Question

In: Statistics and Probability

Microtronic produces scan engines for a bar code systems. They must meet a minimum joint strength...

Microtronic produces scan engines for a bar code systems. They must meet a minimum joint strength of 13 lbs between the brackets and clamps. The team conducts a study of two different welding methods. The company measures joint strength by utilizing a destructive weld test whereby the clamp is pulled until it separates from the bracket (weld joint breaks). Given the study results below. (Assume the alternative hypotheses is not equal to zero and alpha =0.05.

Method A
19
10.3
20.6
10
14.4
11.5
17.1
21.9
20.5
14.7
16.9
21.4

Method B
9.7
10.9
18.6
13.7
11.7
10.6
10.3
17.7
19.9
20.9
15.3
19.0

A) means are statistically different, variances are not.
B) neither means nor variances are statistically different.
C) both means and variances are statistically different
D) variances are statistically different

Solutions

Expert Solution

ANSWER-

Here we have to test

1).Hypothesis for mean-

H0: mean of method A and B are statistically same.

V/s

H1:mean of method A and B are statistically not same.

2).Hypothesis for variance.

H0:variance of method A and B are statistically same.

V/s

H1:variance of method A and B are statistically not same.

# R-code.

x=c(19,10.3,20.6,10,14.4,11.5,17.1,21.9,20.5,14.7,16.9,21.4);x

y=c(9.7,10.9,18.6,13.7,11.7,10.6,10.3,17.7,19.9,20.9,15.3,19);y

a=t.test(x,y);a

b=var.test(x,y);b  

# output -

 [1] 19.0 10.3 20.6 10.0 14.4 11.5 17.1 21.9 20.5 14.7 16.9 21.4
 [1]  9.7 10.9 18.6 13.7 11.7 10.6 10.3 17.7 19.9 20.9 15.3 19.0

        Welch Two Sample t-test

data:  x and y
t = 0.95751, df = 21.979, p-value = 0.3487
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -1.943370  5.276704
sample estimates:
mean of x mean of y 
 16.52500  14.85833 
 
here p-value is greater than 0.05 thus we accept H0 and conclude that mean for method A and B are same.

        F test to compare two variances

data:  x and y
F = 1.0631, num df = 11, denom df = 11, p-value = 0.921
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
 0.3060448 3.6929155
sample estimates:
ratio of variances 
   1.063107 

here p-value is greater than 0.05 thus we accept H0 and conclude that variance for method A and B are same.

Thus final answer is Option

B) neither means nor variances are statistically different.


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