Perform ANOVA to test the significance at 1% level of significance.

Answer to Problem 37E
The ANOVA for the given data is shown below:
Source |
Degrees of freedom |
Sum of squares |
Mean sum of squares | F-ratio |
Fabric A | 2 | 4,414.658 | 2207.329 | 2259.293 |
Type of exposure B | 1 | 47.255 | 47.255 | 48.36745 |
Degree of exposure C | 2 | 983.566 | 491.783 | 503.3603 |
Fabric direction D | 1 | 0.044 | 0.044 | 0.045036 |
Interaction AB | 2 | 30.606 | 15.303 | 15.66325 |
Interaction AC | 2 | 1,101.754 | 275.446 | 281.9304 |
Interaction AD | 2 | 0.94 | 0.47 | 0.481064 |
Interaction BC | 2 | 4.282 | 2.141 | 2.191402 |
Interaction BD | 1 | 0.273 | 0.273 | 0.279427 |
Interaction CD | 2 | 0.494 | 0.247 | 0.252815 |
Interaction ABC | 4 | 14.856 | 3.714 | 3.801433 |
Interaction ABD | 2 | 8.144 | 4.072 | 4.167861 |
Interaction ACD | 4 | 3.068 | 0.767 | 0.785056 |
Interaction BCD | 2 | 0.56 | 0.28 | 0.286592 |
Interaction ABCD | 4 | 1.389 | 0.347 | 0.355 |
Error | 36 | 35.172 | 0.977 | |
Total | 71 | 6,647.091 | 9.621 |
There is sufficient of evidence to conclude that there is an effect of fabric on the extent of color change at 1% level of significance.
There is sufficient of evidence to conclude that there is an effect exposure type on the extent of color change at 1% level of significance.
There is sufficient of evidence to conclude that there is an effect of exposure level on the extent of color change at 1% level of significance.
There is no sufficient of evidence to conclude that there is an effect of fabric direction on the extent of color change at 1% level of significance.
There is sufficient of evidence to conclude that there is an interaction effect of fabric and exposure type on the extent of color change at 1% level of significance.
There is sufficient of evidence to conclude that there is an interaction effect of fabric and exposure level on the extent of color change at 1% level of significance.
There is no sufficient of evidence to conclude that there is an interaction effect of fabric and fabric direction on the extent of color change at 1% level of significance.
There is no sufficient of evidence to conclude that there is an interaction effect of exposure type and exposure level on the extent of color change at 1% level of significance.
There is no sufficient of evidence to conclude that there is an interaction effect of exposure type and fabric direction on the extent of color change at 1% level of significance.
There is no sufficient of evidence to conclude that there is an interaction effect of exposure level and fabric direction on the extent of color change at 1% level of significance.
There is no sufficient of evidence to conclude that there is an interaction effect of fabric, exposure type and exposure level on the extent of color change at 1% level of significance.
There is no sufficient of evidence to conclude that there is an interaction effect of fabric, exposure type and fabric direction on the extent of color change at 1% level of significance.
There is no sufficient of evidence to conclude that there is an interaction effect of fabric, exposure level and fabric direction on the extent of color change at 1% level of significance.
There is no sufficient of evidence to conclude that there is an interaction effect of exposure type, exposure level and fabric direction on the extent of color change at 1% level of significance.
There is no sufficient of evidence to conclude that there is an interaction effect of fabric, exposure type, exposure level and fabric direction on the extent of color change at 1% level of significance.
Explanation of Solution
Given info:
An experiment was conducted to test the effect of fabric, type of exposure, level of exposure and fabric direction on the color change of the fabric. Two observation were noted for each of the four factors.
Calculation:
The general ANOVA table is given below:
Source | Degrees of freedom | Sum of squares | Mean sum of squares | F-ratio |
Factor A | ||||
Factor B | ||||
Factor C | ||||
Factor D | ||||
Interaction AB | ||||
Interaction ABC | ||||
Error | ||||
Total |
The sum of squares for each factor and interaction is calculated by multiplying the mean sum of squares with its corresponding degrees of freedom.
Sum of squares excluding ABCD:
Source | Sum of squares |
A | 4,414.658 |
B | 47.255 |
C | 983.566 |
D | 0.044 |
AB | 30.606 |
AC | 1,101.784 |
AD | 0.94 |
BC | 4.282 |
BD | 0.273 |
CD | 0.494 |
ABC | 14.856 |
ABD | 8.144 |
ACD | 3.068 |
BCD | 0.56 |
Error | 35.172 |
Total | 6,647.091 |
Using the above table SSABCD can be calculated:
The mean sum of squares for the interaction ABCD is given below:
Thus, the mean sum of squares for the interaction ABCD is 0.347.
