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confidence level is that if many samples of the same size were taken from the population, about 95% of the resulting confidence intervals would capture the true population parameter (assuming the conditions are met and the probability model is true). Note that this is a relative frequency interpretation. • We cannot use... |
a normal model may reasonably be used in this setting. Create a 95% confidence interval for the proportion of U.S. adults who live with one or more chronic conditions. Also interpret the confidence interval in the context of the study. 5.8 Twitter users and news, Part I. A poll conducted in 2013 found that 52% of U.S. a... |
from 1,151 US residents, the survey reported a 95% confidence interval of 3.40 to 4.24 days in 2010. (a) Interpret this interval in context of the data. (b) What does “95% confident” mean? Explain in the context of the application. (c) Suppose the researchers think a 99% confidence level would be more appropriate for thi... |
interval, determine if the claim of the Dean from part (b) would be supported based on a 99% confidence interval? 14Pew Research Center, A Majority of Teens Have Experienced Some Form of Cyberbullying. September 27, 2018. 5.3. INTRODUCING HYPOTHESIS TESTING 273 5.3 Introducing hypothesis testing In an experiment, one t... |
Calculate the best estimate for p using the data. Label the point estimate as ˆp. The sample proportion for the complication rate is 9 complications divided by the 142 surgeries the consultant has worked on: ˆp = 9/142 = 0.063. EXAMPLE 5.22 Is it possible to prove that the consultant’s work reduces complications? No. ... |
ect. Our job as data scientists is to play the skeptic: before we buy into the alternative hypothesis, we need to see strong supporting evidence. EXAMPLE 5.24 Identify the null and alternative claim regarding the consultant’s complication rate. H0: The true complication rate for the consultant’s clients is the same as ... |
and the data notably disagree, then we will reject the null hypothesis in favor of the alternative hypothesis. Don’t worry if you aren’t a master of hypothesis testing at the end of this section. We’ll discuss these ideas and details many times in this chapter and the two chapters that follow. The null claim is always... |
. Why did we conduct a two-sided hypothesis test for this setting? The setting was framed in the context of the consultant being helpful, but what if the consultant actually performed worse than the US complication rate? Would we care? More than ever! Since we care about a finding in either direction, we should run a tw... |
-value is represented by the lower tail. When it takes the form p > null value, the p-value is represented by the upper tail. When using p = null value, then the p-value is represented by both tails. FINDING AND INTERPRETING THE P-VALUE We find and interpret the p-value according to the nature of the alternative hypothe... |
the result is not statistically significant. We do not reject H0, and we do not have strong evidence for HA. Recall that the null claim is the claim of no difference. If we reject H0, we are asserting that there is a real difference. If we do not reject H0, we are saying that the null claim is reasonable, but we are not ... |
. If this was the case, the sample may have been too small to reliable detect this effect. EXAMPLE 5.31 An experiment was conducted where study participants were randomly divided into two groups. Both were given the opportunity to purchase a DVD, but one half was reminded that the money, if not spent on the DVD, could b... |
value). 5.3. INTRODUCING HYPOTHESIS TESTING 279 5.3.4 Calculating the p-value by simulation (special topic) When conditions for the applying a normal model are met, we use a normal model to find the p-value of a test of hypothesis. In the complication rate example, the distribution is not normal. It is, however, binomi... |
so we should double the one-tail area to get the p-value:18 p-value = 2 × left tail = 2 × 0.0877 = 0.1754 Figure 5.9: The null distribution for ˆp, created from 10,000 simulated studies. The left tail contains 8.77% of the simulations. For a two-sided test, we double the tail area to get the p-value. This doubling acc... |
which would be the null hypothesis and which the alternative? The jury considers whether the evidence is so convincing (strong) that there is evidence beyond a reasonable doubt of the person’s guilt. That is, the starting assumption (null hypothesis) is that the person is innocent until evidence is presented that conv... |
