From fac5c35122563823b6876ae811da251b82c8a896 Mon Sep 17 00:00:00 2001 From: hardin47 Date: Wed, 11 Oct 2023 15:27:08 -0700 Subject: [PATCH 1/8] discernible --- 11-foundations-randomization.qmd | 21 ++++++++++--------- 17-inference-two-props.qmd | 2 +- 22-inference-many-means.qmd | 2 +- 25-inf-model-mlr.qmd | 10 ++++----- 26-inf-model-logistic.qmd | 2 +- exercises/_01-ex-data-hello.qmd | 2 +- exercises/_02-ex-data-design.qmd | 4 ++-- exercises/_13-ex-foundations-mathematical.qmd | 4 ++-- exercises/_14-ex-foundations-errors.qmd | 2 +- exercises/_16-ex-inference-one-prop.qmd | 4 ++-- exercises/_16-sa-inference-one-prop.qmd | 2 +- exercises/_17-ex-inference-two-props.qmd | 4 ++-- exercises/_17-sa-inference-two-props.qmd | 2 +- exercises/_20-ex-inference-two-means.qmd | 2 +- exercises/_20-sa-inference-two-means.qmd | 4 ++-- exercises/_21-ex-inference-paired-means.qmd | 2 +- exercises/_22-ex-inference-many-means.qmd | 2 +- exercises/_25-ex-inf-model-mlr.qmd | 14 ++++++------- exercises/_26-ex-inf-model-logistic.qmd | 4 ++-- exercises/_26-sa-inf-model-logistic.qmd | 2 +- 20 files changed, 46 insertions(+), 45 deletions(-) diff --git a/11-foundations-randomization.qmd b/11-foundations-randomization.qmd index 7aa4dc94..f689e3ab 100644 --- a/11-foundations-randomization.qmd +++ b/11-foundations-randomization.qmd @@ -716,8 +716,8 @@ The test statistic in the opportunity cost study was the difference in the propo In each of these examples, the **point estimate** of the difference in proportions was used as the test statistic. ::: -When the p-value is small, i.e., less than a previously set threshold, we say the results are **statistically significant**\index{statistically significant}. -This means the data provide such strong evidence against $H_0$ that we reject the null hypothesis in favor of the alternative hypothesis. +When the p-value is small, i.e., less than a previously set threshold, we say the results are **statistically discernible**\index{statistically significant}\index{statistically discernible}. +This means the data provide such strong evidence against $H_0$ that we reject the null hypothesis in favor of the alternative hypothesis.^[Many texts use the phrase "statistically significant" instead of "statistically discernible". We have chosen to use "discernible" to indicate that a precise statistical event has happened, as opposed to an interesting effect which may or may not fit the statistical definition of discernible or significant.] The threshold is called the **significance level**\index{hypothesis testing!significance level}\index{significance level} and often represented by $\alpha$ (the Greek letter *alpha*). The value of $\alpha$ represents how rare an event needs to be in order for the null hypothesis to be rejected. Historically, many fields have set $\alpha = 0.05,$ meaning that the results need to occur less than 5% of the time, if the null hypothesis is to be rejected. @@ -725,37 +725,38 @@ The value of $\alpha$ can vary depending on the the field or the application. ```{r} #| include: false -terms_chp_11 <- c(terms_chp_11, "significance level", "statistically significant") +terms_chp_11 <- c(terms_chp_11, "significance level", "statistically significant", "statistically discernible") ``` +Note that you may have heard the phrase "statistically significant" as a way to describe "statistically discernible." Although in everyday language "significant" would indicate that a difference is large or meaningful, that is not necessarily the case here. -The term "statistically significant" only indicates that the p-value from a study fell below the chosen significance level. +The term "statistically discernible" indicates that the p-value from a study fell below the chosen significance level. For example, in the sex discrimination study, the p-value was found to be approximately 0.02. -Using a significance level of $\alpha = 0.05,$ we would say that the data provided statistically significant evidence against the null hypothesis. +Using a significance level of $\alpha = 0.05,$ we would say that the data provided statistically discernible evidence against the null hypothesis. However, this conclusion gives us no information regarding the size of the difference in promotion rates! ::: {.important data-latex=""} -**Statistical significance.** +**Statistical discernibility.** -We say that the data provide **statistically significant**\index{hypothesis testing!statistically significant} evidence against the null hypothesis if the p-value is less than some predetermined threshold (e.g., 0.01, 0.05, 0.1). +We say that the data provide **statistically discernible**\index{hypothesis testing!statistically discernible.