Is this right?-Testing the mean

#1
Hey all,
Have a homework problem that I want to see if I did right
Problem is:

Suppose a friend tells you that the trout has a mean (mu) of 19 inches. However, the survey reports that for a random sample of 51 fish caught the mean length was x bar = 18.5 inches with an estimated S.D. of s=3.2 inches. Do these data indicate that the average length of a trout caught in the lake is les than mu=19 inches? Use 0.05

Work:

18.5-19/(3.2/sq rt 51)

=-.500/.448
=-1.116 =-1.12

P value = .1314

Since .1314 > .05, we cannot reject H0
 
Last edited:

vinux

Dark Knight
#2
No. The You didn't answer the question. And the first sentence is confusing (Suppose a friend tells you that the trout has a mean (mu) of 19 inches.)
 
#3
No. The You didn't answer the question. And the first sentence is confusing (Suppose a friend tells you that the trout has a mean (mu) of 19 inches.)
I apologize

***Suppose a friend tells you that tje average length of trout caught in pyramid lake has a mean (mu) of 19 inches.

Maybe I am confused on what to do, but I thought I answered the question?
 

Mean Joe

TS Contributor
#6
Hey all,
Have a homework problem that I want to see if I did right
Problem is:

Suppose a friend tells you that the trout has a mean (mu) of 19 inches. However, the survey reports that for a random sample of 51 fish caught the mean length was x bar = 18.5 inches with an estimated S.D. of s=3.2 inches. Do these data indicate that the average length of a trout caught in the lake is les than mu=19 inches? Use 0.05

Work:

18.5-19/(3.2/sq rt 51)

=-.500/.448
=-1.116 =-1.12

P value = .1314

Since .1314 > .05, we cannot reject H0
Looks good enough