A Type II Error is failing to reject H0 when HA is actually true. When we dod not reject the null hypothesis, it is possible that we make a Type II Error. EXAMPLE 5.33 In a US court, the defendant is either innocent (H0) or guilty (HA). What does a Type I Error represent in this context? What does a Type II Error repr... |
does not necessarily mean something was wrong with the data or that we made a computational mistake. Sometimes data simply point us to the wrong conclusion, which is why scientific studies are often repeated to check initial findings. 282 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.3.7 Choosing a significance level If H0 is t... |
Type II Error is called power. It is common for researchers to perform a power analysis to ensure their study collects enough data to detect the effects they anticipate finding. As you might imagine, if the effect they care about is small or subtle, then if the effect is real, the researchers will need to collect a large ... |
fference from the null value. She also would obtain some reasonable estimate for the standard deviation. With these important pieces of information, she would choose a sufficiently large sample size so that the power for the meaningful difference is perhaps 80% or 90%. While larger sample sizes may still be used, she might... |
.3-1.5 for basic principles of data collection. 21 The problem is even greater than p-hacking. In what has been called the “reproducibility crisis”, researchers have failed to reproduce a large proportion of results that were found significant and were published in scientific journals. This problem highlights the importa... |
met, we can calculate the test statistic. The test statistic tells us how many standard errors the point estimate (sample value) is from the null value (i.e. the value hypothesized for the parameter in the null hypothesis). When investigating a single mean or proportion or a difference of means or proportions, the test... |
true. We commit a Type II Error if we call a result not significant when there is a real difference or effect. P(Type II Error) = β. 5.3. INTRODUCING HYPOTHESIS TESTING 285 – The probability of a Type I Error (α) and a Type II Error (β) are inversely related. Decreasing α makes β larger; increasing α makes β smaller. – O... |
ENCE Exercises 5.13 Identify hypotheses, Part I. Write the null and alternative hypotheses in words and then symbols for each of the following situations. (a) A tutoring company would like to understand if most students tend to improve their grades (or not) after they use their services. They sample 200 of the students... |
< 0.6 HA : ˆp > 0.7 5.16 Married at 25. A study suggests that the 25% of 25 year olds have gotten married. You believe that this is incorrect and decide to collect your own sample for a hypothesis test. From a random sample of 25 year olds in census data with size 776, you find that 24% of them are married. A friend of... |
ance level is (I) 0.05 or (II) 0.10. 5.19 Testing for Fibromyalgia. A patient named Diana was diagnosed with Fibromyalgia, a long-term syndrome of body pain, and was prescribed anti-depressants. Being the skeptic that she is, Diana didn’t initially believe that anti-depressants would help her symptoms. However after a ... |
”, we are asserting that the null value is reasonable, not that the parameter is exactly equal to the null value. • For a 95% confidence interval, 95% is not the probability that the true value lies inside the confidence interval (it either does or it doesn’t). Likewise, for a hypothesis test, α is not the probability th... |
a corresponding 99% confidence interval. 5.21 Chronic illness, Part II. In 2013, the Pew Research Foundation reported that “45% of U.S. adults report that they live with one or more chronic conditions”, and the standard error for this estimate is 1.2%. Identify each of the following statements as true or false. Provide... |
stated above? 24National Opinion Research Center, General Social Survey, 2018. 290 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.23 Testing for food safety. A food safety inspector is called upon to investigate a restaurant with a few customer reports of poor sanitation practices. The food safety inspector uses a hypothesis ... |
back to your notes and discover that you made a careless mistake, the sample size should have been n = 500. Will your p-value increase, decrease, or stay the same? Explain. 291 Chapter 6 Inference for categorical data 6.1 Inference for a single proportion 6.2 Inference for the difference of two proportions 6.3 Testing... |