The ANOVA for the given data is shown below:
Source | Degrees of freedom |
Sum of squares |
Mean sum of squares | F-ratio |
Fabric A | 4,414.658 | 2207.329 | 2,259.293 | |
Type of exposure B | 47.255 | 47.255 | 48.36745 | |
Degree of exposure C | 983.566 | 491.783 | 503.3603 | |
Fabric direction D | 0.044 | 0.044 | 0.045036 | |
Interaction AB | 30.606 | 15.303 | 15.66325 | |
Interaction AC | 1,101.754 | 275.446 | 281.9304 | |
Interaction AD | 0.94 | 0.47 | 0.481064 | |
Interaction BC | 4.282 | 2.141 | 2.191402 | |
Interaction BD | 0.273 | 0.273 | 0.279427 | |
Interaction CD | 0.494 | 0.247 | 0.252815 | |
Interaction ABC | 14.856 | 3.714 | 3.801433 | |
Interaction ABD | 8.144 | 4.072 | 4.167861 | |
Interaction ACD | 3.068 | 0.767 | 0.785056 | |
Interaction BCD | 0.56 | 0.28 | 0.286592 | |
Interaction ABCD | 1.389 | 0.347 | 0.355 | |
Error | 35.172 | 0.977 | ||
Total | 6,647.091 | 9.621 |
Where,
The F statistic for each factor is obtained by dividing the mean sum of squares with the mean sum of squares due to error.
Testing the main effects:
Testing the Hypothesis for the factor A:
Null hypothesis:
That is, there is no significant difference in the extent of color change due to the three levels of fabrics.
Alternative hypothesis:
That is, there is significant difference in the extent of color change due to the three levels of fabrics.
Testing the Hypothesis for the factor B:
Null hypothesis:
That is, there is no significant difference in the extent of color change due to the two levels of exposure type.
Alternative hypothesis:
That is, there is a significant difference in the extent of color change due to the two levels of exposure type.
Testing the Hypothesis for the factor C:
Null hypothesis:
That is, there is no significant difference in the extent of color change due to the three levels of exposure level.
Alternative hypothesis:
That is, there is a significant difference in the extent of color change due to the three levels of exposure level.
Testing the Hypothesis for the factor D:
Null hypothesis:
That is, there is no significant difference in the extent of color change due to the two levels of fabric direction.
Alternative hypothesis:
That is, there is a significant difference in the extent of color change due to the two levels of fabric direction.
Testing the Hypothesis for the interaction effect of AB:
Null hypothesis:
That is, there is no significant difference in the extent of color change due to the interaction between fabric and exposure type.
Alternative hypothesis:
That is, there is significant difference in the extent of color change due to the interaction between fabric and exposure type.
Testing the Hypothesis for the interaction effect AC:
Null hypothesis:
That is, there is no significant difference in the extent of color change due to the interaction between fabric and exposure level.
Alternative hypothesis:
That is, there is a significant difference in the extent of color change due to the interaction between fabric and exposure level.
Testing the Hypothesis for the interaction effect AD:
Null hypothesis:
That is, there is no significant difference in the extent of color change due to the interaction between fabric and fabric direction.
Alternative hypothesis:
That is, there is a significant difference in the extent of color change due to the interaction between fabric and fabric direction.
Testing the Hypothesis for the interaction effect BC:
Null hypothesis:
That is, there is no significant difference in the extent of color change due to the interaction between exposure type and exposure level.
Alternative hypothesis:
That is, there is significant difference in the extent of color change due to the interaction between exposure type and exposure level.
Testing the Hypothesis for the interaction effect BD:
Null hypothesis:
That is, there is no significant difference in the extent of color change due to the interaction between exposure type and fabric direction.
Alternative hypothesis:
That is, there is significant difference in the extent of color change due to the interaction between exposure type and fabric direction.
Testing the Hypothesis for the interaction effect CD:
Null hypothesis:
That is, there is no significant difference in the extent of color change due to the interaction between exposure level and fabric direction.
Alternative hypothesis:
That is, there is significant difference in the extent of color change due to the interaction between exposure level and fabric direction.
Testing the Hypothesis for the interaction effect ABC:
Null hypothesis:
That is, there is no significant difference in the extent of color change due to the interaction between fabric, exposure type and exposure level.
Alternative hypothesis:
That is, there is a significant difference in the extent of color change due to the interaction between fabric, exposure type and exposure level.
