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Predicting software blights. Relate to the Pledge Software Engineering Repository data on 498 modules of software law written in 鈥淐鈥� language for a NASA spacecraft instrument, saved in the train. (See Exercise 3.132, p. 209). Recall that the software law in each module was estimated for blights; 49 were classified as 鈥渢rue鈥� (i.e., the module has imperfect law), and 449 were classified as 鈥渇alse鈥� (i.e., the module has corrected law). Consider these to be Arbitrary independent samples of software law modules. Experimenters prognosticated the disfigurement status of each module using the simple algorithm, 鈥淚f the number of lines of law in the module exceeds 50, prognosticate the module to have a disfigurement.鈥� The accompanying SPSS printout shows the number of modules in each of the two samples that were prognosticated to have blights (PRED_LOC = 鈥測es鈥�) and prognosticated to have no blights (PRED_LOC = 鈥渘o鈥�). Now, define the delicacy rate of the algorithm as the proportion of modules. That was rightly prognosticated. Compare the delicacy rate of the algorithm when applied to modules with imperfect law with the delicacy rate of the algorithm when applied to modules with correct law. Use a 99-confidence interval.

DEFECT*PRED_LOC crosstabulation

Step by step solution

01

Find the value of P1 and P2

Consider n₁=449, n₂=49, x₁ = 400, x₂=20.

Find the value of p₁.

$$\begin{gathered} P_1=\frac{x_1}{n_1} \\ =\frac{400}{499} \\ =0.89 \end{gathered}$$

= 0.89

Find the value of p₂.

$$\begin{gathered} P_2=\frac{x_2}{n_2} \\ =\frac{22}{49} \\ =0.41 \end{gathered}$$

02

Let the confidence level be 0.99

1- $伪$= 0.99

$伪$= 1 鈥� 0.99

= 0.01

$\frac{伪}{2}$= 0.005

From appendix table-2, the value of Za_/2 is given below.

$$\begin{gathered} Z_{\frac{a}{2}}=Z_{0.005} \\ =2.58 \end{gathered}$$

=2.58

So, the value of z_{a 2} is 2.58.

03

The formula for 100(1α) % confidence intervals for (P1-P2)

The formula for 100(1伪) % confidence intervals for (P₁-P₂) is below.

$\left(P_1-P_2\right)卤Z_{\frac{a}{2}}\sqrt{\frac{p_1{q}_1}{n_1}+\frac{p_2{q}_2}{n_2}}$

In this, q₁ = 1 鈥� p₁and q₂ = 1 鈥� p_2.

04

Find the 99% confidence interval for (P1-P2)

Substitute P₁ = 0.89 P₂ = .041, n₁ = 449, and n₂ = 49 in the above formula.

= 0.89 鈥� 0.41 卤 2.58$\sqrt{\frac{0.89-\left(1-0.89\right)}{{\displaystyle \frac{4490.41\left(1-0.41\right)}{49}}}}$

=0.48卤2.58($\sqrt{\frac{0.0979}{449}+\frac{0.2419}{49}}$)

=0.48卤2.58(0.0718)

=.48卤0.1852

= (0.2948,0.6652)

Thus, the 99% confidence interval for (P₁ 鈥� P₂) is (0.2948,0.6652).

05

Final answer

The difference in accuracy rate between the modules with correct code and modules with incorrect lies between 0.2948 and 0.6652 at 99% confidence.

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