} evidence against the null hypothesis if the p-value is less than some predetermined threshold (e.g., 0.01, 0.05, 0.1). ::: ::: {.workedexample data-latex=""} In the opportunity cost study in Section \@ref(caseStudyOpportunityCost), we analyzed an experiment where study participants had a 20% drop in likelihood of continuing with a video purchase if they were reminded that the money, if not spent on the video, could be used for other purchases in the future. We determined that such a large difference would only occur 6-in-1,000 times if the reminder actually had no influence on student decision-making. What is the p-value in this study? -Would you classify the result as "statistically significant"? +Would you classify the result as "statistically discernible"? ------------------------------------------------------------------------ The p-value was 0.006. -Since the p-value is less than 0.05, the data provide statistically significant evidence that US college students were actually influenced by the reminder. +Since the p-value is less than 0.05, the data provide statistically discernible evidence that US college students were actually influenced by the reminder. ::: ::: {.important data-latex=""} **What's so special about 0.05?** -We often use a threshold of 0.05 to determine whether a result is statistically significant. +We often use a threshold of 0.05 to determine whether a result is statistically discernible. But why 0.05? Maybe we should use a bigger number, or maybe a smaller number. If you're a little puzzled, that probably means you're reading with a critical eye -- good job! diff --git a/17-inference-two-props.qmd b/17-inference-two-props.qmd index d0efe6dd..87d86e80 100644 --- a/17-inference-two-props.qmd +++ b/17-inference-two-props.qmd @@ -406,7 +406,7 @@ for(i in 1:k){ } ``` -The choice of 95% or 90% or even 99% as a confidence level is admittedly somewhat arbitrary; however, it is related to the logic we used when deciding that a p-value should be declared as "significant" if it is lower than 0.05 (or 0.10 or 0.01, respectively). +The choice of 95% or 90% or even 99% as a confidence level is admittedly somewhat arbitrary; however, it is related to the logic we used when deciding that a p-value should be declared as "discernible" if it is lower than 0.05 (or 0.10 or 0.01, respectively). Indeed, one can show mathematically, that a 95% confidence interval and a two-sided hypothesis test at a cutoff of 0.05 will provide the same conclusion when the same data and mathematical tools are applied for the analysis. A full derivation of the explicit connection between confidence intervals and hypothesis tests is beyond the scope of this text. diff --git a/22-inference-many-means.qmd b/22-inference-many-means.qmd index 850b71b9..f574fc01 100644 --- a/22-inference-many-means.qmd +++ b/22-inference-many-means.qmd @@ -218,7 +218,7 @@ Consider again the original hypotheses: - $H_0:$ $\mu_{OF} = \mu_{IF} = \mu_{C}$ - $H_A:$ The average on-base percentage $(\mu_i)$ varies across some (or all) groups. -Why might it be inappropriate to run the test by simply estimating whether the difference of $\mu_{C}$ and $\mu_{OF}$ is "statistically significant" at a 0.05 significance level? +Why might it be inappropriate to run the test by simply estimating whether the difference of $\mu_{C}$ and $\mu_{OF}$ is "statistically discernible" at a 0.05 significance level? ------------------------------------------------------------------------ diff --git a/25-inf-model-mlr.qmd b/25-inf-model-mlr.qmd index dcf429bb..4ffda6e1 100644 --- a/25-inf-model-mlr.qmd +++ b/25-inf-model-mlr.qmd @@ -9,7 +9,7 @@ source("_common.R") ::: {.chapterintro data-latex=""} In Chapter \@ref(model-mlr), the least squares regression method was used to estimate linear models which predicted a particular response variable given more than one explanatory variable. -Here, we discuss whether each of the variables individually is a statistically significant predictor of the outcome or whether the model might be just as strong without that variable. +Here, we discuss whether each of the variables individually is a statistically discernible predictor of the outcome or whether the model might be just as strong without that variable. That is, as before, we apply inferential methods to ask whether a variable could have come from a population where the particular coefficient at hand was zero. If one of the linear model coefficients is truly zero (in the population), then the estimate of the coefficient (using least squares) will vary around zero. The inference task at hand is to decide whether the coefficient's difference from zero is large enough to decide that the data cannot possibly have come from a model where the true population coefficient is zero. @@ -62,7 +62,7 @@ lm(interest_rate ~ debt_to_income + term + credit_checks, data = loans) %>% mutate(p.value = ifelse(p.value < 0.0001, "<0.0001", round(p.value, 4))) %>% kbl( linesep = "", booktabs = TRUE, - caption = caption_helper("Summary of a linear model for predicting interest rate based on `debt_to_income`, `term`, and `credit_checks`. Each of the variables has its own coefficient estimate as well as p-value significance."), + caption = caption_helper("Summary of a linear model for predicting interest rate based on `debt_to_income`, `term`, and `credit_checks`. Each of the variables has its own coefficient estimate as well as a resulting p-value."), digits = 2, align = "lrrrr" ) %>% kable_styling( @@ -126,7 +126,7 @@ Sometimes a set of predictor variables can impact the model in unusual ways, oft In practice, there will almost always be some degree of correlation between the explanatory variables in a multiple regression model. For regression models, it is important to understand the entire context of the model, particularly for correlated variables. -Our discussion will focus on interpreting coefficients (and their signs) in relationship to other variables as well as the significance (i.e., the p-value) of each coefficient. +Our discussion will focus on interpreting coefficients (and their signs) in relationship to other variables as well as the discernibility (i.e., the p-value) of each coefficient. Consider an example where we would like to predict how much money is in a coin dish based only on the number of coins in the dish. We ask 26 students to tell us about their individual coin dishes, collecting data on the total dollar amount, the total number of coins, and the total number of low coins.