The 20 second cooling time gives a cosmetically better housing. The data are as follows: 5. The normality assumption is much more important when analyzing variances then when analyzing means. A moderate departure from normality could cause problems with both statistical tests and confidence intervals.
Specifically, it will cause the reported significance levels to be incorrect. The normal probability plot indicates that there is not any serious problem with the normality assumption.
P-Value: 0. As in any t-test, the assumption of normality is of only moderate importance. In the paired t-test, the assumption of normality applies to the distribution of the differences. That is, the individual sample measurements do not have to be normally distributed, only their difference.
Do these plots support assumptions of normality and equal variance for both samples? The etch rate is an important characteristic of this process. Two different etching solutionsare being evaluated. See the Minitab output below. From the Minitab output, Specifically, tablet 1 is claimed to be absorbed twice as fast as tablet 2. Assume that and are known. How should we allocate the N observations between the two populations to obtain the most powerful test?
Thus n1 and n2 are assigned proportionally to the ratio of the standard deviations. This has intuitive appeal, as it allocates more observations to the population with the greatest variability. In general, describe how you created the data. Does this give you any insight regarding how the paired t-test works?
A B delta 7. The fact that the difference between pairs is large makes the pooled estimate of the standard deviation large and the two-sample t-test statistic small. Therefore the fairly small difference between the means of the two treatments that is present when they are applied to the same experimental unit cannot be detected.
Generally, if the blocks are very different, then this will occur. Blocking eliminates the variabiliy associated with the nuisance variable that they represent. If the mean burning times of the two flames differ by as much as 2 minutes, find the power of the test. What sample size would be required to detect an actual difference in mean burning time of 1 minute with a power of at least 0.
Rework this problem assuming that the two population variances are unknown but equal. There is no difference in the machines. The P-value for this anlysis is 0. The confidence interval is This interval contains 0. There is no difference in machines. If the mean fill volume of the two machines differ by as much as 0.
What sample size could result in a power of at least 0. Four different mixing techniques can be used economically. There is only a 0.
Mixing technique has an effect. The results agree with the graphical method for this experiment. What conclusion would you draw about the validity of the normality assumption? There is nothing unusual about the normal probability plot of residuals. Comment on the plot. There is nothing unusual about this plot. Predicted Design-Expert automatically generates the scatter plot.
The plot below also shows the sample average for each treatment and the 95 percent confidence interval on the treatment mean. The mean of Treatment 4 is different than the means of Treatments 1, 2, and 3.
However, the mean of Treatment 2 is not different from the means of Treatments 1 and 3 according to the Tukey method, they were found to be different using the graphical method and the Fisher LSD method. Find a 95 percent confidence interval on the mean tensile strength of the portland cement produced by each of the four mixing techniques.
Also find a 95 Does this aid in interpreting the results of the experiment? Strength is usually affected by the percentage of cotton used in the blend of materials for the fiber.
The engineer conducts an experiment with five levels of cotton content and replicated the experiment five times. The data are shown in the following table. Cotton Weight Percentage Observations 15 7 7 15 11 9 20 12 17 12 18 18 25 14 19 19 18 18 30 19 25 22 19 23 35 7 10 11 15 11 a Is there evidence to support the claim that cotton content affects the mean tensile strength?
The percentage of cotton in the fiber appears to have an affect on the tensile strength. What conclusions can you draw? Suppose that 30 percent cotton content is a control. A completely randomized single-factor experiment was conducted with three dosage levels, and the following results were obtained. Dosage Observations 20g 24 28 37 30 30g 37 44 31 35 Because there appears to be a difference in the dosages, the comparison of means is appropriate. An experiment is run for one week at a particular location, and 10 rental contracts are selected at random for each car type.
The results are shown in the following table. If so, which types of cars are responsible for the difference? Should the cause any potential concerns about the validity of the analysis of variance? Because the data is count data, a square root transformation could be applied.
The analysis is shown below. It does not change the interpretation of the data. Data from the last year are shown in the following table. Season Observations Summer 83 85 85 87 90 88 88 84 91 90 Shoulde r 91 87 84 87 85 86 83 Winter 94 91 87 85 87 91 92 86 The 24 potential sponsors were randomly divided into three groups of eight, and one approach was used for each group.
The dollar amounts of the resulting contributions are shown in the following table. The Tukey test will indicate which are different. Approach 2 is different than approach 3. The experiment led to the following data: Temperature Density No, firing temperature does not affect the density of the bricks.
Refer to the Design-Expert output below. There is a Treatment Means Adjusted, If Necessary The analysis of variance tells us that there is no difference in the treatments. Are the analysis of variance assumptions satisfied? There is nothing unusual about the residual plots. Does this graph adequately summarize the results of the analysis of variance in part b.
Explain carefully how you modified the procedure to account for unequal sample sizes. Notice that there are different critical values for the comparisons depending on the sample sizes of the two groups being compared.
Because we could not reject the hypothesis of equal means using the analysis of variance, we should never have performed the Tukey test or any other multiple comparison procedure, for that matter. If you ignore the analysis of variance results and run multiple comparisons, you will likely make type I errors. The following conductivity data are obtained: Coating Type Conductivity 1 2 3 4 a Is there a difference in conductivity due to coating type?
Yes, there is a difference in means. Refer to the Design-Expert output below.. Compute a 99 percent interval estimate of the mean difference between coating types 1 and 4. Treatment 4: 4 Refer to the Design-Expert output above. The Fisher LSD procedure is automatically included in the output.