error no greater than a certain value. 5. Recognize that margin of error calculations only measure sampling error, and that other types of errors may be present. 294 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA 6.1.1 Distribution of a sample proportion (review) The distribution of a sample proportion, such as the distrib... |
both at least 10. 1When sampling without replacement and sampling greater than 10% of the population, a modified standard error formula should be used. 6.1. INFERENCE FOR A SINGLE PROPORTION 295 6.1.3 Confidence intervals for a proportion The Gallup organization began measuring the public’s view of the Supreme Court’s j... |
test the second condition (the success-failure condition) to ensure that the sample size is large enough for the central limit theorem to apply. The successfailure condition is met when np and n(1 − p) are at least 10. Since p is always unknown when constructing a confidence interval for p, we use the sample proportion... |
ed that ˆp can be modeled using a normal distribution, the critical value is a z. The z value can be found in the t-table on page 514, using the bottom row (∞), where the column corresponds to the confidence level. Here the confidence level is 90%, so z=1.65. We can now construct the 90% confidence interval as follows. po... |
a different conclusion based on different confidence levels, which may feel a little jarring. However, this will happen with real data, and it highlights why it is important to be explicit in identifying the confidence level being used. Having worked through this example, we now summarize the steps for constructing a confi... |
��dence level that greater than half of all U.S. adults think there is intelligent life on other planets? Carry out a confidence interval procedure to answer this question. Use the five step framework to organize your work. Identify: First we identify the parameter of interest. Here the parameter is the true proportion o... |
% and 70.9% of U.S. adults think that there is intelligent life on other planets.2 2False. The true percent of U.S. adults that think there is intelligent life on other planets either falls in that interval or it doesn’t. A correct interpretation of the confidence level would be that if we were to repeat this process ov... |
sample percent was 68% and the sample size was 1,033.3 3Navigate to the 1-proportion Z-interval on the calculator. To find x, the number of yes responses in the sample, we multiply the sample proportion by the sample size. Here 0.68 × 1033 = 702.44. We must round this to an integer, so we use x = 702. Also, n =1033 and... |
could use that value. If we have no such estimate, we must use some other value for p. It turns out that the margin of error is largest when p is 0.5, so we typically use this worst case estimate of p = 0.5 if no other estimate is available. 1.96 × 1.962 × 1.962 × 0.5(1 − 0.5) n 0.5(1 − 0.5) n 0.5(1 − 0.5) 0.042 ≤ 0.0... |
we have to do to the sample size in order to halve the margin of error (decrease it by a factor of 2)?4 GUIDED PRACTICE 6.13 A manager is about to oversee the mass production of a new tire model in her factory, and she would like to estimate the proportion of these tires that will be rejected through quality control. ... |
claims she has more than 50% support from the district’s electorate. A newspaper collects a random sample of 500 likely voters in the district and estimates Toohey’s support to be 52%. (a) Identify the null and the alternative hypothesis. What value should we use as the null value, p0? (b) Can we model ˆp using a norm... |
0 ≥ 10 and n(1 − p0) ≥ 10 SE = ˆp(1 − ˆp) n p0(1 − p0) n 6.1. INFERENCE FOR A SINGLE PROPORTION 303 EXAMPLE 6.15 (Continues previous example). Deborah Toohey’s campaign manager claimed she has more than 50% support from the district’s electorate. A newspaper poll finds that 52% of 500 likely voters who were sampled supp... |
hypothesis. We cannot ever prove the null hypothesis directly. The value 0.5 is reasonable, but many other values are reasonable as well. There are many values that would not get rejected by this test. We now summarize the steps for carrying out a hypothesis test for a proportion using the five step framework introduce... |
telephone interviews from a random sample of 1,019 adults in the United States. Does this poll provide evidence that greater than half of U.S. adults oppose nuclear energy? Carry out an appropriate test at the 0.10 significance level. Use the five step framework to organize your work. Identify: We will test the followin... |
