Exercises - Hypothesis Tests for Proportions (One Sample)

  1. 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?

    1. Identify the claim
    2. State the null and alternative hypotheses
    3. Check any relevant assumptions
    4. Find the value of the relevant test statistic
    5. Determine if this is a one-tailed or two-tailed test
    6. Identify the significance level
    7. Find any relevant critical value(s)
    8. Make a sketch of the relevant distribution, indicate the location of the rejection region, the test statistic, and any critical values for the test
    9. Find the p-value for the test
    10. Conclude whether or not to reject the null hypothesis
    11. 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%.

  2. 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: $$z = \frac{\frac{58}{150} - 0.45}{\sqrt{\frac{\left(0.45\right)\left(0.55\right)}{150}}} \doteq -1.5591$$

    $p$-value = $0.0595 \gt \alpha = 0.05$

    Conclusion: fail to reject the null hypothesis

    Inference: there is no statistically significant evidence the politician is correct

    R:
    # Note: the following provides the p.value associated with the "exact" 
    # binomial test, which will slightly differ from the test using a 
    # normal approximation of the binomial distribution given above
    
    > binom.test(58,n=150,p=0.45,alternative="less",conf.level=0.95)$p.value
    [1] 0.06925327
    

  3. 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 $\rightarrow n = 882$

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

    $H_0 : p = 0.5$
    $H_1 : p \neq 0.5$

    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.

    R:
    # Note: the following provides the p.value associated with the "exact"
    # binomial test, which will slightly differ from the test using a
    # normal approximation of the binomial distribution given above
    
    > binom.test(481,n=882,p=0.5,alternative="two.sided",conf.level=0.5)$p.value
    [1] 0.007777031
    

  4. The sample proportion $\widehat{p}$ that answered a particular math question correctly is found for $280$ randomly selected college students. When testing that the claim that the majority of students will get this question correct, a $p$-value of $0.979$ is obtained. Find $\widehat{p}$. Also, what should one conclude about the claim?

    $H_1 : p > 0.5$, so right-tailed test. Area right is $0.979$, so area left is $0.021$. Thus, $z = -2.033$ for the test statistic. Substituting this into the formula for the test statistic and solving for $\widehat{p}$, we have $\widehat{p} = 123/280$.
    R:
    > z = qnorm(1-0.979)
    > p.hat = 0.5 + z*sqrt(0.5*0.5/280)
    > p.hat*280
    [1] 122.9863
    # Suggesting the true p.hat was 123/280
    

  5. Explain the difference between $p$ and the $p$-value.

    $p$ is the true probability of an event or true proportion of the population with a given trait, depending on the context.

    The $p$-value is the probability of seeing what was seen in the sample, or something more compelling (in the sense of the alternative hypothesis) under the assumption of the null hypothesis. When the $p$-value is small enough (i.e., less than the threshold established by the significance level $\alpha$) the sample is unusual enough to constitute evidence that our null hypothesis is probably incorrect. Thus, in these situations, we reject the null hypothesis. If the $p$-value is not smaller than $\alpha$, we haven't seen anything unusual and thus fail to reject the null hypothesis.

  6. Suppose one believes that exactly half of current college applications are submitted online, with the rest still being submitted via regular mail. A simple random sample of one thousand applications is examined, showing that $530$ were submitted online. At a $0.01$ level of significance, test the claim.

    Verify assumptions: $np = nq = 500 \ge 5$; $z = 1.90$; $z_{crit} = \pm 2.5756$; $p\textrm{-value} = 0.0578$; fail to reject; there is no statistically significant evidence that the percentage of applications submitted online is not $50\%$.
    R:
    # Note: the following provides the p.value associated with the "exact"
    # binomial test, which will slightly differ from the test using a
    # normal approximation of the binomial distribution given above
    
    > binom.test(530,n=1000,p=0.50,alternative="two.sided",conf.level=0.99)$p.value
    [1] 0.0620232
    

  7. In the 1980s it was generally believed that autism affected about $5\%$ of the nation's children. Some people believe that the increase in the number of chemicals in the environment has led to an increase in the incidence of autism. A recent study examined 384 children and found that 46 of them showed signs of some form of autism. Is this strong evidence that the level of autism has increased? Perform the appropriate hypothesis test with a significance level of $\alpha = 0.05$.

    Verify assumptions: $np = 19.2, nq = 364.8$ both $\ge 5$; $H_0: p \gt 0.05$; $n = 384$; $\widehat{p} \doteq 0.1198$; $z = 6.275$; $p\textrm{-value} = 1.754 \times 10^{-10}$; reject the null hypothesis; there is very strong evidence that the level of autism has increased.

  8. A company with a fleet of 150 cars found that the emissions system of 7 out of the 22 they tested failed to meet pollution control guidelines. Is this strong evidence that more than 20% of the fleet might be out of compliance? Test an appropriate hypothesis and state your conclusion.

    $H_0 : p \lt 0.20$; The sample size is more than $10\%$ of the $150$ cars in the fleet, so the binomial distribution at the heart of the one-sample proportion test is not appropriate (hypergeometric model instead). It is not clear that the sample is a simple random sample. $np = (22)(0.20) = 4.4 \not\ge 5$. Do not proceed with the test.

  9. In 2001 a national vital statistics report indicated that about $3\%$ of all births produced twins. Data from a large city hospital found only 7 sets of twins were born to 469 teenage girls. Does that suggest that mothers under the age 20 may be less likely to have twins? Test an appropriate hypothesis and state your conclusion.

    $H_0 : p \lt 0.03$; $np = (469)(0.03) \ge 5, nq = (469)(0.97) \ge 5$; $z = -1.91$, $p$-value of $0.0278$; reject the null hypothesis; we have evidence that the rate of twin births to teenagers at this hospital is less than $3\%$. Be careful not to generalize to all teenagers. There may be something special about this hospital.

  10. In a rural area only about $30\%$ of the wells drilled find adequate water at a depth of 100 feet or less. A local man claims to be able to find water by "dowsing" -- using a forked stick to indicate where the well should be drilled. You check with 80 of his customers and find that 27 have wells less than 100 feet deep. What do you conclude about the claim? Use a significance level of $\alpha = 0.05$.

    $H_0 : p = 0.30; H_1 : p \gt 0.30$; $np = (0.3)(80) \ge 5; nq = (0.7)(80) \ge 5$; $z = 0.73$, $p$-value of $0.232 \gt \alpha$; fail to reject; there is no evidence the dowser's chance of finding water is any better than normal drilling.