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Suppose we drew a sample of size 10 from a normal population with unknown mean and standard deviation and got x = 18.87 . Two questions arise: (1) what does this sample tell us about the population from which the sample was drawn, and (2) what would happen if we drew more samples Suppose we drew 5 more samples of size 10 from this population and got x = 20.35, x = 20.04, x = 19.20, x = 19.02, and x = 20.35. In answer to question (1), we might believe that the population from which these samples was drawn had a mean around 20 because these averages tend to group there (in fact, the six samples were drawn from a normal population whose mean is 20 and whose standard deviation is 4). The mean of the 6 samples is 19.64, which supports our feeling that the mean of the original population might have been 20. The standard deviation of the 6 samples is 0.68 and you might not have any intuitive sense about how that relates to the population standard deviation, although you might suspect that the standard deviation of the samples should be less than the standard deviation of the population because the chance of an extreme value for an average should be less than that for an individual term (it just doesn t seem very likely that we would draw a lot of extreme values in a single sample).
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6. You ve just added the event that the EventHandlingScope activity will wait for to cease execution (conveniently named Stop ). Next you need to add the child activity that EventHandlingScope will execute while waiting for the Stop activity to fire. To do this, you need to return to eventHandlingScopeActivity1 s event handling scope view by repeating the first part of step 4. However, instead of selecting the right icon, select the left icon.
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MySQL is available in both source and binary forms for both 32-bit and 64-bit versions of Microsoft Windows. Most often, you will want to use either the Essentials or Complete binary distributions, which include an automated installer to get MySQL up and running in just a few minutes. To install MySQL from a binary distribution, use the following steps:
8. Selecting the method name causes Visual Studio to examine the method and add the parameters the method supports to the Properties pane. Select the (Return Value) property to activate the familiar browse (...) button. Click the browse button to activate the Bind (Return Value) To An Activity s Property dialog box. Click the Bind To A New Member tab, and type StockValue in the New Member Name field, making sure the Create Property option is selected. Click OK to add the StockValue dependency property to your workflow.
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b. The residual plot indicates that a line is a good model for the data for all years. c. There appears to be an outlier in the data at about 28 years of experience. d. The variability of salaries increases as years of experience increases. e. For each additional year of experience, salary is predicted to increase by about $2141. 27. A wine maker advertises that the mean alcohol content of the wine produced by his winery is 11%. A 95% confidence interval, based on a random sample of 100 bottles of wine yields a confidence interval for the true alcohol content of (10.5, 10.9) Could this interval be used as part of a hypothesis test of the null hypothesis H0: p = 0.11 versus the alternative hypothesis HA: p 0.11 at the 0.05 level or confidence a. No, you cannot use a confidence interval in a hypothesis test. b. Yes, because 0.11 is not contained in the 95% confidence interval, a two-sided test at the 0.05 level of significance would provide good evidence that the true mean content is different from 11%. c. No, because we do not know that the distribution is approximately normally distributed. d. Yes, because 0.11 is not contained in the 95% confidence interval, a two-sided test at the 0.05 level of significance would fail to reject the null hypothesis. e. No, confidence intervals can only be used in one-sided significance tests. 28. Tom s career batting average is 0.265 with a standard deviation of 0.035. Larry s career batting average is 0.283 with a standard deviation of 0.029. The distribution of both averages is approximately normal. They play for different teams and there is reason to believe that their career averages are independent of each other. For any given year, what is the probability that Tom will have a higher batting average than Larry a. b. c. d. e. 0.389 0.345 0.589 0.655 You cannot answer this question since the distribution for the difference between their averages cannot be determined from the data given.
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< php class Sandbox_ExampleController extends Zend_Controller_Action { public function failAction() { // try accessing a non-existent array element // generates an OutOfBoundsException $cities = array( 'London', 'Washington', 'Paris', 'Delhi' ); $iterator = new ArrayIterator($cities); $iterator->seek(10); } }
The correct answer is (a). A 95% confidence interval at 12 2 = 10 degrees of freedom has a critical value of t* = 2.228 (from Table B; if you have a TI-84 with the invT function, invT(0.975, 10)= 2.228) The required interval is 0.74 . (2.228)(0.11) = 0.74 0.245.
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Notice that all of the other interconnect signal names have names such as modgen_0_11_xx or ix123. There is no corresponding signal name in the source file to specify the signal name; therefore, the synthesis tool generates names for these signals. The netlist can be used to figure out how well the synthesizer implemented a part of the design, or to track down a problem net. It can be very useful to find out why a critical path was implemented too slowly. When the netlist meets the designer s timing, area, power, and other constraints, the next step is to pass the netlist to the gate level simulator. This simulator checks the functionality of the synthesized design.
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