of U.S. adults oppose nuclear energy (as of March 2016). GUIDED PRACTICE 6.18 In context, interpret the p-value of 0.006 from the previous example.7 7Assuming the normal model is accurate and assuming the null hypothesis is true, i.e. that the true proportion of U.S. adults that oppose nuclear energy really is 0.5, th... |
• Enter the null value, p0. • Enter the number of successes, x. • Enter the sample size, n. 6. Hit the EXE button, which returns z Z-statistic p-value p the sample proportion ^p the sample size n GUIDED PRACTICE 6.19 Using a calculator, find the test statistic and p-value for the earlier Example 6.17. Recall that we we... |
one proportion Z-test and Z-interval require the sampling distribution for ˆp to be nearly normal. For this reason we must check that the following conditions are met. 1. Independence: The data should come from a random sample or random process. When sampling without replacement, check that the sample size is less tha... |
a sample size that is 4 times as large will reduce the standard error of the sample proportion by one-half. (c) The distribution of sample proportions of random samples of size 140 is approximately normal. (d) The distribution of sample proportions of random samples of size 280 is approximately normal. 6.2 Young Ameri... |
’s responsibility to promote equality between men and women. (d) In order to decrease the margin of error to 1%, we would need to quadruple (multiply by 4) the sample size. (e) Based on this confidence interval, there is sufficient evidence to conclude that a majority of Americans think it’s the government’s responsibilit... |
ering”. (d) Without doing any calculations, describe what would happen to the confidence interval if we decided to use a higher confidence level. (e) Without doing any calculations, describe what would happen to the confidence interval if we used a larger sample. 6.7 Study abroad. A survey on 1,509 high school seniors who... |
��ed? 6.9 National Health Plan, Part I. A Kaiser Family Foundation poll for a random sample of US adults in 2019 found that 79% of Democrats, 55% of Independents, and 24% of Republicans supported a generic “National Health Plan”. There were 347 Democrats, 298 Republicans, and 617 Independents surveyed.16 (a) A politica... |
asked each person to take one sip from their cup and identify the soda as diet or regular. 53 participants correctly identified the soda. (a) Do these data provide strong evidence that these people are able to detect the difference between diet and regular soda, in other words, are the results significantly better than j... |
for a heart attack? • How different is the approval of the 2010 healthcare law under two different question phrasings? • Does the use of fish oils reduce heart attacks better than a placebo? Learning objectives 1. State and verify whether or not the conditions for inference on the difference of two proportions using a nor... |
the sampling distribution for each sample proportion must be nearly normal. Second, the observations must be independent, both within and between groups. We cover these conditions in greater detail next. 312 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA 6.2.2 Checking conditions for inference using a normal distribution W... |
�p2, follows a normal distribution. Because the patients were randomly assigned to one of the two groups and one heart attack patient is unlikely to influence the next that was in the study, the observations are considered independent, both within the samples and between the samples (since there is no sampling, there is... |
. That is, we are 90% confident that the treatment of blood thinners changes survival rate for patients like those in the study by -2.7% to +28.7% percentage points. Because this interval contains both negative and positive values, we do not have enough information to say with confidence whether blood thinners harm or he... |
that the true difference in the proportion of [...] is between and. If applicable, draw a conclusion based on whether the interval is entirely above, is entirely below, or contains the value 0. 314 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA EXAMPLE 6.20 A remote control car company is considering a new manufacturer for ... |
�p1 − ˆp2 = 0.879 − 0.958 = −0.079. The SE of the difference of sample proportions is: ˆp1(1− ˆp1) 0.879(1−0.879) 1000 = + ˆp2(1− ˆp2) n2 + 0.958(1−0.958) 1000 n1 = 0.0121 So the 95% confidence interval is given by: 0.879 − 0.958 ± 1.96 × 0.879(1 − 0.879) 1000 + 0.958(1 − 0.958) 1000 −0.079 ± 1.96 × 0.0121 (−0.103, −0.05... |
ate and hit ENTER, which returns: the confidence interval sample 1 proportion sample 2 proportion (, ) ^p1 ^p2 n1 n2 size of sample 1 size of sample 2 CASIO FX-9750GII: 2-PROPORTION Z-INTERVAL 1. Navigate to STAT (MENU button, then hit the 2 button or select STAT). 2. Choose the INTR option (F4 button). 3. Choose the Z ... |
