
Concept explainers
If the null hypothesis is rejected in Exercises 1 through 8, use the Scheffé test when the
a. State the hypotheses and identify the claim.
b. Find the critical value(s).
c. Compute the test value.
d. Make the decision.
e. Summarize the results.
Use the traditional method of hypothesis testing unless otherwise specified.
1. Lengths of Various Types of Bridges The data represent the lengths in feet of three types of bridges in the United States. At a = 0.01, test the claim that there is no significant difference in the means of the lengths of the types of bridges.
Simple truss | Segmented concrete | Continuous plate |
745 | 820 | 630 |
716 | 750 | 573 |
700 | 790 | 525 |
650 | 674 | 510 |
647 | 660 | 480 |
625 | 640 | 460 |
608 | 636 | 451 |
598 | 620 | 450 |
550 | 520 | 450 |
545 | 450 | 425 |
534 | 392 | 420 |
528 | 370 | 360 |
a.

To state: The hypothesis
To identify: The claim.
Answer to Problem 12.1.1RE
The hypothesis
The claim is that, all the means are same.
Explanation of Solution
Given info:
The data shows the lengths in feet of three types of bridges in the United States. The level of significance is 0.01.
Calculation:
The hypotheses are given below:
Null hypothesis:
Alternative hypothesis:
Here, all the means are same is tested. Hence, the claim is that, all the means are same.
b.

To find: The critical value.
Answer to Problem 12.1.1RE
The critical value is 5.285.
Explanation of Solution
Given info:
The level of significance is 0.01. The number of samples k is 3, the sample sizes
Calculation:
The degrees of freedom are
Where
Substitute 3 for k in
Substitute 36 for N and 3 for k in
Critical value:
The critical F-value is obtained using the Table H: The F-Distribution with the level of significance
Procedure:
- Locate 30 and 40 in the degrees of freedom, denominator row of the Table H.
- Obtain the value in the corresponding degrees of freedom, numerator column below 2.
That is,
That is, the critical value is 5.285.
Rejection region:
The null hypothesis would be rejected if
c.

To compute: The test value.
Answer to Problem 12.1.1RE
The test value is 6.94.
Explanation of Solution
Calculation:
Software procedure:
Step-by-step procedure to obtain the test statistic using the MINITAB software:
- Choose Stat > ANOVA > One-Way.
- In Response, enter the Temperatures.
- In Factor, enter the Factor.
- Click OK.
Output using the MINITAB software is given below:
From the MINITAB output, the test value F is 6.94.
d.

To make: The decision.
Answer to Problem 12.1.1RE
The null hypothesis is rejected.
Explanation of Solution
Conclusion:
From the result of part (c), the test value is 6.94.
Here, the F-statistic value is greater than the critical value.
That is,
Thus, it can be concluding that, the null hypothesis is rejected.
e.

To explain: The results.
Answer to Problem 12.1.1RE
The result concludes that, there is a significant difference between the means
Explanation of Solution
Calculation:
From the results, it can be observed that the null hypothesis is rejected. Thus, it can be concluding that there is evidence to reject the claim that all means are same.
Consider,
Step-by-step procedure to obtain the test mean and standard deviation using the MINITAB software:
- Choose Stat > Basic Statistics > Display Descriptive Statistics.
- In Variables enter the columns Florida, Pennsylvania and Maine.
- Choose option statistics, and select Mean, Variance and N total.
- Click OK.
Output using the MINITAB software is given below:
The sample sizes
The means are
The sample variances are
Here, the samples of sizes of three states are equal. So, the test used here is Tukey test.
Tukey test:
Critical value:
Here, k is 3 and degrees of freedom
Substitute 36 for N and 3 for k in v
The critical F-value is obtained using the Table N: Critical Values for the Tukey test with the level of significance
Procedure:
- Locate nearest value of 33 in the column of v of the Table H.
- Obtain the value in the corresponding row below 3.
That is, the critical value is 4.45.
Comparison of the means:
The formula for finding
That is,
Comparison between the means
The hypotheses are given below:
Null hypothesis:
Alternative hypothesis:
Rejection region:
The null hypothesis would be rejected if absolute value greater than the critical value.
Absolute value:
The formula for comparing the means
Substitute 620.5 and 610.2 for
Thus, the value of
Hence, the absolute value of
Conclusion:
The absolute value is 0.34.
Here, the absolute value is lesser than the critical value.
That is,
Thus, the null hypothesis is not rejected.
Hence, there is no significant difference between the means
Comparison between the means
The hypotheses are given below:
Null hypothesis:
Alternative hypothesis:
Rejection region:
The null hypothesis would be rejected if absolute value greater than the critical value.
Absolute value:
The formula for comparing the means
Substitute 620.5 and 477.8 for
Thus, the value of
Hence, the absolute value of
Conclusion:
The absolute value is 4.72.
Here, the absolute value is greater than the critical value.
That is,
Thus, the null hypothesis is rejected.
Hence, there is significant difference between the means
Comparison between the means
The hypotheses are given below:
Null hypothesis:
Alternative hypothesis:
Rejection region:
The null hypothesis would be rejected if absolute value greater than the critical value.
Absolute value:
The formula for comparing the means
Substitute 610.2 and 477.8 for
Thus, the value of
Hence, the absolute value of
Conclusion:
The absolute value is 4.38.
Here, the absolute value is lesser than the critical value.
That is,
Thus, the null hypothesis is not rejected.
Hence, there is no significant difference between the means
Want to see more full solutions like this?
Chapter 12 Solutions
Elementary Statistics: A Step By Step Approach
- Discuss and explain in the picturearrow_forwardBob and Teresa each collect their own samples to test the same hypothesis. Bob’s p-value turns out to be 0.05, and Teresa’s turns out to be 0.01. Why don’t Bob and Teresa get the same p-values? Who has stronger evidence against the null hypothesis: Bob or Teresa?arrow_forwardReview a classmate's Main Post. 1. State if you agree or disagree with the choices made for additional analysis that can be done beyond the frequency table. 2. Choose a measure of central tendency (mean, median, mode) that you would like to compute with the data beyond the frequency table. Complete either a or b below. a. Explain how that analysis can help you understand the data better. b. If you are currently unable to do that analysis, what do you think you could do to make it possible? If you do not think you can do anything, explain why it is not possible.arrow_forward
- 0|0|0|0 - Consider the time series X₁ and Y₁ = (I – B)² (I – B³)Xt. What transformations were performed on Xt to obtain Yt? seasonal difference of order 2 simple difference of order 5 seasonal difference of order 1 seasonal difference of order 5 simple difference of order 2arrow_forwardCalculate the 90% confidence interval for the population mean difference using the data in the attached image. I need to see where I went wrong.arrow_forwardMicrosoft 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 COLORADOarrow_forward
- 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
- Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw HillCollege Algebra (MindTap Course List)AlgebraISBN:9781305652231Author:R. David Gustafson, Jeff HughesPublisher:Cengage Learning