[^25-inf-model-mlr-2] @@ -307,7 +307,7 @@ terms_chp_25 <- c(terms_chp_25, "multicollinearity") ``` Although diving into the details are beyond the scope of this text, we will provide one more reflection about multicollinearity. -If the predictor variables have some degree of correlation, it can be quite difficult to interpret the value of the coefficient or evaluate whether the variable is a statistically significant predictor of the outcome. +If the predictor variables have some degree of correlation, it can be quite difficult to interpret the value of the coefficient or evaluate whether the variable is a statistically discernible predictor of the outcome. However, even a model that suffers from high multicollinearity will likely lead to unbiased predictions of the response variable. So if the task at hand is only to do prediction, multicollinearity is likely to not cause you substantial problems. @@ -393,7 +393,7 @@ We refer to the model given with only `bill_lengh_mm` as the **smaller** model. It is seen in Table \@ref(tab:peng-lm-bill) with coefficient estimates of the parameters as well as standard errors and p-values. We refer to the model given with `bill_lengh_mm`, `bill_depth_mm`, `flipper_length_mm`, `sex`, and `species` as the **larger** model. It is seen in Table \@ref(tab:peng-lm-all) with coefficient estimates of the parameters as well as standard errors and p-values. -Given what we know about high correlations between body measurements, it is somewhat unsurprising that all of the variables have low p-values, suggesting that each variable is a statistically significant predictor of `body_mass_g`, given all other variables in the model. +Given what we know about high correlations between body measurements, it is somewhat unsurprising that all of the variables have low p-values, suggesting that each variable is a statistically discernible predictor of `body_mass_g`, given all other variables in the model. However, in this section, we will go beyond the use of p-values to consider independent predictions of `body_mass_g` as a way to compare the smaller and larger models. **The smaller model:** diff --git a/26-inf-model-logistic.qmd b/26-inf-model-logistic.qmd index 883af7fb..1d80b7a5 100644 --- a/26-inf-model-logistic.qmd +++ b/26-inf-model-logistic.qmd @@ -271,7 +271,7 @@ Using the example above and focusing on each of the variable p-values (here we w - $H_0: \beta_3 = 0$ given `to_multiple`, `cc`, and `urgent_subj` are included in the model - $H_0: \beta_4 = 0$ given `to_multiple`, `dollar`, and `dollar` are included in the model -The very low p-values from the software output tell us that three of the variables (that is, not `cc`) act as statistically significant predictors in the model at the significance level of 0.05, despite the inclusion of any of the other variables. +The very low p-values from the software output tell us that three of the variables (that is, not `cc`) act as statistically discernible predictors in the model at the significance level of 0.05, despite the inclusion of any of the other variables. Consider the p-value on $H_0: \beta_1 = 0$. The low p-value says that it would be extremely unlikely to observe data that yield a coefficient on `to_multiple` at least as far from 0 as -1.91 (i.e. $|b_1| > 1.91$) if the true relationship between `to_multiple` and `spam` was non-existent (i.e., if $\beta_1 = 0$) **and** the model also included `cc` and `dollar` and `urgent_subj`. Note also that the coefficient on `dollar` has a small associated p-value, but the magnitude of the coefficient is also seemingly small (0.07). diff --git a/exercises/_01-ex-data-hello.qmd b/exercises/_01-ex-data-hello.qmd index dff064a4..74f842d4 100644 --- a/exercises/_01-ex-data-hello.qmd +++ b/exercises/_01-ex-data-hello.qmd @@ -239,7 +239,7 @@ They randomly selected 6 of these daycare centers and instituted a monetary fine In the remaining 4 daycare centers no fine was introduced. The study period was divided into four: before the fine (weeks 1–4), the first 4 weeks with the fine (weeks 5-8), the last 8 weeks with fine (weeks 9–16), and the after fine period (weeks 17-20). Throughout the study, the number of kids who were picked up late was recorded each week for each daycare. -The study found that the number of late-coming parents increased significantly when the fine was introduced, and no reduction occurred after the fine was removed.