The means of Coating Type 3 and Coating Type 4 are not different. However, Coating Types 1 and 2 produce higher mean conductivity that does Coating Types 3 and 4. Which coating produces the highest conductivity? We wish to minimize conductivity. Since coatings 3 and 4 do not differ, and as they both produce the lowest mean values of conductivity, use either coating 3 or 4.
As type 4 is currently being used, there is probably no need to change. Analyze the residuals and draw conclusions about model adequacy. There is nothing unusual in the normal probability plot. A funnel shape is seen in the plot of residuals versus predicted conductivity indicating a possible non-constant variance. Coating Type The resulting compressive strength of the concrete specimen is the response.
Rodding Level Compressive Strength 10 15 20 25 a Is there any difference in compressive strength due to the rodding level? There are no differences. Design Expert Output What conclusions can you draw about the underlying model assumptions? Rodding Level Radon enriched water was used in the experiment and six different orifice diameters were tested in shower heads. The data from the experiment are shown in the following table.
Orifice Dia. There is at least one treatment mean that is different. There is nothing unusual about the residuals. Orifice Diameter Circuit Type Response Time 1 9 12 10 8 15 2 20 21 23 17 30 3 6 5 8 16 7 a Test the hypothesis that the three circuit types have the same response time. How do they compare with the conclusions from part a. Notice that the results indicate that the mean of treatment 2 differs from the means of both treatments 1 and 3, and that the means for treatments 1 and 3 are the same.
In this and the other two treatments is very obvious. Either type 1 or type 3 as they are not different from each other and have the lowest response time. Are the basic analysis of variance assumptions satisfied? The normal probability plot has some points that do not lie along the line in the upper region. This may indicate potential outliers in the data.
Circuit Type Test data have been obtained for four types of fluids. Treatment 3. There is nothing unusual in the residual plots. Fluid Type The following data have been obtained: Circuit Design Noise Observed 1 19 20 19 30 8 2 80 61 73 56 80 3 47 26 25 35 50 4 95 46 83 78 97 a Is the amount of noise present the same for all four designs?
No, at least one treatment mean is different. There is nothing too unusual about the residual plots, although there is a mild outlier present. Circuit Design Low noise is best. Unless there are other reasons for choosing Type 3, Type 1 would be selected. Each chemist makes three determinations, and the results are the following: Chemist Percentage of Methyl Alcohol 1 Chemist Chemists Total C1 C2 C3 1 It is s suspected that the lives in weeks of the three brands are different.
Five batteries of each brand are tested with the following results: Weeks of Life Brand 1 Brand 2 Brand 3 76 96 80 92 75 96 96 84 98 92 82 a Are the lives of these brands of batteries different? Yes, at least one of the brands is different. Brand Construct a 99 percent interval estimate on the mean difference between the lives of battery brands 2 and 3. If the manufacturer will replace without charge any battery that fails in less than 85 weeks, what percentage would the company expect to replace?
Chose brand 3 for longest life. Mean life of this brand in Assuming normality, then the probability of failure before 85 weeks is: The following concentrations are obtained: Catalyst 1 2 3 4 No, their means are different.
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Solutions Manual for: Communications Systems, 5 th edition. Muhammad Talha. A short summary of this paper. Download Download PDF. Translate PDF. All Rights Reserved. Chapter 2 2. We need to find a function with the stated properties. These sums must always be zero. The resulting filter is non-causal and therefore not realizable in practice.
These high-frequency components are responsible for producing the sharper edges. However, this accuracy also depends on the sampling rate being high enough to include the higher frequencies.
Amplitude 5 1. This results in greater overshoot. However, as the frequency of the pulse train continues to increase, the centre frequency is no longer in the pass band, and the resulting output will also be attenuated. However, this does not extend the bandwidth of the filter, so the reduction in overshoot is minimal. This improves with higher sampling rates. This is because a greater percentage of the signal energy is concentrated at low frequencies. Musical instruments create notes that have significant energy in frequencies beyond the human vocal range.
This is particularly true of instruments whose notes have sharp attack times. Chapter 3 3. The circuit can be rearranged as follows: a b Let the voltage Vb-Vd be the voltage across the output resistor, with Vb and Vd being the voltages at each node. However, the distortion factor increases with decreasing a. Higher frequency terms will be filtered out, and so can be ignored for the purposes of determining the output of the detector. Assume that the mixer performs a multiplication of the two signals.
Assuming that no branch can be zeroed, the narrowest resolution occurs with a modulation frequency of kHz. The widest bandwidth occurs when there is a modulation frequency of kHz. All of the other product components can be discarded. This is twice the period of the carrier wave, and should provide some smoothing capability. Using this approximation, the voltage will Rl C decay by a factor of 0. From the code, it can be seen that the voltage decay is close to this figure.
However, it is somewhat slower than what was calculated using the linear approximation. In a real circuit, it would also be expected that the decay would be slower, as the voltage does not simply turn off, but rather decreases over time.
Problem 3. Problem 4. This doubles the frequency deviation. Consider the slope circuit response: The response of X1 f after the resonant peak is the same as for a single pole low-pass filter. Because the filters are symmetric about the central frequency, the contribution of the second filter is identical.
Once to form the transmitted signal and once by the envelope detector. In addition, the signal also has a DC offset, which results from the action of the envelope detector. The change in amplitude is the result of the modulation process and filters used in detection.
The above signal has been multiplied by a constant gain factor in order to highlight the differences with the original message signal. As the message frequency may no longer lie wholly within the bandwidth of either the differentiator or the low-pass filter.
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