In a presidential election, 308 out of 611 voters surveyed said that they voted for the candidate who won. Use a 0.01 significance level to test the claim that among all voters, the percentage who believe that they voted for the winning candidate is equal to 43%, which is the actual percentage of votes for the winning candidate. What does the result suggest about voter perceptions?

- Identify the claim
- State the null and alternative hypotheses
- Check any relevant assumptions
- Find the value of the relevant test statistic
- Determine if this is a one-tailed or two-tailed test
- Identify the significance level
- Find any relevant critical value(s)
- Make a sketch of the relevant distribution, indicate the location of the rejection region, the test statistic, and any critical values for the test
- Find the p-value for the test
- Conclude whether or not to reject the null hypothesis
- Make an appropriate inference

(a) The claim is that among all voters, the percentage who believe that they voted for the winning candidate is 43%.

(b) The null hypothesis: $H_0 : p = 0.43$

The alternative hypothesis: $H_1 : p \neq 0.43$(c) Assumptions: $np = (611)(0.43) = 262.73 \ge 5$, $nq = (611)(0.57) = 348.27 \ge 5$, so the requirements/assumptions of the test are met.

(d) Test Statistic: $$z = \frac{\frac{308}{611} - 0.43}{\sqrt{\frac{(.43)(.57)}{611}}} \doteq 3.699$$

This is a two-tailed test (as the alternative hypothesis involves "$\neq$")

Significance level: $\alpha = 0.01$

Critical values: $\pm2.5758$ (on a TI-calculator, find $\pm$

`invNorm(0.005)`

)$p$-value $\doteq 0.000216$ (on a TI-calculator, find

`2*normalcdf(3.699,999999999)`

)Reject the null as $p$-value $\lt \alpha$ (or as the test statistic falls in the rejection region)

There is significant evidence that the percentage of voters that believed they voted for the winning candidate was not 43%.

The Gallup Poll reported that 45% of individuals felt they were worse off than 1 year ago. A politician feels that this is too high for her district, so she commissions her own survey and finds that, out of 150 randomly selected citizens, 58 feel they are worse off today than 1 year ago. At significance level 0.05, is the politician correct about her district?

$n = 150$

$H_0 : p = 0.45$

$H_1 : p \lt 0.45$Check assumptions:

$np = (150)(0.45) = 67.5 \ge 5$ and

$nq = (150)(0.55) = 82.5 \ge 5$,

so requirements are met.$\widehat{p} = 58\,/\,150$

Test statistic: $$z = \frac{\frac{58}{150} - 0.45}{\sqrt{\frac{\left(\frac{58}{100}\right)\left(\frac{42}{100}\right)}{150}}} \doteq -2.357$$

$p$-value = 0.0092 \lt \alpha = 0.05

Conclusion: reject the null hypothesis

Inference: there is statistically significant evidence the politician is correct

Adults were randomly selected for a Newsweek poll. They were asked if they "favor or oppose using federal tax dollars to fund medical research using stem cells obtained from human embryos." Of those polled, 481 were in favor, 401 were opposed, and 120 were unsure. A politician claims that people don't really understand the stem cell issue and their responses to such questions are random responses equivalent to a coin toss. Exclude the 120 subjects who said that they were unsure, and use a 0.01 significance level to test the claim that the proportion of subjects who respond in favor is equal to 0.5. What does the result suggest about the politician's claim?

481 in favor, 401 opposed $\rightarrow n = 882$

$\widehat{p} = 481\,/\,882$

$H_0 : p = 0.5$

$H_1 : p \neq 0.55$Checking assumptions: $np = nq = (882)(0.5) = 441 \ge 5$, so the requirements are met.

Test statistic:

$$z = \frac{\frac{481}{882} - 0.5}{\sqrt{\frac{(0.5)(0.5)}{882}}} \doteq 2.6937$$$p$-value $= 0.00706 \lt \alpha = 0.01$, so reject the null hypothesis.

There is statistically significant evidence the politician is wrong.