I need to solve thise without using any coding ::: AppsFaculty Directory | M D N

Question

I need to solve thise without using any coding

::: AppsFaculty Directory | M D Nrw Tab Cassinpria Ru1dGuil 7. A hardness testing machine presses a rod with a pointed tip into a metal specimen with a known force. By measuring the depth of the depression caused by the tip, the hardness of the specimen is determined. Two different tips are available for this machine. Although the precision of the measurements produced by the two tips is the same, it is suspected that one tip produces different mean hardness ratings than the other a) At first, you're unsure whether the variances should be assumed same or different Assume they are the same and estimate the difference in population means to a 5% significance level b) Try it the other way c) Describe the differences in the results of a) and b) d) Discuss which approach you believe is more correct and why 4 4 4 10 sum 48 49 652 PM Type here to search · ": 10/11/2011

Explanation / Answer

Q7.
PART A.
When assuming population sd are equal
Given that,
mean(x)=4.8
standard deviation , s.d1=2.3944
number(n1)=10
y(mean)=4.9
standard deviation, s.d2 =2.2336
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.101
since our test is two-tailed
reject Ho, if to < -2.101 OR if to > 2.101
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (9*5.7332 + 9*4.989) / (20- 2 )
s^2 = 5.3611
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=4.8-4.9/sqrt((5.3611( 1 /10+ 1/10 ))
to=-0.1/1.0355
to=-0.0966
| to | =0.0966
critical value
the value of |t | with (n1+n2-2) i.e 18 d.f is 2.101
we got |to| = 0.0966 & | t | = 2.101
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != -0.0966 ) = 0.9241
hence value of p0.05 < 0.9241,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -0.0966
critical value: -2.101 , 2.101
decision: do not reject Ho
p-value: 0.9241

PART B.
When assuming population sd are not equal
Given that,
mean(x)=4.8
standard deviation , s.d1=2.3944
number(n1)=10
y(mean)=4.9
standard deviation, s.d2 =2.2336
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.262
since our test is two-tailed
reject Ho, if to < -2.262 OR if to > 2.262
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =4.8-4.9/sqrt((5.73315/10)+(4.98897/10))
to =-0.097
| to | =0.097
critical value
the value of |t | with min (n1-1, n2-1) i.e 9 d.f is 2.262
we got |to| = 0.09657 & | t | = 2.262
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -0.0966 ) = 0.925
hence value of p0.05 < 0.925,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -0.097
critical value: -2.262 , 2.262
decision: do not reject Ho
p-value: 0.925

PART C.
both the methods provide valid results it depends on the criteria given in the context, if we have a chance
to look in population data we choose equal variance method as it has more control on the data when you consider
population standard deviation rather sample satandard deviation

PART D.
for the above, we choose PART A, as the reason it has minimal chance of getting type I error

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---