you may know, by 2014 nearly all Americans will be required to have health insurance. [People who do not buy insurance will pay a penalty] while [People who cannot afford it will receive financial help from the government]. Do you approve or disapprove of this policy? For each randomly sampled respondent, the statements... |
difference of proportions: p1 − p2. In the last section, we used a 2-proportion Z-interval to estimate the parameter p1 − p2; here, we will use a 2-proportion Z-test to test the null hypothesis that p1 − p2 = 0, i.e. that p1 = p2. Recall that the test statistic Z has the form: Z = point estimate − null value SE of esti... |
p2. Here: ˆpc = 771(0.47) + 732(0.34) 771 + 732 = 0.407 POOLED SAMPLE PROPORTION When the null hypothesis is p1 = p2, it is useful to find the pooled sample proportion: ˆpc = number of “successes” number of cases = x1 + x2 n1 + n2 = n1 ˆp1 + n2 ˆp2 n1 + n2 Here x1 represents the number of successes in sample 1. If x1 i... |
6.25 Complete the hypothesis test using a significance level of 0.01. We have already set up the hypotheses and verified that the difference of proportions can be modeled using a normal distribution. We can now calculate the test statistic and p-value. Z = point estimate − null value SE of estimate = (0.47 − 0.34) − 0 0.... |
2 1 SE of estimate: ˆpc(1 − ˆpc) n1 null value: 0 + 1 n2, where ˆpc is the pooled proportion p-value = (based on the Z-statistic and the direction of HA) Conclude: Compare the p-value to α, and draw a conclusion in context. If the p-value is < α, reject H0; there is sufficient evidence that [HA in context]. If the p-valu... |
) ≥ 10, 12933(1 − 0.0133) ≥ 10, 12938(0.0133) ≥ 10, and 12938(1 − 0.0133) ≥ 10, so both conditions are met. Calculate: We will calculate the Z-statistic and the p-value. Z = point estimate − null value SE of estimate The point estimate is the difference of sample proportions: ˆp1− ˆp2 = 0.0112−0.0155 = −0.0043. The valu... |
, <, or > to correspond to HA. 7. Choose Calculate and hit ENTER, which returns: z ^p1 ^p2 Z-statistic sample 1 proportion sample 2 proportion p ^p p-value pooled sample proportion CASIO FX-9750GII: 2-PROPORTION Z-TEST 1. Navigate to STAT (MENU button, then hit the 2 button or select STAT). 2. Choose the TEST option (F... |
, less than, or equal to p2. • When comparing two proportions to each other, the parameter of interest is the difference of proportions, p1 − p2, and we use the difference of sample proportions, ˆp1 − ˆp2, as the point estimate. • The sampling distribution for ˆp1 − ˆp2 is nearly normal when the success-failure condition... |
�p2) ≥ 10. Success-failure for Test: n1 ˆpc ≥ 10, n1(1 − ˆpc) ≥ 10, n2 ˆpc ≥ 10, and n2(1 − ˆpc) ≥ 10. • When the conditions are met, we calculate the confidence interval and the test statistic using the same structure as in the previous section. Confidence interval: point estimate ± z × SE of estimate Test statistic: Z ... |
not. The table below displays how many patients survived and died in each group.21 survived died control 4 30 treatment 24 45 Suppose we are interested in estimating the difference in survival rate between the control and treatment groups using a confidence interval. Explain why we cannot construct such an interval usin... |
FERENCE FOR CATEGORICAL DATA 6.18 The Daily Show. A Pew Research foundation poll indicates that among a random sample of 1,099 college graduates, 33% watch The Daily Show. Meanwhile, 22% of the 1,110 people with a high school degree but no college degree in the poll watch The Daily Show. A 95% confidence interval for (p... |
sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,69... |
fferent for the two states. (Reminder: Check conditions) (b) It is possible the conclusion of the test in part (a) is incorrect. If this is the case, what type of error was made? 6.23 Prenatal vitamins and Autism. Researchers studying the link between prenatal vitamin use and autism surveyed the mothers of a random samp... |
fit using chi-square In this section, we develop a method for assessing a null model when the data take on more than two categories, such as yes/no/maybe instead of simply yes/no. This allows us to answer questions such as the following: • Are juries representative of the population in terms of race/ethnicity, or is th... |
If the individuals are randomly selected to serve on a jury, about how many of the 275 people would we expect to be White? How many would we expect to be Black? About 72% of the population is White, so we would expect about 72% of the jurors to be White: 0.72 × 275 = 198. Similarly, we would expect about 7% of the jur... |
