In semiconductor manufacturing, wet chemical etching is often used to remove silicon from the backs of wafers prior to metallization. The etch rate is an important characteristic of this process and known to follow a normal distribution. Two different etching solutions have been compared using two random samples of 10 wafers for each solution. Assume the variances are equal. The etch rates are as follows (in ml per minute): Solution 1 Solution 2
9.810.410.210.0
9.410.310.610.2
9.310.010.710.7
9.610.310.410.4
10.210.110.510.3
(a) Calculate the sample mean for solution 1.

(b) Calculate the sample standard deviation for solution 1.

(c) Calculate the sample mean for solution 2.

(d) Calculate the sample standard deviation for solution 2.

(e) Test the hypothesis
H
0
:
m
1
=
m
2
versus
H
1
:
m
1
â 
m
2
.

Calculate
t
0
.

(f) Calculate a 95% two-sided confidence interval on the difference in mean etch rate.

(e) There is sufficient evidence to reject the claim that the mean etch rate is the same for both solutions

Explanation:

Given

Solutions 1 and 2

Solving (a): Mean of solution 1

For Solution 1, we have the following data:

The sample mean is calculated as:

Where

So, we have:

Solving (b): Mean of Solution 2

For Solution 1, we have the following data:

The sample mean is calculated as:

Where

So, we have:

Solving (c): Sample Standard Deviation of solution 1

This is calculated as:

This gives:

--- approximated

Solving (d): Sample Standard Deviation of solution 2

This is calculated as:

So, we have:

Solving (e): Test the hypothesis

Start by calculating pooled standard deviation

Calculate test statistic

Calculate the degree of freedom

The p value is the in the column title of the student t distribution in row 18

The p value is less than the significance level (0.05).

i.e.

So: Reject the null hypothesis

Solving (f): 95% two-sided confidence interval

Calculate the degree of freedom

Calculate

Using student t distribution

Calculate margin of error (E)

The boundaries of the confidence are:

The confidence interval is:

a. formula: f=ma or f/m=a

10,000n/1,267kg≈7.9m/

b. formula: a= and s=d/t

speed= 394.6/15

s=26.3m/s

a=

a=1.75m/

c. 7.9-1.75=difference of 6.15m/

d. the force that most likely caused this difference is friction forces

answer; when one group has an extremely high score at pretest;