^[The [`daycare_fines`](http://openintrostat.github.io/openintro/reference/daycare_fines.html) data used in this exercise can be found in the [**openintro**](http://openintrostat.github.io/openintro) R package.] [@Gneezy:2000] +The study found that the number of late-coming parents increased discernibly when the fine was introduced, and no reduction occurred after the fine was removed.^[The [`daycare_fines`](http://openintrostat.github.io/openintro/reference/daycare_fines.html) data used in this exercise can be found in the [**openintro**](http://openintrostat.github.io/openintro) R package.] [@Gneezy:2000] ```{r} library(openintro) diff --git a/exercises/_02-ex-data-design.qmd b/exercises/_02-ex-data-design.qmd index eff3e50d..efae5841 100644 --- a/exercises/_02-ex-data-design.qmd +++ b/exercises/_02-ex-data-design.qmd @@ -226,7 +226,7 @@ To assess the effectiveness of taking large doses of vitamin C in reducing the d A quarter of the patients were assigned a placebo, and the rest were evenly divided between 1g Vitamin C, 3g Vitamin C, or 3g Vitamin C plus additives to be taken at onset of a cold for the following two days. All tablets had identical appearance and packaging. The nurses who handed the prescribed pills to the patients knew which patient received which treatment, but the researchers assessing the patients when they were sick did not. -No significant differences were observed in any measure of cold duration or severity between the four groups, and the placebo group had the shortest duration of symptoms. [@Audera:2001] +No discernible differences were observed in any measure of cold duration or severity between the four groups, and the placebo group had the shortest duration of symptoms. [@Audera:2001] a. Was this an experiment or an observational study? Why? @@ -283,7 +283,7 @@ In one 2009 study, a team of researchers recruited 38 men and divided them rando They also recruited 38 women, and they randomly placed half of these participants into the treatment group and the other half into the control group. One group was given 25 grams of chia seeds twice a day, and the other was given a placebo. The subjects volunteered to be a part of the study. -After 12 weeks, the scientists found no significant difference between the groups in appetite or weight loss. [@Nieman:2009] +After 12 weeks, the scientists found no discernible difference between the groups in appetite or weight loss. [@Nieman:2009] a. What type of study is this? diff --git a/exercises/_13-ex-foundations-mathematical.qmd b/exercises/_13-ex-foundations-mathematical.qmd index 991cdb66..943c96f1 100644 --- a/exercises/_13-ex-foundations-mathematical.qmd +++ b/exercises/_13-ex-foundations-mathematical.qmd @@ -9,7 +9,7 @@ In 2013, the Pew Research Foundation reported that "45% of U.S. adults report th ii. If we repeated this study 1,000 times and constructed a 95% confidence interval for each study, then approximately 950 of those confidence intervals would contain the true fraction of U.S. adults who suffer from chronic illnesses. - iii. The poll provides statistically significant evidence (at the $\alpha = 0.05$ level) that the percentage of U.S. adults who suffer from chronic illnesses is below 50%. + iii. The poll provides statistically discernible evidence (at the $\alpha = 0.05$ level) that the percentage of U.S. adults who suffer from chronic illnesses is below 50%. iv. Since the standard error is 1.2%, only 1.2% of people in the study communicated uncertainty about their answer. @@ -20,7 +20,7 @@ A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least b. Identify each of the following statements as true or false. Provide an explanation to justify each of your answers. - i. The data provide statistically significant evidence that more than half of U.S. adult Twitter users get some news through Twitter. Use a significance level of $\alpha = 0.01$. + i. The data provide statistically discernible evidence that more than half of U.S. adult Twitter users get some news through Twitter. Use a significance level of $\alpha = 0.01$. ii. Since the standard error is 2.4%, we can conclude that 97.6% of all U.S. adult Twitter users were included in the study. diff --git a/exercises/_14-ex-foundations-errors.qmd b/exercises/_14-ex-foundations-errors.qmd index f26fe84a..1843389d 100644 --- a/exercises/_14-ex-foundations-errors.qmd +++ b/exercises/_14-ex-foundations-errors.qmd @@ -42,7 +42,7 @@ Determine if the following statements are true or false, and explain your reason c. Suppose the null hypothesis is $p = 0.5$ and we fail to reject $H_0$. Under this scenario, the true population proportion is 0.5. - d. With large sample sizes, even small differences between the null value and the observed point estimate, a difference often called the effect size, will be identified as statistically significant. + d. With large sample sizes, even small differences between the null value and the observed point estimate, a difference often called the effect size, will be identified as statistically discernible. \clearpage diff --git a/exercises/_16-ex-inference-one-prop.qmd b/exercises/_16-ex-inference-one-prop.qmd index c5065b68..3337ef8d 100644 --- a/exercises/_16-ex-inference-one-prop.qmd +++ b/exercises/_16-ex-inference-one-prop.qmd @@ -61,7 +61,7 @@ A Survey USA poll conducted in Seattle, WA in May 2021 reports that of the 650 r 1. **Assisted reproduction.