A sample is randomly selected from a population with a mean

A sample is randomly selected from a population with a mean of = 50, and a treatment is administered to the individuals in the sample. After treatment, the sample is found to have a mean of M = 56 with a standard deviation of s = 8.

(a) If there are n = 4 individuals in the sample, are the data sufficient to reject H_o and conclude that the treatment has a significant effect using a two-tailed test with = .05? (Use 3 decimal places.)

t =______

b) If there are n = 16 individuals in the sample, are the data sufficient to conclude that the treatment has a significant effect? Again, assume a two-tailed test with = .05. (Use 3 decimal places.)

Conclusion

-Fail to reject the null hypothesis, there is a significant treatment effect.

Reject the null hypothesis, there is a significant treatment effect.   

Reject the null hypothesis, there is not a significant treatment effect.

Fail to reject the null hypothesis, there is not a significant treatment effect.

Solution

Set Up Hypothesis
Null, H0: U=50
Alternate, H1: U!=50
Test Statistic
Population Mean(U)=50
Sample X(Mean)=56
Standard Deviation(S.D)=8
Number (n)=4
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =56-50/(8/Sqrt(3))
to =1.5
| to | =1.5
Critical Value
The Value of |t | with n-1 = 3 d.f is 3.182
We got |to| =1.5 & | t | =3.182
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value :Two Tailed ( double the one tail ) -Ha : ( P != 1.5 ) = 0.2306
Hence Value of P0.05 < 0.2306,Here We Do not Reject Ho

ANS.
t-critical = ±3.182
to =1.5
Fail to reject the null hypothesis, there is not a significant treatment effect.