A politician claims that he is supported by a clear majority

A politician claims that he is supported by a clear majority of voters. In a recent survey, 21 out of 40 randomly selected voters indicated that they would vote for the politician. Use a 1% significance level for the test. Use Table 1.

Select the null and the alternative hypotheses.


  

Calculate the sample proportion. (Round your answer to 3 decimal places.)


  


  

Calculate the value of test statistic. (Round intermediate calculations to 4 decimal places. Round your answer to 2 decimal places.)


  


  

Calculate the p-value of the test statistic. (Round intermediate calculations to 4 decimal places. Round \"z\" value to 2 decimal places and final answer to 4 decimal places.)

A politician claims that he is supported by a clear majority of voters. In a recent survey, 21 out of 40 randomly selected voters indicated that they would vote for the politician. Use a 1% significance level for the test. Use Table 1.

Solution

Set Up Hypothesis
Under The Null Hypothesis H0:P=0.5
Under The Alternate Hypothesis H1: P>0.5
Test Statistic
No. Of Success chances Observed (x)=21
Number of objects in a sample provided(n)=40
No. Of Success Rate ( P )= x/n = 0.525
Success Probability   ( Po )=0.5
Failure Probability ( Qo) = 0.5
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.525-0.5/(Sqrt(0.25)/40)
Zo =0.3162
| Zo | =0.3162
Critical Value
The Value of |Z ?| at LOS 0.01% is 2.33
We got |Zo| =0.316 & | Z ? | =2.33
Make Decision
Hence Value of |Zo | < | Z ? | and Here we Do not Reject Ho
P-Value: Right Tail - Ha : ( P > 0.31623 ) = 0.37591
Hence Value of P0.01 < 0.37591,Here We Do not Reject Ho

a)
   H0: p = 0.50; HA: p > 0.50

b)
P = x/n = 0.525

c)
Zo =0.3162

d)
P-Value: Right Tail - Ha : ( P > 0.31623 ) = 0.37591

A politician claims that he is supported by a clear majority of voters. In a recent survey, 21 out of 40 randomly selected voters indicated that they would vote
A politician claims that he is supported by a clear majority of voters. In a recent survey, 21 out of 40 randomly selected voters indicated that they would vote

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