** Assisted Reproductive Technology (ART) is a collection of techniques that help facilitate pregnancy (e.g., in vitro fertilization). The 2018 ART Fertility Clinic Success Rates Report published by the Centers for Disease Control and Prevention reports that ART has been successful in leading to a live birth in 48.8% of cases where the patient is under 35 years old. [@web:art2018] A new fertility clinic claims that their success rate is higher than average for this age group. A random sample of 30 of their patients yielded a success rate of 60%. A consumer watchdog group would like to determine if this provides strong evidence to support the company's claim. - a. Write the hypotheses to test if the success rate for ART at this clinic is significantly higher than the success rate reported by the CDC. + a. Write the hypotheses to test if the success rate for ART at this clinic is discernibly higher than the success rate reported by the CDC. b. Describe a setup for a simulation that would be appropriate in this situation and how the p-value can be calculated using the simulation results. @@ -534,7 +534,7 @@ A researcher wanting to test this claim randomly sampled 80 such people. He then filled 80 plain white cups with soda, half diet and half regular through random assignment, and 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 just random guessing? Your response should use a mathematical model. + 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 discernibly better than just random guessing? Your response should use a mathematical model. b. Interpret the p-value in this context. diff --git a/exercises/_16-sa-inference-one-prop.qmd b/exercises/_16-sa-inference-one-prop.qmd index 9f6f45e2..275ef0fc 100644 --- a/exercises/_16-sa-inference-one-prop.qmd +++ b/exercises/_16-sa-inference-one-prop.qmd @@ -34,7 +34,7 @@ 1. \(a)  We want to check for a majority (or minority), so we use the following hypotheses: $H_0: p = 0.5$ and $H_A: p \neq 0.5$. We have a sample proportion of $\hat{p} = 0.55$ and a sample size of $n = 617$ independents. Since this is a random sample, independence is satisfied. The success-failure condition is also satisfied: $617 \times 0.5$ and $617 \times (1 - 0.5)$ are both at least 10 (we use the null proportion $p_0 = 0.5$ for this check in a one-proportion hypothesis test). Therefore, we can model $\hat{p}$ using a normal distribution with a standard error of $SE = \sqrt{\frac{p(1 - p)}{n}} = 0.02$. (We use the null proportion $p_0 = 0.5$ to compute the standard error for a one-proportion hypothesis test.) Next, we compute the test statistic: $Z = \frac{0.55 - 0.5}{0.02} = 2.5.$ This yields a one-tail area of 0.0062, and a p-value of $2 \times 0.0062 = 0.0124.$ Because the p-value is smaller than 0.05, we reject the null hypothesis. We have strong evidence that the support is different from 0.5, and since the data provide a point estimate above 0.5, we have strong evidence to support this claim by the TV pundit. (b) No. Generally we expect a hypothesis test and a confidence interval to align, so we would expect the confidence interval to show a range of plausible values entirely above 0.5. However, if the confidence level is misaligned (e.g., a 99% confidence level and a $\alpha = 0.05$ significance level), then this is no longer generally true. \addtocounter{enumi}{1} -1. \(a)  $H_0: p = 0.5$. $H_A: p > 0.5$. Independence (random sample, $<10\%$ of population) is satisfied, as is the success-failure conditions (using $p_0 = 0.5$, we expect 40 successes and 40 failures). $Z = 2.91$ $\to$ p- value $= 0.0018$. Since the p-value $< 0.05$, we reject the null hypothesis. The data provide strong evidence that the rate of correctly identifying a soda for these people is significantly better than just by random guessing. (b) If in fact people cannot tell the difference between diet and regular soda and they randomly guess, the probability of getting a random sample of 80 people where 53 or more identify a soda correctly would be 0.0018. +1. \(a)  $H_0: p = 0.5$. $H_A: p > 0.5$. Independence (random sample, $<10\%$ of population) is satisfied, as is the success-failure conditions (using $p_0 = 0.5$, we expect 40 successes and 40 failures). $Z = 2.91$ $\to$ p- value $= 0.0018$. Since the p-value $< 0.05$, we reject the null hypothesis. The data provide strong evidence that the rate of correctly identifying a soda for these people is discernibly better than just by random guessing. (b) If in fact people cannot tell the difference between diet and regular soda and they randomly guess, the probability of getting a random sample of 80 people where 53 or more identify a soda correctly would be 0.0018. \addtocounter{enumi}{1} 1. \(a) The sample is from all computer chips manufactured at the factory during the week of production. We might be tempted to generalize the population to represent all weeks, but we should exercise caution here since the rate of defects may change over time. (b) The fraction of computer chips manufactured at the factory during the week of production that had defects. (c) Estimate the parameter using the data: $\hat{p} = \frac{27}{212} = 0.127$. (d) *Standard error* (or $SE$). (e) Compute