data. In this example we have four categories: White, Black, Hispanic, and other. Because we have four values rather than just one or two, we need a new tool to analyze the data. Our strategy will be to find a test statistic that measures the overall deviation between the observed and the expected counts. We first find t... |
distribution, we will be able to obtain a p-value to evaluate whether there appears to be racial/ethnic bias in the juries for the city we are considering. 6.3.3 The chi-square distribution and finding areas The chi-square distribution is sometimes used to characterize data sets and statistics that are always positive ... |
. One important difference from the t-table is that the chi-square table only provides upper tail values. Upper tail 1 df 2 3 4 5 6 7 0.3 1.07 2.41 3.66 4.88 6.06 7.23 8.38 0.2 1.64 3.22 4.64 5.99 7.29 8.56 9.80 0.1 2.71 4.61 6.25 7.78 9.24 10.64 12.02 0.05 3.84 5.99 7.81 9.49 11.07 12.59 14.07 0.02 5.41 7.82 9.84 11.67... |
.1 and 0.2. Using a calculator or statistical software allows us to get more precise areas under the chi-square curve than we can get from the table alone. 0510152025Degrees of Freedom249 330 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA (a) (c) (e) (b) (d) (f) Figure 6.8: (a) Chi-square distribution with 3 degrees of free... |
value). 7. Enter the Upper bound (use a large number, such as 1000). 8. Enter the degrees of freedom, df. 9. Hit the EXE button. GUIDED PRACTICE 6.33 Figure 6.8(c) shows an upper tail for a chi-square distribution with 5 degrees of freedom and a cutoff of 5.1. Find the tail area using a calculator.29 GUIDED PRACTICE 6.... |
statistic χ2 follows a chi-square distribution with k − 1 degrees of freedom, where k is the number of bins or categories of the variable. EXAMPLE 6.37 How many categories were there in the juror example? How many degrees of freedom should be associated with the chi-square distribution used for χ2? In the jurors examp... |
� because we test whether or not the proposed or expected distribution is a good fit for the observed data. 051015 6.3. TESTING FOR GOODNESS OF FIT USING CHI-SQUARE 333 CHI-SQUARE GOODNESS OF FIT TEST FOR ONE-WAY TABLE Suppose we are to evaluate whether there is convincing evidence that a set of observed counts O1, O2,.... |
with the specified distribution of color, Identify: Identify the hypotheses and the significance level, α. H0: The distribution of [...] matches the specified or population distribution. HA: The distribution of [...] doesn’t match the specified or population distribution. Choose: Choose the correct test procedure and iden... |
lin’s sample of size 712. (See the paragraph before this example for more background.) website percentages (2008): observed percentages: Brown Blue 24% 13% 18.7% 18.7% 19.5% 14.5% 15.1% 13.5% Orange Green Yellow 16% Red 13% 14% 20% Is there evidence at the 5% significance level that the distribution of M&M’s in 2016 wer... |
counts as well as the expected counts. To find the observed counts, we use the observed percentages. For example, 18.7% of 712 = 0.187(712) = 133. observed counts: expected counts: Blue Orange Green Yellow Red Brown 133 170.9 139 113.9 133 142.4 108 92.6 96 92.6 103 99.6 χ2 = (observed − expected)2 expected = (133 − 17... |
CHI-SQUARE GOODNESS OF FIT TEST Use STAT, TESTS, χ2GOF-Test. 1. Enter the observed counts into list L1 and the expected counts into list L2. 2. Choose STAT. 3. Right arrow to TESTS. 4. Down arrow and choose D:χ2GOF-Test. 5. Leave Observed: L1 and Expected: L2. 6. Enter the degrees of freedom after df: 7. Choose Calcul... |
: is L1 and Expected: is L2. Let df: be 5. You should find that χ2 = 17.36 and p-value = 0.004. 338 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA Section summary The inferential procedures we saw in the first two sections of this chapter are based on the test statistic following a normal distribution. In this section, we int... |
ed distribution of color, the hypotheses can often be written as: H0: The distribution of [...] matches the specified or population distribution. HA: The distribution of [...] doesn’t match the specified or population distribution. We test these hypotheses at the α significance level using a χ2χ2χ2 goodness of fit test. • ... |
at the 5% significance level. (c) When finding the p-value of a chi-square test, we always shade the tail areas in both tails. (d) As the degrees of freedom increases, the variability of the chi-square distribution decreases. A professor using an open source introductory statistics book predicts 6.27 Open source textboo... |
