Homework SourseThe program office budgets 30000 per program review at the c
| The program office budgets $30,000 per program review at the contractor’s site. Your concern for “end of year” spending drills is that you have budgeted enough for the reviews (i.e. you are only concerned if the actual costs are higher than the $30,000 target). A sample of 16 trips yielded a mean of $32,000 and a standard deviation of $3,500. Test the budgeted amount at the 80% level of confidence. Select the correct answer out of each pair of choices. (Carry intermediate calculations to three decimal places.) |
| The tp is 1.341 |
| The tp is 0.866 |
| The tc is 11.429 |
| The tc is 2.286 |
| We would reject the null hypothesis |
| We would fail to reject the null hypothesis |
| We would conclude that it is reasonable to use the $30,000 budget figure |
| We would recommend revising the budget figure |
| The program office budgets $30,000 per program review at the contractor’s site. Your concern for “end of year” spending drills is that you have budgeted enough for the reviews (i.e. you are only concerned if the actual costs are higher than the $30,000 target). A sample of 16 trips yielded a mean of $32,000 and a standard deviation of $3,500. Test the budgeted amount at the 80% level of confidence. Select the correct answer out of each pair of choices. (Carry intermediate calculations to three decimal places.) |
| The tp is 1.341 |
| The tp is 0.866 |
| The tc is 11.429 |
| The tc is 2.286 |
| We would reject the null hypothesis |
| We would fail to reject the null hypothesis |
| We would conclude that it is reasonable to use the $30,000 budget figure |
| We would recommend revising the budget figure |
Solution
t-test For Single Mean
Set Up Hypothesis
Null, H0: U<30000
Alternate, H1: U>30000
Test Statistic
Population Mean(U)=30000
Sample X(Mean)=32000
Standard Deviation(S.D)=3500
Number (n)=16
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =32000-30000/(3500/Sqrt(15))
to =2.286
| to | =2.286
Critical Value
The Value of |t | with n-1 = 15 d.f is 0.866
We got |to| =2.286 & | t | =0.866
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Right Tail - Ha : ( P > 2.2857 ) = 0.01862
Hence Value of P0.2 > 0.01862,Here we Reject Ho
ANS:
The tp is 0.866
The tc is 2.286
We would reject the null hypothesis
We would conclude that it is reasonable to use the $30,000 budget figure