the $SE$ using $\hat{p} = 0.127$ in place of $p$: $SE \approx \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.127(1 - 0.127)}{212}} = 0.023$. (f) The standard error is the standard deviation of $\hat{p}$. A value of 0.10 would be about one standard error away from the observed value, which would not represent a very uncommon deviation. (Usually beyond about 2 standard errors is a good rule of thumb.) The engineer should not be surprised. (g) Recomputed standard error using $p = 0.1$: $SE = \sqrt{\frac{0.1(1 - 0.1)}{212}} = 0.021$. This value isn't very different, which is typical when the standard error is computed using relatively similar proportions (and even sometimes when those proportions are quite different!). diff --git a/exercises/_17-ex-inference-two-props.qmd b/exercises/_17-ex-inference-two-props.qmd index d77a01b3..9829b427 100644 --- a/exercises/_17-ex-inference-two-props.qmd +++ b/exercises/_17-ex-inference-two-props.qmd @@ -387,7 +387,7 @@ Based on this information, determine if the following statements are true or fal c. 95% of random samples will produce 95% confidence intervals that include the true difference between the population proportions of 18-29 year olds and 30-49 year olds who have personally experienced harassment online. - d. We can conclude that there is a significant difference between the proportions of 18-29 year olds and 30-49 year olds who have personally experienced harassment online is too large to plausibly be due to chance, if in fact there is no difference between the two proportions. + d. We can conclude that there is a discernible difference between the proportions of 18-29 year olds and 30-49 year olds who have personally experienced harassment online is too large to plausibly be due to chance, if in fact there is no difference between the two proportions. e. The 90% confidence interval for $(p_{18-29} - p_{30-49})$ cannot be calculated with only the information given in this exercise. @@ -461,4 +461,4 @@ A Gallup poll surveyed Americans about their employment status and whether they b. State appropriate hypotheses to test for difference in proportions of diabetes between employed and unemployed Americans. - c. The sample difference is about 1%. If we completed the hypothesis test, we would find that the p-value is very small (about 0), meaning the difference is statistically significant. Use this result to explain the difference between statistically significant and practically significant findings. + c. The sample difference is about 1%. If we completed the hypothesis test, we would find that the p-value is very small (about 0), meaning the difference is statistically discernible Use this result to explain the difference between statistically discernible and practically different findings. diff --git a/exercises/_17-sa-inference-two-props.qmd b/exercises/_17-sa-inference-two-props.qmd index 1d7ca0ea..b246245a 100644 --- a/exercises/_17-sa-inference-two-props.qmd +++ b/exercises/_17-sa-inference-two-props.qmd @@ -16,7 +16,7 @@ 1. \(a) In effect, we are checking whether men are paid more than women (or vice-versa), and we would expect these outcomes with either chance under the null hypothesis: $H_0: p = 0.5$ and $H_A: p \neq 0.5.$ We'll use $p$ to represent the fraction of cases where men are paid more than women. (b) There isn't a good way to check independence here since the jobs are not a simple random sample. However, independence does not seem unreasonable, since the individuals in each job are different from each other. The success-failure condition is met since we check it using the null proportion: $p_0 n = (1 - p_0) n = 10.5$ is greater than 10. We can compute the sample proportion, $SE$, and test statistic: $\hat{p} = 19 / 21 = 0.905$ and $SE = \sqrt{\frac{0.5 \times (1 - 0.5)}{21}} = 0.109$ and $Z = \frac{0.905 - 0.5}{0.109} = 3.72.$ The test statistic $Z$ corresponds to an upper tail area of about 0.0001, so the p-value is 2 times this value: 0.0002. Because the p-value is smaller than 0.05, we reject the notion that all these gender pay disparities are due to chance. Because we observe that men are paid more in a higher proportion of cases and we have rejected $H_0$, we can conclude that men are being paid higher amounts in ways not explainable by chance alone. If you're curious for more info around this topic, including a discussion about adjusting for additional factors that affect pay, please see the following video by Healthcare Triage: youtu.be/aVhgKSULNQA. \addtocounter{enumi}{1} -1. \(a) $H_0: p = 0.5$. $H_A: p \neq 0.5$. Independence (random sample) is satisfied, as is the success-failure conditions (using $p_0 = 0.5$, we expect 40 successes and 40 failures). $Z = 2.91$ $\to$ the one tail area is 0.0018, so the p-value is 0.0036. Since the p-value $< 0.05$, we reject the null hypothesis. Since we rejected $H_0$ and the point estimate suggests people are better than random guessing, we can conclude the rate of correctly identifying a soda for these people is significantly better than just by random guessing. (b) If in fact people cannot tell the difference between diet and regular soda and they were randomly guessing, the probability of getting a random sample of 80 people where 53 