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4.57 Influencing Voters: Is There Difterence in Effectiveness between a Phone Call and Flyer? Exercise 4.37 on page 235 describes a study to inves- tigate which m...

Question

4.57 Influencing Voters: Is There Difterence in Effectiveness between a Phone Call and Flyer? Exercise 4.37 on page 235 describes a study to inves- tigate which method, recorded phone call or flyer, is more effective in persuading voters to vote for particular candidate. Since in this case, the alternative hypothesis is not specified in par- ticular direction the hypotheses are Ho Pc Pf VS Ha Pc # Pf: AlI else is as in Exercise 4.56_ including the randomization distribution shown in Figure 4

4.57 Influencing Voters: Is There Difterence in Effectiveness between a Phone Call and Flyer? Exercise 4.37 on page 235 describes a study to inves- tigate which method, recorded phone call or flyer, is more effective in persuading voters to vote for particular candidate. Since in this case, the alternative hypothesis is not specified in par- ticular direction the hypotheses are Ho Pc Pf VS Ha Pc # Pf: AlI else is as in Exercise 4.56_ including the randomization distribution shown in Figure 4.18 (a) Sketch smooth curves that roughly approximate the distribution in Figure 4.18 and shade in the proportion of the area corresponding to the p-value for each of D = 0.2 and D = -0.4 b Two possible sample statistics are given below, along with several possible p-values. Select the most accurate p-value for each sample statistic Statistics D=02,D =-0.4 P-values 0.008,0.066,0.150,0.392,0.842 Ofall five p-values given in part (b) , which prO- vides the strongest evidence that the methods are not equally effective?



Answers

A sample survey contacted an SRS of 663 registered voters in Oregon shortly after an
clection and asked respondents whether they had voted. Voter records show that 56$\%$ of registered
voters had actually voted. We will see later that in repeated random samples of size 663 , the proportion in the sample who voted (call this proportion With vary according to the Normal distribution
with mean $\mu=0.56$ and standard deviation $\sigma=0.019$

(a) If the respondents answer truthfully, what is $\mathrm{P}(0.52 \leq V \leq 0.60)$ ? This is the probability that the sample proportion $V$ estimates the population proportion 0.56 within $\pm 0.04$
(b) In fact, 72$\%$ of the respondents said they had voted $(V=0.72) .$ If respondents answer truthfully, what is $P(V \geq 0.72) ?$ This probability is so small that it is good evidence that some people who did not vote claimed that they did vote.

We have a sample with n equals 493 members. And of those 493 members. 136 of them satisfying certain conditions want to meet our equal 136. We want to test to clean the population proportion P is greater than .214 With a confidence level of one of alpha equals 0.01. Now that we've got to find the confidence level, we have to proceed to the following steps to conduct this hypothesis test first is appropriate to the normal distribution. Yes, it is. Because both MP and QPR greater than five. What are prophecies that we're testing I know is that people's 0.214 are alternative. Is that P is greater than 0.214 Ak We're conducting a one tailed test. Next compute P hot. And our test statistic P hot remember is just all over end 10.276 Z is equal to p hot minus pee. All over route PQ over end. As we see on the formula on the right, giving Z equals 3.35 Next compute the P value associated this test statistic the P value. We can identify an easy table corresponds to the area under the normal curve to the right of Z equals 3.35 or P equals 0.4 We've graphed this also on the right to demonstrate how this looks on the distribution. Next we reject asian all based on this P value. Yes, we do. He is listening to alpha and we interpret that to mean that we have evidence that P is greater than .214.

All right. We are observing a population with mean 4.8 and we want to calculate the sample mean and standard deviation for the following sample data obtain to do this. We we know the definition of expire and S for a sample which are given here, X is equal to some of the data divided by n or 4.4, and s is equal to the sum of the deviations about the mean square, divided by n minus one or 0.28 Next we want to implement a left tailed test on this particular population, so we want to test the population is actually less than 4.8. Using a significance level output equals 0.5 Where we are told that X is approximately normal, that is this distribution is approximately normal. So we have to answer the following questions. In order to complete this test. To start off with, what are the significance of hypotheses? Alpha equals 0.5 No hypothesis H is new at this 0.4 point eight and H. A. Is that me was less than 4.8. It's the left tail test. What distribution When we used to be the test statistic, we're going to use a student's T distribution because sigma is unknown. We know that we can do so because the shape is symmetrical and bell curved. Given this, we want to calculate the T stat which is given by this formula. The T stat here equates a negative 3.499 Now, given this T stat, we want to compute the P P interval and then sketch the associated students distribution for P. Since we have degree of freedom, uh six minus one equals five. We have the R. P interval is between 50.5 point 01 We can grasp this as the area to the left of our T stat highlighted here in yellow and Marcus P. From this, we can conclude since P is less than legal to alpha, we have statistically significant findings and we can reject R. H. Saw, which we can interpret to ultimately mean that we have sufficient evidence suggesting our population means is less than no means for pointing.

We have a sample with N equals 83. And of that 83 members of the sample 64 satisfy a certain condition. Want to meet our equal 64. Now we want to use the sample data to test the claim that p does not equal 2.75 for the population at a confidence level of 5% or alpha equals 0.5 Now that we've identified the confidence level, we can go to the following procedural steps to conduct this hypothesis test first. Is the normal distribution appropriate to use? Yes, it is. Because both N. P and Q. P. R greater than five. B. One of the hypotheses were testing we're testing whether or not H not P equals 10.75 or H a p does not equal 0.75 Ak We're conducting a two tailed test. Next compute P hot. And the test statistic P hot is all over. N equals 20.77 Z is given by the formula on the right, taking in as input P hat P Q n N producing down to Z equals 0.42 Next we compute the P value. We use the table to see the area outside Z equals plus and minus 0.42 as we shaded in yellow and the graph on the right. This P value corresponds to 0.6745 Next we reject H not, no, we do not. P is greater than alpha and we interpret this signing to suggest that we lack evidence that P does not equal .75.

What we want to implement the right tail test that is. We want to test whether or not the population mean is greater than the known mean of new equals 1.75 Were given a sample with n equals 46 objects in the sample sample. Mean 2.5 and sample standard deviation 0.82 And we want to implement this test at the alpha equals 0.1 significant level. So start off with the state with the significance and hypothesis are alpha equals 0.1 H. Not null hypothesis is music was 1.75 and H. A alternative hypothesis is much was greater than 1.75 Next one distribution. Are we using Let's compute the statistic. We're gonna use a student's T distribution because stigma is unknown for the population, but we have met the requirements to use this distribution because N is greater than 30 so we can use it. We're gonna compete this statistic T. Using this formula for this problem, this solves down to 2.48 equals T. Next let's compute p using a statistic in this problem we have degree of freedom and minus when it was 45. So using a one tail tea table, we identify R. P. Interval is falling between 10.5 point 01 We can graft this as follows. Our T value 2.48 is marching the T axis and R. P value is the area under the student's T distribution. To the right of that P value. Next, we can conclude from this, P is less than the alpha. So we have statistically significant findings, which means we can reject the null hypothesis. H not we can interpret this to mean that we have evidence suggesting new is actually greater than the known population means 17.5.


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