Chicken Delight claims that 89 percent of its orders are del

Chicken Delight claims that 89 percent of its orders are delivered within 10 minutes of the time the order is placed. A sample of 70 orders revealed that 59 were delivered within the promised time. At the .02 significance level, can we conclude that less than 89 percent of the orders are delivered in less than 10 minutes? What is the decision rule? (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.) 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.) What is your decision regarding the null hypothesis?

Solution

Set Up Hypothesis
Null, H0:P>=0.89
Alternate, H1: P<0.89
Test Statistic
No. Of Success chances Observed (x)=59
Number of objects in a sample provided(n)=70
No. Of Success Rate ( P )= x/n = 0.8429
Success Probability ( Po )=0.89
Failure Probability ( Qo) = 0.11
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.84286-0.89/(Sqrt(0.0979)/70)
Zo =-1.2606
| Zo | =1.2606
Critical Value
The Value of |Z | at LOS 0.02% is 2.05
We got |Zo| =1.261 & | Z | =2.05
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value: Left Tail -Ha : ( P < -1.26059 ) = 0.10373
Hence Value of P0.02 < 0.10373,Here We Do not Reject Ho

[ANSWERS]
1. Reject Ho, if Z<-2.05
2. Zo =-1.2606
3. Do not Reject Ho