Question

A newspaper is conducting a statewide survey concerning the race for governor. The newspaper will take a simple random sample of n registered voters and determine X = the number of voters that will vote for the Democratic candidate. Is there evidence that a clear majority of the population will vote for the Democratic candidate? To answer this, they will test the hypotheses H₀: p = 0.50 versus H_a: p > 0.50. Consider the two scenarios, where in Scenario 1, n = 1200 and X = 640. In Scenario 2, n = 120 and X = 64. Even though the values for $download?verifier=Il2VgIGbfYwBEU70LG85E9$ are the same in the two scenarios, we come to opposite decisions (we reject H₀ in one scenario and we do not in the other). What is the reason for these contrasting decisions?

We must have made a mistake. With equal values of $download?verifier=sXfuzZH1gILcM8W6cSrjNj$, we must come to equal decisions.

The 64 people in the small sample must also be part of the 640 people in the big sample. This means we are counting them twice, causing the results of the test with the larger sample to be wrong.

When the sample size is larger, the margin of error is larger. This may cause the results to be biased.

The sample size in the first scenario is much larger, leading to a test with higher power.

Answer 1

SCENARIO 1:

GIVEN:

Sample size of registered voters

(n) = 1200

Number of voters that will vote for the Democratic candidate

(X) = 640

HYPOTHESIS:

We claim "clear majority of the population will vote for the Democratic candidate" which indicates that "More than 50% of the population will vote for the Democratic candidate".

(That is, the true proportion of voters who will vote for the Democratic candidate is less than or equal to 50%)

Ho :p < 0.50

(That is, the true proportion of voters who will vote for the Democratic candidate is more than 50%)

LEVEL OF SIGNIFICANCE: Let us consider

g= 0,05

TEST STATISTIC:

z = (Ô-po)/ (po(1 - po))/n

which follows standard normal distribution

CALCULATION:

The sample proportion of voters that will vote for the Democratic candidate is,

p= x/n

  $=640/1200$

  

=0.53

Now

z = (Ô-po)/ (po(1 - po))/n

  

= (0.53 -0.50)/ (0.50(1 -0.50))/1200

  

= 0.03/0.0144

$z=2.08$

P VALUE:

The right tailed p value for test statistic $z=2.08$ is,

P2 > 2.08 = 1-P z < 2.08
{Since
P > a = 1- Plz<a
}

Using the z table, the probability value is the value with corresponding row 2.0 and column 0.08.

  

=1-0.9812

  

= 0.0188

DECISION RULE:

$Reject\, \, H_{0}\, \, if\, \, p-value\leq \alpha$

CONCLUSION:

Since the calculated p value is less than the significance level , we reject the null hypothesis and conclude that the true proportion of voters who will vote for the Democratic candidate is more than 50%. That is, there is sufficient evidence to prove that a clear majority of the population will vote for the Democratic candidate.

g= 0,05

SCENARIO 2:

GIVEN:

Sample size of registered voters $(n)=120$

Number of voters that will vote for the Democratic candidate

(X) = 64

HYPOTHESIS:

We claim "clear majority of the population will vote for the Democratic candidate" which indicates that "More than 50% of the population will vote for the Democratic candidate".

(That is, the true proportion of voters who will vote for the Democratic candidate is less than or equal to 50%)

Ho :p < 0.50

(That is, the true proportion of voters who will vote for the Democratic candidate is more than 50%)

LEVEL OF SIGNIFICANCE: Let us consider

g= 0,05

TEST STATISTIC:

z = (Ô-po)/ (po(1 - po))/n

which follows standard normal distribution

CALCULATION:

The sample proportion of voters that will vote for the Democratic candidate is,

p= x/n

  

= 64/120

  

=0.53

Now

z = (Ô-po)/ (po(1 - po))/n

  $=(0.53-0.50)/\sqrt{(0.50(1-0.50))/120}$

  

= 0.03/0.0456

= 0.66

P VALUE:

The right tailed p value for test statistic is,

= 0.66

$P[z>0.66]=1-P[z<0.66]$ {Since }

P > a = 1- Plz<a

Using the z table, the probability value is the value with corresponding row 0.6 and column 0.06.

  

=1-0.7454

  

= 0.2546

DECISION RULE:

$Reject\, \, H_{0}\, \, if\, \, p-value\leq \alpha$

CONCLUSION:

Since the calculated p value is greater than the significance level , we fail to reject the null hypothesis and conclude that the true proportion of voters who will vote for the Democratic candidate is less than or equal to 50%. That is, there is no sufficient evidence to prove that a clear majority of the population will vote for the Democratic candidate.

g= 0,05

REASON FOR CONTRACTING DECISIONS:

The sample size in the first scenario is much larger, leading to a test with higher power.