4.
The Pew Research Center reported that 57% of Americans favored the Wall Street bailout of 2008. A random sample of 300 people showed that 155 people favored the bailout. If appropriate, test whether the population proportion of Americans is different from 57%. Use level of significance a = 0.01. What if we cut the sample size down to between 150-160 people, and we cut the sample number of successes as well, so that the sample proportion p stayed the same? Suppose that everything else stayed the same. Describe what would happen to and the conclusion.
A)
would remain the same and so would the conclusion.
B)
would increase and the conclusion would change.
C)
would decrease and the conclusion would remain the same.
D)
would increase and the conclusion would remain the same.
To test whether the population proportion of Americans is different from 57%, we can use a hypothesis test. Let's define the null and alternative hypotheses:
Null Hypothesis (H₀): The population proportion of Americans is 57%.
Alternative Hypothesis (H₁): The population proportion of Americans is different from 57%.
To perform the hypothesis test, we can use the Z-test for a single proportion.
First, let's calculate the test statistic, which is the Z-score. The formula for the Z-score is:
Z = (p - P₀) / √[(P₀ * (1 - P₀)) / n]
Where:
p is the sample proportion (155/300 = 0.5167),
P₀ is the hypothesized population proportion (0.57),
n is the sample size (300).
Using these values, we can calculate the Z-score:
Z = (0.5167 - 0.57) / √[(0.57 × (1 - 0.57)) / 300] ≈ -1.225
Next, we need to determine the critical value. Since the significance level is given as α = 0.01, we need to find the critical Z-value(s) for a two-tailed test at α/2 = 0.005. Using a Z-table or calculator, we find the critical Z-value to be approximately ±2.576.
Since the calculated Z-score (-1.225) falls within the range of -2.576 to +2.576, we fail to reject the null hypothesis. This means our evidence is not strong enough to conclude that the population proportion is different from 57% at the 1% significance level.
Now, let's consider the scenario where we have a smaller sample size (between 150-160) and a reduced number of successes, but the sample proportion (p) remains the same. In this case, the standard error of the proportion would increase because the formula for the standard error is √[(p̂ * (1 - p̂)) / n], where p̂ is the sample proportion.
If we lower the sample size, the standard error would increase, widening the confidence interval. As a result, the range of values considered acceptable for the population proportion would become larger. This means that the likelihood of rejecting the null hypothesis (concluding that the population proportion is different from 57%) would decrease.
Therefore, the correct answer is C) The p-value would decrease, and the conclusion would remain the same.