or more identify a soda correctly (or 53 or more identify a soda incorrectly) would be 0.0036. +1. \(a) $H_0: p = 0.5$. $H_A: p \neq 0.5$. Independence (random sample) is satisfied, as is the success-failure conditions (using $p_0 = 0.5$, we expect 40 successes and 40 failures). $Z = 2.91$ $\to$ the one tail area is 0.0018, so the p-value is 0.0036. Since the p-value $< 0.05$, we reject the null hypothesis. Since we rejected $H_0$ and the point estimate suggests people are better than random guessing, we can conclude the rate of correctly identifying a soda for these people is discernibly better than just by random guessing. (b) If in fact people cannot tell the difference between diet and regular soda and they were randomly guessing, the probability of getting a random sample of 80 people where 53 or more identify a soda correctly (or 53 or more identify a soda incorrectly) would be 0.0036. \addtocounter{enumi}{1} 1. Before we can calculate a confidence interval, we must first check that the conditions are met. There aren't at least 10 successes and 10 failures in each of the four groups (treatment/control and yawn/not yawn), $(\hat{p}_C - \hat{p}_T)$ is not expected to be approximately normal and therefore cannot calculate a confidence interval for the difference between the proportions of participants who yawned in the treatment and control groups using large sample techniques and a critical Z score. diff --git a/exercises/_20-ex-inference-two-means.qmd b/exercises/_20-ex-inference-two-means.qmd index 495364fa..cacf0672 100644 --- a/exercises/_20-ex-inference-two-means.qmd +++ b/exercises/_20-ex-inference-two-means.qmd @@ -423,7 +423,7 @@ An experiment was conducted to measure and compare the effectiveness of various Newly hatched chicks were randomly allocated into six groups, and each group was given a different feed supplement. In this exercise we consider chicks that were fed casein and soybean. Assume that the conditions for conducting inference using mathematical models are met, and using the data provided below, test the hypothesis that the average weight of chickens that were fed casein is different than the average weight of chickens that were fed soybean. -If your hypothesis test yields a statistically significant result, discuss whether the higher average weight of chickens can be attributed to the casein diet. [@data:chickwts] +If your hypothesis test yields a statistically discernible result, discuss whether the higher average weight of chickens can be attributed to the casein diet. [@data:chickwts] ```{r} library(tidyverse) diff --git a/exercises/_20-sa-inference-two-means.qmd b/exercises/_20-sa-inference-two-means.qmd index b34a4ab2..ccc1a1ec 100644 --- a/exercises/_20-sa-inference-two-means.qmd +++ b/exercises/_20-sa-inference-two-means.qmd @@ -7,7 +7,7 @@ 1. \(a) We are 95% confident that the population average price per carat of 0.99 carat diamonds is \$2 to \$23 lower than the population average price per carat of 1 carat diamonds. (b) We are 95% confident that the population average price per carat of 0.99 carat diamonds is \$2.91 to \$21.10 lower than the population average price per carat of 1 carat diamonds. \addtocounter{enumi}{1} -1. The difference is not zero (statistically significant), but there is no evidence that the difference is large (practically significant), because the interval provides values as low as 1 lb. +1. The difference is not zero (statistically discernible), but there is no evidence that the difference is large (practically different), because the interval provides values as low as 1 lb. \addtocounter{enumi}{1} 1. $H_0: \mu_{0.99} = \mu_{1}$ and $H_A: \mu_{0.99} \ne \mu_{1}$. Independence: Both samples are random and represent less than 10% of their respective populations. Also, we have no reason to think that the 0.99 carats are not independent of the 1 carat diamonds since they are both sampled randomly. Normality: The distributions are not extremely skewed, hence we can assume that the distribution of the average differences will be nearly normal as well. $T_{22} = 2.23$, p-value = 0.0131. Since p-value less than 0.05, reject $H_0$. The data provide convincing evidence that the difference in population averages of price per carat of 0.99 carats and 1 carat diamonds are different. @@ -19,7 +19,7 @@ 1. \(a) $\mu_{\bar{x}_1} = 15$, $\sigma_{\bar{x}_1} = 20 / \sqrt{50} = 2.8284.$ (b) $\mu_{\bar{x}_2} = 20$, $\sigma_{\bar{x}_1} = 10 / \sqrt{30} = 1.8257.$ (c) $\mu_{\bar{x}_2 - \bar{x}_1} = 20 - 15 = 5$, $\sigma_{\bar{x}_2 - \bar{x}_1} = \sqrt{\left(20 / \sqrt{50}\right)^2 + \left(10 / \sqrt{30}\right)^2} = 3.3665.$ (d) Think of $\bar{x}_1$ and $\bar{x}_2$ as being random variables, and we are considering the standard deviation of the difference of these two random variables, so we square each standard deviation, add them together, and then take the square root of the sum: $SD_{\bar{x}_2 - \bar{x}_1} = \sqrt{SD_{\bar{x}_2}^2 + SD_{\bar{x}_1}^2}.