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?
(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: H0:p=0.43
The alternative hypothesis: H1:p≠0.43
(c) Assumptions: np=(611)(0.43)=262.73≥5, nq=(611)(0.57)=348.27≥5, so the requirements/assumptions of the test are met.
(d) Test Statistic: z=308611−0.43√(.43)(.57)611≐3.699
This is a two-tailed test (as the alternative hypothesis involves "≠")
Significance level: α=0.01
Critical values: ±2.5758 (on a TI-calculator, find ±invNorm(0.005)
)
p-value ≐0.000216 (on a TI-calculator, find 2*normalcdf(3.699,999999999)
)
Reject the null as p-value <α (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
H0:p=0.45
H1:p<0.45
Check assumptions:
np=(150)(0.45)=67.5≥5 and
nq=(150)(0.55)=82.5≥5,
so requirements are met.
ˆp=58/150
Test statistic: z=58150−0.45√(0.45)(0.55)150≐−1.5591
p-value = 0.0595>α=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)$p.value [1] 0.06925327
Adults were randomly selected for a Newsweek poll. They were asked if they "favor or oppose using federal tax dollars to fund medical research using stem cells obtained from human embryos." Of those polled, 481 were in favor, 401 were opposed, and 120 were unsure. A politician claims that people don't really understand the stem cell issue and their responses to such questions are random responses equivalent to a coin toss. Exclude the 120 subjects who said that they were unsure, and use a 0.01 significance level to test the claim that the proportion of subjects who respond in favor is equal to 0.5. What does the result suggest about the politician's claim?
481 in favor, 401 opposed →n=882
ˆp=481/882
H0:p=0.5
H1:p≠0.5
Checking assumptions: np=nq=(882)(0.5)=441≥5, so the requirements are met.
Test statistic:
z=481882−0.5√(0.5)(0.5)882≐2.6937p-value =0.00706<α=0.01, so reject the null hypothesis.
There is statistically significant evidence the politician is wrong.
R: # Note: the following provides the p.value associated with the "exact" # binomial test, which will slightly differ from the test using a # normal approximation of the binomial distribution given above > binom.test(481,n=882,p=0.5,alternative="two.sided",conf.level=0.5)$p.value [1] 0.007777031
The sample proportion ˆp 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 ˆp. Also, what should one conclude about the claim?
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
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 α) 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 α, we haven't seen anything unusual and thus fail to reject the null hypothesis.
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.
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
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 α=0.05.
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.
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.
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.