The mean should not be used on account of the extreme value 95, and the mode is not desirable because the sample size is too small. Therefore, the variance of the sample mean is reduced from 0. Therefore, the variance of the sample mean is increased from 0. P X Therefore, about 0.
So, 36! Therefore, the mean amount to be 0. This means that the assumption of the equality of the population means are not reasonable.
No, this is not very strong evidence that the population mean of the process exceeds the government limit. Such a small probability means that the difference of 4 is not likely if the two population means are equal.
Hence W follows, approximately, a normal distribution when n is large. Since n n! Conclusion values are not valid. Since the value P X So, the result is inconclusive. Since, from Table A. Hence the variances may not be equal. Hence, by solving 2. Note that the value n can be affected by the z values picked 2. Hence, by equating 2. Note that Table A. However, one can deduce the conclusion based on the values in the last line of the table. Also, computer software gives the value of 0.
Also, z0. Solutions for Exercises in Chapter 9 9. So, So, by calculating 3. The tolerance interval is Hence the claim is valid. The tolerance interval is 1. So, the upper limit is 3. Since t0. So, the interval is So, the tolerance interval is So, the tolerance limit is Since 62 exceeds the lower bound of the interval, yes, this is a cause of concern.
Since 6. The lower bound of the one-sided tolerance interval is Their claim is not necessarily correct. It is known that z0. So p The treatment appears to reduce the mean amount of metal removed.
So, p 0. So, 4. So, p So, p 4. So, p 1. So, r 0. Solutions for Exercises in Chapter 9 80 40 9. From this study we conclude that there is a significantly higher proportion of women in electrical engineering than there is in chemical engineering. So, 4 0. Hence, 19 6. Hence, 8 0. Hence, 11 2. So, 1. Then solve them to obtain the maximum likelihood estimates.
So, 2. Since 0 is not in the interval, the claim appears valid. Since z0. Since the interval contains 0. Hence, polishing does increase the average endurance limit. There is a significant difference in salaries between the two regions. Solutions for Exercises in Chapter 9 b Since the sample sizes are large enough, it is not necessary to assume the normality due to the Central Limit Theorem. It is known that f0. Hence, the sample sizes in Review Exercise 9. Using Table A. So, the limit of the one-sided tolerance interval is 6.
Since this interval contains 10, the claim by the union leaders appears valid. Apparently, the mean operating costs of type A engines are higher than those of type B engines. We assumed normality in the calculation. Still, we need to assume normality in the distribution. So, the upper bound is 3. Solutions for Exercises in Chapter 9 b z0. We then obtain Also, from Exercise The White Cheddar Popcorn, on average, weighs less than 5. Decision: reject H0. Solutions for Exercises in Chapter 10 Decision: Reject H0 and conclude that men who use TM, on average, mediate more than 8 hours per week.
Decision: Fail to reject H0. Since 1. Decision: Reject H0 and conclude that it takes less than 35 minutes, on the average, to take the test. So, fail to reject H0 and conclude that 6. Decision: Do not reject H0. Degrees of freedom is calculated as Degrees of freedom is calculated as 0. Decision: Do not reject H0 and conclude that the two methods are not significantly different.
Decision: Reject H0 when a significance level is above 0. Decision: Reject H0. Decision: Reject H0 ; length of storage influences sorbic acid residual concentrations.
The sample size needed is 6. The sample size would be 5. Hence, fail to reject H0. So, 0. Reject H0 ; breathing frequency significantly higher in the presence of CO. Denote by X for those who choose lasagna. Decision: Reject H0 at level 0. The proportion of urban residents who favor the nuclear plant is larger than the proportion of suburban residents who favor the nuclear plant.
The proportion of couples married less than 2 years and planning to have children is significantly higher than that of couples married 5 years and planning to have children. It cannot be shown that breast cancer is more prevalent in the urban community.
There is no significant evidence to conclude that the new medicine is more effective. Decision: Fail to reject H0 ; the variance of aflotoxins is not significantly different from 4. Using the table, 0. There is sufficient evidence to conclude that the standard deviation is different from Decision: Reject H0 ; the variance of the distance achieved by the diesel model is less than the variance of the distance achieved by the gasoline model.
Since f0. So, the variability of the time to assemble the product is not significantly greater for men. Decision: Fail to reject H0 ; the data is from a distribution not significantly different from h y; 8, 3, 5. Computation: A goodness-of-fit test with 5 degrees of freedom is based on the following data: oi 5 4 13 8 5 5 ei 6.
