.4 - Under what circumstances is it permissible to use...Ch. 7.4 - In a certain binomial distribution, =0.25 and n=40...Ch. 7.4 - In a certain binomial distribution, =0.30 and n=20...Ch. 7.4 - Prob. 7.43ECh. 7.4 - Prob. 7.44ECh. 7.4 - Prob. 7.45ECh. 7.4 - Prob. 7.46ECh. 7.4 - Prob. 7.47ECh. 7.4 - Repeat Exercise 7.46, but without using the...Ch. 7.5 - What is the relationship between the Poisson...Ch. 7.5 - Every day, drivers arrive at a tollbooth. If the...Ch. 7.5 - The main switchboard at the Home Shopping Network...Ch. 7.5 - A random variable is Poision distributed with =1.5...Ch. 7.5 - A random variable is Poision distributed with...Ch. 7.5 - A random variable is Poision distributed with...Ch. 7.5 - The owner of a self-service carwash has found that...Ch. 7.5 - A taxi dispatcher has found that successive calls...Ch. 7.5 - During 2008, U.S. general aviation pilots had 1.20...Ch. 7.5 - The Bureau of Labor Statistics reports the...Ch. 7.6 - A computer statistical package has simulated 1000...Ch. 7.6 - Prob. 7.60ECh. 7.6 - Prob. 7.61ECh. 7.6 - Prob. 7.62ECh. 7 - According to a recent survey, 35% of the adult...Ch. 7 - Prob. 7.64CECh. 7 - Prob. 7.65CECh. 7 - During an annual heating season, the average gas...Ch. 7 - It has been reported that the average monthly cell...Ch. 7 - During 2004, the average number of flying hours...Ch. 7 - During fiscal 2008, the average daily volume for...Ch. 7 - For itemized tax returns in the $60,000—$75,000...Ch. 7 - Of the 1.0 million persons of voting age in Maine,...Ch. 7 - A public relations agency tells its client that...Ch. 7 - Given the results of Exercise 7.72, evaluate the...Ch. 7 - A researcher is studying a particular process and...Ch. 7 - A safety researcher has found that fourth to sixth...Ch. 7 - The number of defects in rolls of aluminum sheet...Ch. 7 - The mileage death rate for motorcycle riders has...Ch. 7 - The “20-ounce” package of mozzarella cheese...Ch. 7 - State Police Academy graduates from a large gradua...Ch. 7 - The trainer for a professional football team has...Ch. 7 - The precooked weight of hamburgers at a gourmet...Ch. 7 - Boxes are filled with sugar by a machine that is...Ch. 7 - In Exercise 7.82, the company would like to put in...Ch. 7 - Discount Micros, a computer mail-order company,...Ch. 7 - The U-Drive car rental corporation has found that...Ch. 7 - Ronnie Rogers, the customer relations manager for...Ch. 7 - A storage warehouse in a relatively remote part of...
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- Suppose the Internal Revenue Service reported that the mean tax refund for the year 2022 was $3401. Assume the standard deviation is $82.5 and that the amounts refunded follow a normal probability distribution. Solve the following three parts? (For the answer to question 14, 15, and 16, start with making a bell curve. Identify on the bell curve where is mean, X, and area(s) to be determined. 1.What percent of the refunds are more than $3,500? 2. What percent of the refunds are more than $3500 but less than $3579? 3. What percent of the refunds are more than $3325 but less than $3579?arrow_forwardA 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_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 127arrow_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 8 136 127 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_forward
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