Testing the Hypothesis for the interaction effect ABD:
Null hypothesis:
That is, there is no significant difference in the extent of color change due to the interaction between fabric, exposure type and fabric direction.
Alternative hypothesis:
That is, there is a significant difference in the extent of color change due to the interaction between fabric, exposure type and fabric direction.
Testing the Hypothesis for the interaction effect ACD:
Null hypothesis:
That is, there is no significant difference in the extent of color change due to the interaction between fabric, exposure level and fabric direction.
Alternative hypothesis:
That is, there is a significant difference in the extent of color change due to the interaction between fabric, exposure level and fabric direction.
Testing the Hypothesis for the interaction effect BCD:
Null hypothesis:
That is, there is no significant difference in the extent of color change due to the interaction between exposure type, exposure level and fabric direction.
Alternative hypothesis:
That is, there is a significant difference in the extent of color change due to the interaction between exposure type, exposure level and fabric direction.
Testing the Hypothesis for the interaction effect ABCD:
Null hypothesis:
That is, there is no significant difference in the extent of color change due to the interaction between exposure type, exposure level and fabric direction.
Alternative hypothesis:
That is, there is a significant difference in the extent of color change due to the interaction between fabric, exposure type, exposure level and fabric direction.
P-value for the main effect of A:
Software procedure:
Step-by-step procedure to find the P-value is given below:
- Click on Graph, select View Probability and click OK.
- Select F, enter 2 in numerator df and 36 in denominator df.
- Under Shaded Area Tab select X value under Define Shaded Area By and select right tails.
- Choose X value as 2,259.29.
- Click OK.
Output obtained from MINITAB is given below:
P-value for the main effect of B:
Software procedure:
Step-by-step procedure to find the P-value is given below:
- Click on Graph, select View Probability and click OK.
- Select F, enter 1 in numerator df and 36 in denominator df.
- Under Shaded Area Tab select X value under Define Shaded Area By and select right tails.
- Choose X value as 48.37.
- Click OK.
Output obtained from MINITAB is given below:
P-value for the main effect of C:
Software procedure:
Step-by-step procedure to find the P-value is given below:
- Click on Graph, select View Probability and click OK.
- Select F, enter 2 in numerator df and 36 in denominator df.
- Under Shaded Area Tab select X value under Define Shaded Area By and select right tails.
- Choose X value as 503.36.
- Click OK.
Output obtained from MINITAB is given below:
P-value for the main effect of D:
Software procedure:
Step-by-step procedure to find the P-value is given below:
- Click on Graph, select View Probability and click OK.
- Select F, enter 1 in numerator df and 36 in denominator df.
- Under Shaded Area Tab select X value under Define Shaded Area By and select right tails.
- Choose X value as 0.05.
- Click OK.
Output obtained from MINITAB is given below:
P-value for the interaction effect of A and B:
Software procedure:
Step-by-step procedure to find the P-value is given below:
- Click on Graph, select View Probability and click OK.
- Select F, enter 2 in numerator df and 36 in denominator df.
- Under Shaded Area Tab select X value under Define Shaded Area By and select right tails.
- Choose X value as 15.66.
- Click OK.
Output obtained from MINITAB is given below:
P-value for the interaction effect of A and C:
Software procedure:
Step-by-step procedure to find the P-value is given below:
- Click on Graph, select View Probability and click OK.
- Select F, enter 4 in numerator df and 36 in denominator df.
- Under Shaded Area Tab select X value under Define Shaded Area By and select right tails.
- Choose X value as 281.93.
- Click OK.
Output obtained from MINITAB is given below:
P-value for the interaction effect of A and D:
Software procedure:
Step-by-step procedure to find the P-value is given below:
- Click on Graph, select View Probability and click OK.
- Select F, enter 2 in numerator df and 36 in denominator df.
- Under Shaded Area Tab select X value under Define Shaded Area By and select right tails.
- Choose X value as 0.48.
- Click OK.
Output obtained from MINITAB is given below:
P-value for the interaction effect of B and C:
Software procedure:
Step-by-step procedure to find the P-value is given below:
- Click on Graph, select View Probability and click OK.
- Select F, enter 2 in numerator df and 36 in denominator df.
- Under Shaded Area Tab select X value under Define Shaded Area By and select right tails.
- Choose X value as 2.19.
- Click OK.
Output obtained from MINITAB is given below:
P-value for the interaction effect of B and D:
Software procedure:
Step-by-step procedure to find the P-value is given below:
- Click on Graph, select View Probability and click OK.