Hence, the data was not sufficient to show that the oxygen consumptions was higher when there was little or not CO. Reject H0 ; the speed is increased by using the facilitation tools. Decision: Reject H0 at level higher than 0. There is no significant change in WBC leukograms. First we do the s2 f -test to test equality of the variances.
First we do the f -test to s2 test equality of the variances. This is a two-sided test. Since 4. Decision: Reject H0 ; there is a significant difference in the number of defects. Therefore, i i i i 11 Therefore, students scoring below 59 should be denied admission. A simple linear model seems suitable for the data. Solutions for Exercises in Chapter 11 It appears that attending professional meetings would not result in publishing more papers.
Now Now 3. Level of significance: 0. Sxy Since T1. There is a strong evidence that the slope is not 0. Hence emitter drive-in time influences gain in a positive linear fashion. Hence the linear model is not adequate.
A better model is a quadratic one using emitter drive-in time to explain the variability in gain. Hence, lack-of-fit test is insignificant and the linear model is adequate. The lack-of-fit test is significant and the linear model does not appear to be the best model. If there were replicates, a lack-of-fit test could be performed. The results do not change.
The pure error test is not as sensitive because the loss of error degrees of freedom. It does not appear to be linear. Decision: Reject H0 ; the lack-of-fit test is significant. Decision: Fail to reject H0 at level 0. Hence H0 cannot be rejected.
The linear model is adequate at the level 0. The lack-of-fit test is insignificant. Solutions for Exercises in Chapter 11 c Observations 4, 9, 10, and 17 have the lowest standard errors of prediction. The R2 is only 0. There is still a pattern in the residuals. The residuals appear to be more random. This model is the best of the three models attempted. Perhaps a better model could be found.
Hence, the time it takes to run two miles has a significant influence on maximum oxygen uptake. Therefore, there is no lack of fit and the quadratic model fits the data well.
Solutions for Exercises in Chapter 12 0. Again, fail to reject H0. Hence, Hence, we reject H0. The regression explained by the model is significant.
At level of 0. The partial f -test statistic is Solutions for Exercises in Chapter 12 Hence, the drive ratio variable is important. Hence, variable x3 appears to be unimportant. Hence, the model with interaction and pure quadratic terms is better.
There appears to be little advantage using the full model. Hence, the model without using x1 is very competitive. In comparing the three models, it appears that the model with x2 only is slightly better. So, the difference between the estimates are smaller than one standard error of each.
So, no significant difference in a van and an suv in terms of gas mileage performance. The company would prefer female customers. A: s2 For company B B: s2 1. The value 0. Leaf Frequency 3 35 2 3 4 5 11 14 99 14 89 4. Class Interval 0. The variability for decreased also as seen by looking at the picture in a. The gap represents an increase of over ppm. It appears from the data that hydrocarbon emissions decreased considerably between and and that the extreme large emission over ppm were no longer in evidence.
This trimmed mean is in the middle of the mean and median using the full amount of data. Due to the skewness of the data to the right see plot in c , it is common to use trimmed data to have a more robust result.
It does show certain relationship. The relationship between load and wear in c is not as strong as the case in a , especially for the load at Low High Using Table A. So, the limit of the one-sided tolerance interval is 6. Since this interval contains 10, the claim by the union leaders appears valid. Apparently, the mean operating costs of type A engines are higher than those of type B engines. We assumed normality in the calculation. Still, we need to assume normality in the distribution.
So, the upper bound is 3. We then obtain Also, from Exercise Comparing results from Exercises Solutions for Exercises in Chapter 10 0. OC curve 0. Decision: reject H0. The White Cheddar Popcorn, on average, weighs less than 5. Decision: Reject H0 and conclude that men who use TM, on average, mediate more than 8 hours per week.
Decision: Fail to reject H0. Solutions for Exercises in Chapter 10 Decision: Reject H0 and conclude that it takes less than 35 minutes, on the average, to take the test. So, fail to reject H0 and conclude that the 6. Do not 1. Using Decision: Do not reject H0. Degrees of freedom is calculated as Degrees of freedom is calculated as 0. Decision: Reject H0. Decision: Reject H0 ; length of storage influences sorbic acid residual concentrations. Decision: Reject H0 when a significance level is above 0.
The sample size needed is 6. The sample size would be 5. Hence, fail to reject H0. So, 0. Reject H0 ; breathing frequency significantly higher in the presence of CO.
Denote by X for those who choose lasagna. Decision: Reject H0 at level 0. The proportion of urban residents who favor the nuclear plant is larger than the proportion of suburban residents who favor the nuclear plant. The proportion of couples married less than 2 years and planning to have children is significantly higher than that of couples married 5 years and planning to have children.
