Suppose a recent report states that 58 of the residents in B

Suppose a recent report states that 58% of the residents in Boston feel that the mayor is doing a good job. A local reporter feels otherwise. She believes that the proportion differs from 0.58. In order to test her claim, a sample of 180 residents from her city is taken and 99 feel that the mayor is doing a good job.

State the hypothesis to test if the proportion differs.

What is the value of the test statistic? What is the p-value you would use for the rejection rule test?

At a 5% level of significance, what is your conclusion to the hypothesis test? Do the sample results support the reporter’s claim?

Solution

Set Up Hypothesis
Null, H0:P=0.58
Alternate, H1: P!=0.58
Test Statistic
No. Of Success chances Observed (x)=99
Number of objects in a sample provided(n)=180
No. Of Success Rate ( P )= x/n = 0.55
Success Probability ( Po )=0.58
Failure Probability ( Qo) = 0.42
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.55-0.58/(Sqrt(0.2436)/180)
Zo =-0.8155
| Zo | =0.8155
Critical Value
The Value of |Z | at LOS 0.05% is 1.96
We got |Zo| =0.815 & | Z | =1.96
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -0.81549 ) = 0.41479
Hence Value of P0.05 < 0.4148,Here We Do not Reject Ho

She believes that 58% residents in Boston feel that the mayor is doing a good job

[ANSWERS]

1. Zo =-0.8155

2. ( P != -0.81549 ) = 0.41479

3. results support the reporter’s claim. She believes that 58% residents in Boston feel that the mayor is doing a good job