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Using the component requires telling Visual Studio .NET that the external assembly is to be used with a project. This is done by setting a reference. The reference is added to the Web project by right-clicking the References node in the Solution Explorer and clicking Add Reference. The Add Reference dialog box has three tabs; .NET, COM, and Projects. If the component was written with Visual Studio .NET, the .NET tab could be used to Browse to the desired folder. Selecting the .dll file using this method does not cause the assembly to be rebuilt when the current project is built. The COM tab is used to set references to COM components. If a reference is set to a COM component, Visual Studio .NET creates a COM Callable Wrapper class, that can be used to access the COM component. The Project tab is used to set a reference to a project instead of an assembly. Setting a reference to a project tells Visual Studio .NET that the project must be built prior to building the current project. This essentially sets up the build order for the solution. The only projects that are visible in this tab are the projects that are in the current solution. This means that the DataComponent project must be loaded in the current solution to be able to select it. Setting a reference to the project is generally considered to the best selection, because building the solution builds all of the projects in the correct order. In this example, a reference is set to the DataComponent.dll assembly instead of the project. This allows the projects to be out of sync for the testing purposes.

Null hypothesis: Test statistic:

Alternative Hypotheses H1: H1: H1:

In the new Web project, the following code has been added to the Web page to test the ExecuteScaler functionality.

2 1 2 1 2 1 2 2 2 2 2 2

EXAMPLE 10-11

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Oxide layers on semiconductor wafers are etched in a mixture of gases to achieve the proper thickness. The variability in the thickness of these oxide layers is a critical characteristic of the wafer, and low variability is desirable for subsequent processing steps. Two different mixtures of gases are being studied to determine whether one is superior in reducing the variability of the oxide thickness. Twenty wafers are etched in each gas. The sample standard deviations of oxide thickness are s1 1.96 angstroms and s2 2.13 angstroms, respectively. Is there any evidence to indicate that either gas is preferable Use 0.05. The eight-step hypothesis-testing procedure may be applied to this problem as follows: 1. 2. 3. 4. 5. The parameters of interest are the variances of oxide thickness 2 and 2. We will 1 2 assume that oxide thickness is a normal random variable for both gas mixtures. H0: H1:

Imports DataComponent Imports System.Data.SqlClient Public Class WebForm1 Inherits System.Web.UI.Page Private Sub Page_Load( _

2 1 2 1 2 2 2 2

3 4 5

2 Since n1 n2 20, we will reject H0: 2 f0.025,19,19 2.53 or 1 2 if f0 if f0 f0.975,19,19 1 f0.025,19,19 1 2.53 0.40. 7. Computations: Since s2 (1.96)2 3.84 and s2 (2.13)2 4.54, the test statistic is 1 2

15

f0 8.

3.84 4.54

ByVal sender As System.Object, _ ByVal e As System.EventArgs) Handles MyBase.Load Dim cn As New SqlConnection( _ server=.;database=northwind;trusted_connection=true ) Dim x As Long x = Db.ExecuteScaler( _ cn, Select count(*) from customers ) Response.Write( Count of Customers= & x.ToString()) End Sub End Class

Conclusions: Since f0.975,19,19 0.40 f0 0.85 f0.025,19,19 2.53, we cannot 2 reject the null hypothesis H0: 2 1 2 at the 0.05 level of signi cance. Therefore, there is no strong evidence to indicate that either gas results in a smaller variance of oxide thickness.

We may also nd a P-value for the F-statistic in Example 10-11. Since f0.50,19,19 1.00, the computed value of the test statistic f0 s 2 s 2 3.84 4.54 0.85 is nearer the lower 1 2 tail of the F distribution than the upper tail. The probability that an F-random variable with 19 numerator and denominator degrees of freedom is less than 0.85 is 0.3634. Since it is arbitrary which population is identi ed as one, we could have computed the test statistic as f0 4.54 3.84 1.18. The probability that an F-random variable with 19 numerator and denominator degrees of freedom exceeds 1.18 is 0.3610. Therefore, the P-value for the test statistic f0 0.85 is the sum of these two probabilities, or P 0.3634 0.3610 0.7244. 2 Since the P-value exceeds 0.05, the null hypothesis H0: 2 1 2 cannot be rejected. (The probabilities given above were computed using a hand-held calculator.)

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