The spring for a prosthetic foot you are designing needs to be 2 cm in length. Y
ID: 3273940 • Letter: T
Question
The spring for a prosthetic foot you are designing needs to be 2 cm in length. You randomly sample 36 springs from a supplier and find their lengths to be 2.03 cm on average, with a standard deviation of +/- .1 cm. Is there cause for concern that your supplier is systematically designing springs that are different than your specifications?
Yes because I get a one-sample t value of 1.8. Since the critical value of t (2-tailed) at alpha = .05 is equal to right around 2.03, and 1.8 is less than this critical value, I MUST reject the null hypothesis that the sample comes from the referent population.
No because I get a one-sample t value of 1.8. Since the critical value of t (2-tailed) at alpha = .05 is equal to right around 2.03, and 1.8 is less than this critical value, I CANNOT reject the null hypothesis that the sample comes from the referent population.
Neither A nor B
A & B
A.Yes because I get a one-sample t value of 1.8. Since the critical value of t (2-tailed) at alpha = .05 is equal to right around 2.03, and 1.8 is less than this critical value, I MUST reject the null hypothesis that the sample comes from the referent population.
B.No because I get a one-sample t value of 1.8. Since the critical value of t (2-tailed) at alpha = .05 is equal to right around 2.03, and 1.8 is less than this critical value, I CANNOT reject the null hypothesis that the sample comes from the referent population.
C.Neither A nor B
D.A & B
Explanation / Answer
Given that,
population mean(u)=2
sample mean, x =2.03
standard deviation, s =0.1
number (n)=36
null, Ho: =2
alternate, H1: !=2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.03
since our test is two-tailed
reject Ho, if to < -2.03 OR if to > 2.03
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =2.03-2/(0.1/sqrt(36))
to =1.8
| to | =1.8
critical value
the value of |t | with n-1 = 35 d.f is 2.03
we got |to| =1.8 & | t | =2.03
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 1.8 ) = 0.0805
hence value of p0.05 < 0.0805,here we do not reject Ho
ANSWERS
---------------
null, Ho: =2
alternate, H1: !=2
test statistic: 1.8
critical value: -2.03 , 2.03
decision: do not reject Ho
p-value: 0.0805
B.
No because I get a one-sample t value of 1.8. Since the critical value of t (2-tailed) at alpha = .05 is equal to right around 2.03, and 1.8 is less than this critical value, I CANNOT reject the null hypothesis that the sample comes from the referent population.
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