Can someone please help break this question down by steps? Thank you Huggers Pol
ID: 3387191 • Letter: C
Question
Can someone please help break this question down by steps? Thank you
Huggers Polls contend that an agent conducts a mean of 53 in-depth home survey’s every week. A streamlined survey form has been introduced, and Hugger want to evaluate its effectives. The number of in-depth surveys conducted during a week by a random sample of agents are:
53
57
50
55
58
54
60
52
59
62
60
60
51
59
56
At the .05 significance level can we conclude that the mean number of interviews conducted by their agents is more than 53 per week?
Explanation / Answer
Set Up Hypothesis
Null, H0: U=53
Alternate, H1: U>53
Test Statistic
Population Mean(U)=53
Sample X(Mean)=56.375
Standard Deviation(S.D)=3.6124
Number (n)=16
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =56.375-53/(3.6124/Sqrt(15))
to =3.737
| to | =3.737
Critical Value
The Value of |t | with n-1 = 15 d.f is 1.753
We got |to| =3.737 & | t | =1.753
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Right Tail - Ha : ( P > 3.7371 ) = 0.00099
Hence Value of P0.05 > 0.00099,Here we Reject Ho
we conclude that the mean number of interviews conducted by their agents is more than 53