- Select F, enter 1 in numerator df and 36 in denominator df.
- Under Shaded Area Tab select X value under Define Shaded Area By and select right tails.
- Choose X value as 0.28.
- Click OK.
Output obtained from MINITAB is given below:
P-value for the interaction effect of C and D:
Software procedure:
Step-by-step procedure to find the P-value is given below:
- Click on Graph, select View Probability and click OK.
- Select F, enter 2 in numerator df and 36 in denominator df.
- Under Shaded Area Tab select X value under Define Shaded Area By and select right tails.
- Choose X value as 0.25.
- Click OK.
Output obtained from MINITAB is given below:
P-value for the interaction effect of A, B and C:
Software procedure:
Step-by-step procedure to find the P-value is given below:
- Click on Graph, select View Probability and click OK.
- Select F, enter 4 in numerator df and 36 in denominator df.
- Under Shaded Area Tab select X value under Define Shaded Area By and select right tails.
- Choose X value as 3.80.
- Click OK.
Output obtained from MINITAB is given below:
P-value for the interaction effect of A, B and D:
Software procedure:
Step-by-step procedure to find the P-value is given below:
- Click on Graph, select View Probability and click OK.
- Select F, enter 2 in numerator df and 36 in denominator df.
- Under Shaded Area Tab select X value under Define Shaded Area By and select right tails.
- Choose X value as 4.17.
- Click OK.
Output obtained from MINITAB is given below:
P-value for the interaction effect of A, C and D:
Software procedure:
Step-by-step procedure to find the P-value is given below:
- Click on Graph, select View Probability and click OK.
- Select F, enter 4 in numerator df and 36 in denominator df.
- Under Shaded Area Tab select X value under Define Shaded Area By and select right tails.
- Choose X value as 0.79.
- Click OK.
Output obtained from MINITAB is given below:
P-value for the interaction effect of B, C and D:
Software procedure:
Step-by-step procedure to find the P-value is given below:
- Click on Graph, select View Probability and click OK.
- Select F, enter 2 in numerator df and 36 in denominator df.
- Under Shaded Area Tab select X value under Define Shaded Area By and select right tails.
- Choose X value as 0.29.
- Click OK.
Output obtained from MINITAB is given below:
P-value for the interaction effect of A, B, C and D:
Software procedure:
Step-by-step procedure to find the P-value is given below:
- Click on Graph, select View Probability and click OK.
- Select F, enter 4 in numerator df and 36 in denominator df.
- Under Shaded Area Tab select X value under Define Shaded Area By and select right tails.
- Choose X value as 0.355.
- Click OK.
Output obtained from MINITAB is given below:
Conclusion:
For the main effect of A:
The P- value for the factor A (fabric) is 0.000 and the level of significance is 0.01.
Here, the P- value is lesser than the level of significance.
That is,
Thus, the null hypothesis is rejected,
Hence, there is sufficient of evidence to conclude that there is an effect of fabric on the extent of color change at 1% level of significance.
For main effect of B:
The P- value for the factor B (exposure level) is 0.000 and the level of significance is 0.01.
Here, the P- value is lesser than the level of significance.
That is,
Thus, the null hypothesis is not rejected.
Hence, there is sufficient of evidence to conclude that there is an effect exposure type on the extent of color change at 1% level of significance.
For main effect of C:
The P- value for the factor C (exposure level) is 0.000 and the level of significance is 0.01.
Here, the P- value is lesser than the level of significance.
That is,
Thus, the null hypothesis is rejected.
Hence, there is sufficient of evidence to conclude that there is an effect of exposure level on the extent of color change at 1% level of significance.
For main effect of D:
The P- value for the factor D (fabric direction) is 0.8243 and the level of significance is 0.01.
Here, the P- value is greater than the level of significance.
That is,
Thus, the null hypothesis is not rejected.
Hence, there is no sufficient of evidence to conclude that there is an effect of fabric direction on the extent of color change at 1% level of significance.
Interaction effect of factor A and B:
The P- value for the interaction effect AB (fabric and exposure type) is 0.000 and the level of significance is 0.01.
Here, the P- value is lesser than the level of significance.
That is,
Thus, the null hypothesis is rejected,
Hence, there is sufficient of evidence to conclude that there is an interaction effect of fabric and exposure type on the extent of color change at 1% level of significance.
Interaction effect of factor A and C:
The P- value for the interaction effect AC (fabric and exposure level) is 0.000 and the level of significance is 0.01.
Here, the P- value is lesser than the level of significance.
