Exercise A: Find the sample proportion. n=550, x= 308 n=480, x= 168 n=620, x= 248 n=740, x= 259 n=1,120, x= 616
Recall the following z-score formula:
z=(x-μ)/σ
If the sample means are
z=(x ̅-μ_x ̅ )/σ_x ̅
The Central Limit Theorem is not limited to sample means. It can also be applied to sample proportions. The sample size is sufficiently large if np>5 and nq>5.
The following formula below can be used as an appropriate test statistic to test the hypothesis about population proportion p.
z=(p ̂-p)/√(pq/n) or z=(p ̂-p)/√((p(1-p))/n)
where: p ̂=sample proportion q=1-p
p=population proportion n=sample size
The following are the steps when testing hypothesis concerning a proportion:
Step 1: State the null and alternative hypotheses.
Step 2: Choose a level of significance α.
Step 3: Compute the test-statistic.
Step 4: Determine the critical value.
Step 5: Make a decision.
Null Hypothesis:
The null hypothesis is usually written in the form: H_0 ∶p=p ̂
Where p ̂ = specific numerical value for the population proportion p.
Alternative Hypothesis:
The alternative hypothesis can be any of the following:
H_a ∶p≠p ̂ (two-tailed test)
H_a ∶p>p ̂ (one-tailed test)
H_a ∶p<p ̂ (one-tailed test)
Example: It is believed that in the coming election, 65% of the voters in the province of South Cotabato will vote for the administration candidate for governor. Suppose 713 out of the 1,150 randomly selected voters indicate that they would vote for the administration candidate. At 0.10 level of significance, find out whether the percentage of voters for the administration candidate is different from 65%.
Solution:
Step 1: State the null and alternative hypotheses.
H_o:There is not enough evidence that the percentage of voters for the administration candidate
is different〖 from 65%. H〗_0 ∶p=65
H_a:There is enough evidence that the percentage of voters for the administration candidate
is different〖 from 65%. H〗_0 ∶p≠65
Step 2: Choose a level of significance: α=0.10
Step 3: Compute the test statistic.
p ̂=x/n = 713/1,150 = 0.62
z=(p ̂-p)/√((p(1-p))/n) =(0.62-0.65)/√((0.65(1-0.65))/1,150) = -2.133
Step 4: Determine the critical value.
The alternative hypothesis is non-directional. Hence, the two-tailed test shall be used. Divide α by 2, and then subtract the quotient from 0.5.
α/2=0.10/2=0.05
0.5-0.05=0.45
Using the Areas Under the Normal Curve Table, z_(α/2)=1.645. At 10% level of significance the critical value is ±1.645.
Step 5: Draw a conclusion.
Conclusion:
Since the computed test statistic z = -2.133 falls in the rejection region, reject the null hypothesis. Conclude that 0.10 level of significance, there is enough evidence that the percentage of voters
for the administration candidate is different from 65%.
Exercise A: Find the sample proportion.
n=550, x= 308
n=480, x= 168
n=620, x= 248
n=740, x= 259
n=1,120, x= 616
Exercise B: Test each of the hypotheses by using the given information:
〖 H〗_0 ∶p=42,
〖 H〗_a ∶p≠42
Sample size (n)= 150
Sample proportion=0.45
α= 0.05
Compute the test-statistic.
Determine the critical value and draw the normal curve.
Conclusion.
〖 H〗_0 ∶p=0.35,
〖 H〗_a ∶p>0.35
Sample size (n)= 180
Sample proportion=0.40
α= 0.05
Compute the test-statistic.
Determine the critical value and draw the normal curve.
Conclusion.
〖 H〗_0 ∶p=0.70,
〖 H〗_a ∶p≠0.70
Sample size (n)= 150
Sample proportion=0.75
α= 0.05
a. Compute the test-statistic.
Determine the critical value and draw the normal curve.
Conclusion.
Exercise C:
A research conducted on a certain company last year showed that 25% of the employees would rather drink coffee than soft drinks during break time. The company has recently decided to give free coffee during break time. In the new research conducted this year, out of the 125 randomly sampled employees 28% said that they would rather drink coffee than soft drinks. At 0.05 level of significance, is there sufficient evidence to suggest that the coffee drinkers have increased since the company has decided to give free coffee during break time?
Exercise D:
Before the Mayweather vs Pacquiao's Fight of the Century, 75 % of the people in Manila said that they preferred boxing over basketball. After the fight, out of the 150 randomly . chosen people in Manila, 105 said they preferred boxing over basketball. Does this indicate that people in Manila are losing interest in boxing? Use 0.05 level of significance.
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