Chicken Delight claims that 85 percent of its orders are del

Chicken Delight claims that 85 percent of its orders are delivered within 10 minutes of the time the order is placed. A sample of 80 orders revealed that 64 were delivered within the promised time. At the .10 significance level, can we conclude that less than 85 percent of the orders are delivered in less than 10 minutes?

Compute the value of the test statistic. (Negative amount should be indicated by a minus sign.Round sample proportion to 2 decimal places. Round your answer to 2 decimal places.)

Solution

Set Up Hypothesis
Null, H0:P=0.85
Alternate, H1: P<0.85
Test Statistic
No. Of Success chances Observed (x)=64
Number of objects in a sample provided(n)=80
No. Of Success Rate ( P )= x/n = 0.8
Success Probability ( Po )=0.85
Failure Probability ( Qo) = 0.15
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.8-0.85/(Sqrt(0.1275)/80)
Zo =-1.2524
| Zo | =1.2524
Critical Value
The Value of |Z | at LOS 0.1% is 1.28
We got |Zo| =1.252 & | Z | =1.28
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value: Left Tail -Ha : ( P < -1.25245 ) = 0.1052
Hence Value of P0.1 < 0.1052,Here We Do not Reject Ho

we don\'t have evidence to conclude that less than 85 percent of the orders are delivered in less than 10 minutes