That is,
Thus, the null hypothesis is rejected.
Hence, there is sufficient of evidence to conclude that there is an interaction effect of fabric and exposure level on the extent of color change at 1% level of significance.
Interaction effect of factor A and D:
The P- value for the interaction effect AD (fabric and fabric direction) is 0.6227 and the level of significance is 0.01.
Here, the P- value is greater than the level of significance.
That is,
Thus, the null hypothesis is not rejected.
Hence, there is no sufficient of evidence to conclude that there is an interaction effect of fabric and fabric direction on the extent of color change at 1% level of significance.
Interaction effect of factor B and C:
The P- value for the interaction effect BC (exposure type and exposure level) is 0.1266 and the level of significance is 0.01.
Here, the P- value is greater than the level of significance.
That is,
Thus, the null hypothesis is not rejected,
Hence, there is no sufficient of evidence to conclude that there is an interaction effect of exposure type and exposure level on the extent of color change at 1% level of significance.
Interaction effect of factor B and D:
The P- value for the interaction effect BD (exposure type and fabric direction) is 0.5999 and the level of significance is 0.01.
Here, the P- value is greater than the level of significance.
That is,
Thus, the null hypothesis is not rejected,
Hence, there is no sufficient of evidence to conclude that there is an interaction effect of exposure type and fabric direction on the extent of color change at 1% level of significance.
Interaction effect of factor C and D:
The P- value for the interaction effect CD (exposure level and fabric direction) is 0.7801 and the level of significance is 0.01.
Here, the P- value is greater than the level of significance.
That is,
Thus, the null hypothesis is not rejected,
Hence, there is no sufficient of evidence to conclude that there is an interaction effect of exposure level and fabric direction on the extent of color change at 1% level of significance.
Interaction effect of factor A,B and C:
The P- value for the interaction effect ABC (fabric, exposure type and exposure level) is 0.01119 and the level of significance is 0.01.
Here, the P- value is greater than the level of significance.
That is,
Thus, the null hypothesis is not rejected.
Hence, there is no sufficient of evidence to conclude that there is an interaction effect of fabric, exposure type and exposure level on the extent of color change at 1% level of significance.
Interaction effect of factor A,B and D:
The P- value for the interaction effect ABD (fabric, exposure type and fabric direction) is 0.0235 and the level of significance is 0.01.
Here, the P- value is greater than the level of significance.
That is,
Thus, the null hypothesis is not rejected.
Hence, there is no sufficient of evidence to conclude that there is an interaction effect of fabric, exposure type and fabric direction on the extent of color change at 1% level of significance.
Interaction effect of factor A,C and D:
The P- value for the interaction effect ACD (fabric, exposure level and fabric direction) is 0.5394 and the level of significance is 0.01.
Here, the P- value is greater than the level of significance.
That is,
Thus, the null hypothesis is not rejected.
Hence, there is no sufficient of evidence to conclude that there is an interaction effect of fabric, exposure level and fabric direction on the extent of color change at 1% level of significance.
Interaction effect of factor B, C and D:
The P- value for the interaction effect BCD (exposure type, exposure level and fabric direction) is 0.7500 and the level of significance is 0.01.
Here, the P- value is greater than the level of significance.
That is,
Thus, the null hypothesis is not rejected.
Hence, there is no sufficient of evidence to conclude that there is an interaction effect of exposure type, exposure level and fabric direction on the extent of color change at 1% level of significance.
Interaction effect of factor A, B, C and D:
The P- value for the interaction effect ABCD (fabric, exposure type, exposure level and fabric direction) is 0.8388 and the level of significance is 0.01.
Here, the P- value is greater than the level of significance.
That is,
Thus, the null hypothesis is not rejected.
Hence, there is no sufficient of evidence to conclude that there is an interaction effect of fabric, exposure type, exposure level and fabric direction on the extent of color change at 1% level of significance.
Therefore, there is significant difference in the extent of color change with respect to the main effect A, B, D and interaction effects AB, AC are significant at 1% level of significance. The remaining second order interactions and third order interaction are not significant at 1% level of significance.
Want to see more full solutions like this?
Chapter 11 Solutions
PROBABILITY & STATS FOR ENGINEERING &SCI
- 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
- If 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_forwardProblem 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_forward
- Solve 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_forwardged 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_forward
- Big Ideas Math A Bridge To Success Algebra 1: Stu...AlgebraISBN:9781680331141Author:HOUGHTON MIFFLIN HARCOURTPublisher:Houghton Mifflin Harcourt
