n 8 m 240 standard deviation 30 Twotailed test with alpha

n= 8 m= 240 standard deviation = 30 Two-tailed test with alpha = .05

How did they find out that the critical boundary is = -1.96 which corresponds with M= 220.4 (How did we get M= 220.4?) Thanks.

Solution

Null, H0: U=220.4
Alternate, H1: U!=220.4
Test Statistic
Population Mean(U)=220.4
Sample X(Mean)=240
Standard Deviation(S.D)=30
Number (n)=8
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =240-220.4/(30/Sqrt(8))
to =1.848
| to | =1.848
Critical Value
The Value of |t | with n-1 = 7 d.f is 2.365
We got |to| =1.848 & | t | =2.365
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value :Two Tailed ( double the one tail ) -Ha : ( P != 1.8479 ) = 0.1071
Hence Value of P0.05 < 0.1071,Here We Do not Reject Ho

[ANSWERS]
a. It is the value we receive from Z-table. We match the table value with the level of significance
provided