Question

A poll done for Newsweek 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 $76$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 Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results.

Short answer

The number of people in the United States who have seen or felt an angel is less than $13\%$.

Step-by-step solution

Introduction 

Null hypotheses is,

${\mathrm{H}}_0:\mathrm{p}\ge 0.13$

Alternative hypotheses is,

${\mathrm{H}}_0:\mathrm{p}<0.13$

The proportion of Americans who have seen or sensed angels is the random variable${\mathrm{P}}^{\text{'}}$.

Normal distribution with parameters $\mathrm{p}=0.13$and $\sqrt{\frac{\mathrm{p}(1-\mathrm{p})}{\mathrm{n}}}=\sqrt{\frac{0.13(1-0.13)}{76}}$.

The test statistic for this test is,

$\mathrm{z}=\frac{\left|{\mathrm{p}}^{\text{'}}-\mathrm{p}\right|}{\sqrt{\mathrm{p}(1-\mathrm{p})/\mathrm{n}}}$

$=\frac{0.026-0.13}{\sqrt{0.113/76}}$

Explanation

When the p-value is bigger than the established alpha value, we do not reject the null hypothesis. Hence,

$\mathrm{p}\text{-value}=0.0036<0.05$

$=\mathrm{\alpha }$

There the data to tell is that the percentage of Americans who have seen or sensed an angel is less than $13\%$at a level of significance of $5\%$.

${\mathrm{p}}^{\text{'}}-{\mathrm{z}}_{\frac{\mathrm{\alpha }}{2}}\sqrt{\frac{{\mathrm{p}}^{\text{'}}\left(1-{\mathrm{p}}^{\text{'}}\right)}{\mathrm{n}}}\le \mathrm{p}\le {\mathrm{p}}^{\text{'}}+{\mathrm{z}}_{\frac{\mathrm{\alpha }}{2}}\sqrt{\frac{{\mathrm{p}}^{\text{'}}\left(1-{\mathrm{p}}^{\text{'}}\right)}{\mathrm{n}}}........1$

percentage point $\frac{\mathrm{\alpha }}{2}$of the standard normal distribution.

$\frac{\mathrm{\alpha }}{2}=0.025$

$\Rightarrow {\mathrm{z}}_{\frac{\mathrm{\alpha }}{2}}={\mathrm{z}}_{0.025}$

localid="1649942557746" $=1.96............2$

From equation $1$and$2$

$0.026-1.96\sqrt{\frac{0.026(1-0.026)}{76}}\le \mathrm{p}\le 0.026+1.96\sqrt{\frac{0.026(1-0.026)}{76}}$

$0.026-1.96\times 0.018\le \mathrm{p}\le 0.026+1.96\times 0.018$

Population proportion is,

$0\le \mathrm{p}\le 0.623$

Conclusion

Test is,

$$\begin{gathered} \text{p-value}=\mathrm{P}\left\{{\mathrm{p}}^{\text{'}}<0.026\right\} \\ =\mathrm{P}\{\mathrm{z}<-2.688\} \\ =0.0036 \end{gathered}$$