Question

A poll done for an organization found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 78 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent's survey, would you agree with the organization's poll? Conduct a hypothesis test at the 5% level.

a) State the null hypothesis. State the null hypothesis.
H0: p > 0.13
H0: p ≤ 0.13
H0: p ≠ 0.13
H0: p ≥ 0.13

Answer 1

Based on the contingent's survey, with 2 out of 78 respondents reporting angelic experiences, the evidence rejects the claim that ≥13% have encountered angels, favoring a proportion less than 13% at a 5% significance level.

The null hypothesis (H0) in this case is typically set up to challenge the organization's claim. Since the organization's poll suggests that 13% of Americans have seen or sensed the presence of an angel, the null hypothesis would aim to challenge this claim.

Given the context, the appropriate null hypothesis is:

H0: p ≤ 0.13

This null hypothesis assumes that the true proportion of Americans who have seen or sensed the presence of an angel is less than or equal to 13%. The contingent's survey, if significantly different, could support rejecting this claim in favor of an alternative hypothesis suggesting that the proportion is indeed less than 13%.

let's conduct a hypothesis test at the 5% level using the contingent's survey results:

Given:

Sample size (n) = 78

Number of people who have seen or sensed an angel in the contingent's survey = 2

Calculate the sample proportion:

`\[ \hat{p} = \frac{\text{Number who have seen or sensed an angel}}{\text{Total surveyed}} = (2)/(78) \approx 0.0256 \]`

Given the null hypothesis (H0: p ≤ 0.13), we'll use this sample proportion to conduct a one-sample z-test to determine if there's enough evidence to reject the null hypothesis in favor of the alternative hypothesis at the 5% significance level.

Calculate the z-score:

`\[ z = \frac{\hat{p} - p_0}{\sqrt{(p_0 \cdot (1 - p_0))/(n)}} \]`

`\[ z = \frac{0.0256 - 0.13}{\sqrt{(0.13 \cdot (1 - 0.13))/(78)}} \approx (-0.1044)/(0.0346) \approx -3.02 \]`

Using a z-table or calculator, the critical z-value for a one-tailed test at a 5% significance level is approximately -1.645.

Since the obtained z-value (-3.02) is less than the critical z-value (-1.645), we reject the null hypothesis. Therefore, based on the contingent's survey, there is sufficient evidence to suggest that the proportion of Americans who have seen or sensed the presence of an angel is less than 13%.