It cannot be shown that breast cancer is more prevalent in the urban community. Using the table, 0. Since f0. So, the variability of the time to assemble the product is not significantly greater for men. Solutions for Exercises in Chapter 10 A goodness-of-fit test with 6 degrees of freedom is based on the following data: oi 8 4 5 11 14 14 4 ei 6. Computation: A goodness-of-fit test with 5 degrees of freedom is based on the following data: oi 5 4 13 8 5 5 ei 6.
Reject H0 ; the speed is increased by using the facilitation tools. There is no significant change in WBC leukograms. First we do the f -test s2 to test equality of the variances.
Decision: H0 cannot be rejected at 0. First we do the f -test to test s2 equality of the variances. Solutions for Exercises in Chapter 10 s2 This is a two-sided test. Decision: Reject H0 at level higher than 0. Chapter 11 Simple Linear Regression and Correlation Be aware: The values you obtain may be slightly different from those listed here, due to rounding. There- i i i i fore, 25 65, Therefore, i i i i 11 Solutions for Exercises in Chapter 11 c Residuals appear to be random as desired.
A simple linear model seems suitable for the data. Therefore, students scoring below 59 should be denied admission. Solutions for Exercises in Chapter 11 d Residuals appear to be random as desired. It appears that attending professional meetings would not result in publishing more papers.
Now Now 3. Solutions for Exercises in Chapter 11 Solutions for Exercises in Chapter 11 80 60 y 40 20 5 10 15 20 25 30 x Level of significance: 0. Hence, lack-of-fit test is insignificant and the linear model is adequate.
There is a strong evidence that the slope is not 0. Hence emitter drive-in time influences gain in a positive linear fashion.
Hence the linear model is not adequate. A better model is a quadratic one using emitter drive-in time to explain the variability in gain. Since T1. The lack-of-fit test is significant and the linear model does not appear to be the best model. If there were replicates, a lack-of-fit test could be performed. The results do not change. The pure error test is not as sensitive because the loss of error degrees of freedom.
It does not appear to be linear. Decision: Reject H0 ; the lack-of-fit test is significant. There does not seem to be a strong linear relationship present. Decision: Fail to reject H0 at level 0.
Hence H0 cannot be rejected. The linear model is adequate at the level 0. The lack-of-fit test is insignificant. The R2 is only 0. There is still a pattern in the residuals. The residuals appear to be more random. This model is the best of the three models attempted. Perhaps a better model could be found. Hence, the time it takes to run two miles has a significant influence on maximum oxygen uptake. Solutions for Exercises in Chapter 11 8 6 4 residual 2 0 -2 -4 Time Therefore, there is no lack of fit and the quadratic model fits the data well.
Solutions for Exercises in Chapter 12 Again, fail to reject H0. Hence, M SR Hence, we reject H0. The regression explained by the model is significant. At level of 0. The partial f -test statistic is Hence, the drive ratio variable is important. Hence, variable x3 appears to be unimpor- tant. Hence, the model with interaction and pure quadratic terms is better. Hence, the model without using x1 is very competitive.
In comparing the three models, it appears that the model with x2 only is slightly better. The company would prefer female customers. Perhaps other variables need to be considered.
Hence only x1 remains in the model and the final model is the same one as in a. Hence the final model is still the same one as in a and b. Fail to reject H0. The residual plot and a normal probability plot are given here. When the Cp method is used, the best model is model with the constant term.
We do not appear to have the normality. Both models appear to better than the full model. Using the s2 criterion, the model with x1 , x3 and x4 has the smallest value of 0. These two models are quite competitive. However, the model with two variables has one less variable, and thus may be more appealing. Note that observations 2 and 14 are beyond the 2 standard deviation lines. Both of those observations may need to be checked. The orthogonality is maintained with the use of interaction terms.
There are no degrees of freedom available for computing the standard error. The two-way interactions are not significant. Therefore, both x12 and x22 may be dropped out from the model.
Therefore, some other models may be explored. It is about Chapter 13 One-Factor Experiments: General However, this decision is very marginal since the P -value is very close to the significance level.
The mean sorption for the solvent Chloroalkanes is the highest. We know that it is significantly higher than the rate of Esters. Therefore, [ 0. Now, Critical region: reject H0 when 9 0. A, C, D A, D Hence 8. Hence 5. Therefore, the mean rate of return are significantly higher for median and high financial leverage than for control.
The means are not all equal. Decision: Reject H0 ; mean percent of foreign additives is not the same for all three brand of jam. The means are: Jam A: 2. Based on the means, Jam A appears to have the smallest amount of foreign additives. Once the data is transformed, it is not necessary that the data would follow a normal distribution. So, it is likely that the transformed data are normally distributed.
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