In semiconductor manufacturing, wet chemical etching is often used to remove sil
ID: 3281969 • Letter: I
Question
In semiconductor manufacturing, wet chemical etching is often used to remove silicon from the backs of wafers prior to metalization. The etch rate is an important characteristic in this process and known to follow a normal distribution. Two different etching solutions have been compared, using two random samples of 10 wafers for etch solution. The observed etch rates are as follows (in mils/min):
Round your answer to 3 decimal places.
Test Statistic = ?
Solution 1 Solution 2 9.5 10.6 10.1 10.0 9.4 10.3 10.6 10.2 9.3 10.0 10.7 10.7 9.6 10.3 10.4 10.4 10.2 10.1 10.5 10.3Explanation / Answer
Concepts and reason
Statistical hypotheses testing: Hypotheses testing is used to make inferences about the population based on the sample data. The hypotheses test consists of the null hypothesis and alternative hypothesis.
Null hypothesis: The null hypothesis states that there is no difference in the test, which is denoted by. Moreover, the sign used in the null hypothesis is equal, greater than or equal and less than or equal.
Alternative hypothesis: The hypothesis that differs from the is called an alternative hypothesis. This signifies that there is a significant difference in the test. The sign used in the alternative hypothesis is less than, greater than, or not equal.
Chi-square test for variance:
For testing the variance or standard deviation of the population is equal to the particular value, the chi-square test is used, which is either a two-sided test or one-sided test.
The probability value related to a statistical test is named as a p-value. If the p-value is less than the significance level, then the null hypothesis is rejected. The smaller p-value gives the stronger evidence to reject the null hypothesis. Thus, it can be concluded that “Reject ” when the p-value is smaller.
Fundamentals
The formula for Chi-square test is shown below:
Where,
n is the sample size
is the sample variance
is the population variance
Degrees of freedom:
Rejection rule for P-value method:
If, then reject the null hypothesis.
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First Step | All Steps | Answer Only
Step-by-step
Step 1 of 4
The hypotheses are given below:
Null hypothesis:
Alternative hypothesis:
Explanation | Common mistakes | Hint for next step
The null hypothesis is “there is no significant difference variation between the two etching solutions.” and the alternative hypothesis is “there is significant difference variation between the two etching solutions”.
Step 2 of 4
Instructions to build a two-sample variance test to find the test statistic:
1.Choose Stat > Basic Statistics > 2 Variance.
2.Under Data, choose Samples in different columns.
3.In First, enter solution 1.
4.In Second, enter solution 2.
5.Check Options and enter the Confidence level as 95.0.
6.In Hypothesized ratio, enter Stdev 1 / Stdev 2.
7.Choose not equal in the alternative.
8.Click OK in all dialog boxes.
Follow the above instructions to get the test statistic value:
From the output, the test statistics is 3.632 and the p-value is 0.069.
Thus, the test statistic is 3.632.
Explanation | Common mistakes | Hint for next step
The test statistics is obtained by substituting variables solution 1 and solution 2, level of significance and alternative hypothesis in MINITAB.
Step 3 of 4
The MINITAB output for the two variances is,
From the MINITAB output, the P-value is 0.069
Explanation | Common mistakes | Hint for next step
The P-value is obtained by the given sample data substituting variables solution 1 and solution 2 and level of significance and alternative hypothesis in MINITAB
Step 4 of 4
The conclusion is stated below:
Use the significance level 0.069.
The P-value is 0.069 and the level of significance is 0.05.
That is,
By the rejection rule, do not reject the null hypothesis.
Therefore, it can be concluded that there is no significant difference between the two etching solutions.
Do not reject: There is no significant difference between the two etching solutions.
Explanation | Common mistakes
Since the P-value is greater than the level of significance, the null hypothesis is do not rejected based on the rejection rule. It can be concluded that there are the solutions significantly different at the 95% confidence level.
Answer
Thus, the test statistic is 3.632.
Do not reject: There is no significant difference between the two etching solutions.
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