p-value in the context of the problem. Photo by Shrikant Rao (http://flic.kr/p/4Xjdkk) CC BY 2.0 license 35Liwei Teng et al. “Forage and bed sites characteristics of Indian muntjac (Muntiacus muntjak) in Hainan Island, China”. In: Ecological Research 19.6 (2004), pp. 675–681. 340 CHAPTER 6. INFERENCE FOR CATEGORICAL DA... |
described as the following: H0: The algorithms each perform equally well. HA: The algorithms do not perform equally well. In this experiment, the explanatory variable is the search algorithm. However, an outcome variable is also needed. This outcome variable should somehow reflect whether the search results align with ... |
that you will have different search results. While the data presented in this section resemble what might be encountered in a real experiment, these data are simulated. 342 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA 6.4.2 Expected counts in two-way tables EXAMPLE 6.43 From the experiment, we estimate the proportion of u... |
78 × (column 2 total) = 1769.5 0.7078 × (column 3 total) = 1769.5 Looking back to how the fraction 0.7078 was computed – as the fraction of users who did not perform a new search (7078/10000) – these three expected counts could have been computed as row 1 total table total row 1 total table total row 1 total table tota... |
2 If the null hypothesis is true (i.e. the algorithms are equally useful), then the test statistic χ2 = 6.12 closely follows a chi-square distribution with 2 degrees of freedom. Using this information, we can compute the p-value for the test, which is depicted in Figure 6.13. COMPUTING DEGREES OF FREEDOM FOR A TWO-WAY... |
is that there is some difference in performance among the algorithms. This chi-square test does not tell us which algorithm performed better than the others. To answer this question, we could compare the relevant proportions or construct bar graphs. The proportion that resulted in the new search can be calculated as cu... |
pected counts: All expected counts are ≥ 5 (calculate and record expected counts). Calculate: Calculate the χ2-statistic, df, and p-value. test statistic: χ2 = (observed − expected)2 expected df = (# of rows − 1) × (# of columns − 1) p-value = (area to the right of χ2-statistic with the appropriate df ) Conclude: Compa... |
for each question type. HA: The likelihood of disclosing the problem is not the same for each question type. Choose: We want to know if the distribution of disclose/hide is the same for each of the three question types, so we want a chi-square test for homogeneity. Check: This is an experiment in which there were thre... |
applies when there is only one random sample and there are two categorical variables. The null claim is always that the two variables are independent, while the alternate claim is that the variables are dependent. EXAMPLE 6.48 Figure 6.14 summarizes the results of a Pew Research poll. A random sample of adults in the ... |
column one is found by multiplying the row one total (2119) and column one total (1458), then dividing by the table total (4223): 2119×1458 = 731.6. Similarly for the first column and the second row: 2104×1458 4223 = 726.4. Repeating this process, we get the expected counts: 4223 Approve Disapprove Obama Congr. Dem. Co... |
political party, Identify: Identify the hypotheses and the significance level, α. H0: [variable 1] and [variable 2] are independent. HA: [variable 1] and [variable 2] are dependent. Choose: Choose the correct test procedure and identify it by name. Here we use a χ2χ2χ2 test for independence. Check: Check that the test ... |
and taking action to help address climate change. Identify: We will test the following hypotheses at the α = 0.05 significance level. H0: Generation and taking action to help address climate change are independent. HA: Generation and taking action to help address climate change are dependent. Choose: Two variables were... |
null is true and the model holds. Because the p-value is so small, we reject the null hypothesis. 6.4. HOMOGENEITY AND INDEPENDENCE IN TWO-WAY TABLES 351 6.4.5 Technology: the chi-square test for two-way tables TI-83/84: ENTERING DATA INTO A TWO-WAY TABLE 1. Hit 2ND x−1 (i.e. MATRIX). 2. Right arrow to EDIT. 3. Hit 1 ... |
n is for columns. • Enter the data. • Return to the test page by hitting EXIT twice. 6. Enter the Observed matrix that was used by hitting MAT (F1 button) and the matrix letter (e.g. C). 7. Enter the Expected matrix where the expected values will be stored (e.g. D). 8. Hit the EXE button, which returns χ2 p df chi-squ... |
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