$ \addtocounter{enumi}{1} -1. \(a) Chicken fed linseed weighed an average of 218.75 grams while those fed horsebean weighed an average of 160.20 grams. Both distributions are relatively symmetric with no apparent outliers. There is more variability in the weights of chicken fed linseed. (b) $H_0: \mu_{ls} = \mu_{hb}$. $H_A: \mu_{ls} \ne \mu_{hb}$. We leave the conditions to you to consider. $T=3.02$, $df = min(11, 9) = 9$ $\to$ p-value $= 0.014$. Since p-value $<$ 0.05, reject $H_0$. The data provide strong evidence that there is a significant difference between the average weights of chickens that were fed linseed and horsebean. (c) Type 1 Error, since we rejected $H_0$. (d) Yes, since p-value $>$ 0.01, we would not have rejected $H_0$. +1. \(a) Chicken fed linseed weighed an average of 218.75 grams while those fed horsebean weighed an average of 160.20 grams. Both distributions are relatively symmetric with no apparent outliers. There is more variability in the weights of chicken fed linseed. (b) $H_0: \mu_{ls} = \mu_{hb}$. $H_A: \mu_{ls} \ne \mu_{hb}$. We leave the conditions to you to consider. $T=3.02$, $df = min(11, 9) = 9$ $\to$ p-value $= 0.014$. Since p-value $<$ 0.05, reject $H_0$. The data provide strong evidence that there is a discernible difference between the average weights of chickens that were fed linseed and horsebean. (c) Type 1 Error, since we rejected $H_0$. (d) Yes, since p-value $>$ 0.01, we would not have rejected $H_0$. \addtocounter{enumi}{1} 1. $H_0: \mu_C = \mu_S$. $H_A: \mu_C \ne \mu_S$. $T = 3.27$, $df=11$ $\to$ p-value $= 0.007$. Since p-value $< 0.05$, reject $H_0$. The data provide strong evidence that the average weight of chickens that were fed casein is different than the average weight of chickens that were fed soybean (with weights from casein being higher). Since this is a randomized experiment, the observed difference can be attributed to the diet. diff --git a/exercises/_21-ex-inference-paired-means.qmd b/exercises/_21-ex-inference-paired-means.qmd index ad16d032..d6478912 100644 --- a/exercises/_21-ex-inference-paired-means.qmd +++ b/exercises/_21-ex-inference-paired-means.qmd @@ -160,7 +160,7 @@ The bootstrap distribution below was produced by bootstrapping from the sample o c. Interpret both confidence intervals using words like "population" and "score". - d. From the confidence intervals calculated above, does it appear that there is a significant difference in reading and writing scores, on average? + d. From the confidence intervals calculated above, does it appear that there is a discernible difference in reading and writing scores, on average? 1. **Possible paired randomized differences.** Data were collected on five people. diff --git a/exercises/_22-ex-inference-many-means.qmd b/exercises/_22-ex-inference-many-means.qmd index 73994327..4e834a78 100644 --- a/exercises/_22-ex-inference-many-means.qmd +++ b/exercises/_22-ex-inference-many-means.qmd @@ -346,7 +346,7 @@ Determine if the following statements are true or false, and explain your reason b. the standardized variability between groups is higher than the standardized variability within groups. - c. the pairwise analysis will identify at least one pair of means that are significantly different. + c. the pairwise analysis will identify at least one pair of means that are discernibly different. d. the appropriate $\alpha$ to be used in pairwise comparisons is 0.05 / 4 = 0.0125 since there are four groups. diff --git a/exercises/_25-ex-inf-model-mlr.qmd b/exercises/_25-ex-inf-model-mlr.qmd index 257565a1..bff0f136 100644 --- a/exercises/_25-ex-inf-model-mlr.qmd +++ b/exercises/_25-ex-inf-model-mlr.qmd @@ -92,9 +92,9 @@ In this exercise we work with data from a survey of 55 Duke University students a. There are three variables described in the figure, and each is paired with each other to create three different scatterplots. Rate the pairwise relationships from most correlated to least correlated. - b. When using only one variable to model `gpa`, is `out` a significant predictor variable? Is `sleepnight` a significant predictor variable? Explain. + b. When using only one variable to model `gpa`, is `out` a discernible predictor variable? Is `sleepnight` a discernible predictor variable? Explain. - c. When using both `out` and `sleepnight` to predict `gpa` in a multiple regression model, are either of the variables significant? Explain. + c. When using both `out` and `sleepnight` to predict `gpa` in a multiple regression model, are either of the variables discernible? Explain. \clearpage @@ -217,9 +217,9 @@ Researchers wanting to understand the relationship between these variables for b a. There are three variables described in the figure, and each is paired with each other to create three different scatterplots. Rate the pairwise relationships from most correlated to least correlated. - b. When using only one variable to model a tree's `volume`, is `diam`eter a significant predictor variable? Is `height` a significant predictor variable? Explain. + b. When using only one variable to model a tree's `volume`, is `diam`eter a discernible predictor variable? Is `height` a discernible predictor variable? Explain. - c. When using both `diam`eter and `height` to predict a tree's `volume`, are both predictor variables still significant? Explain. + c. When using both `diam`eter and `height` to predict a tree's `volume`, are both predictor variables still discernible? Explain. \clearpage @@ -309,7 +309,7 @@ A study was conducted at a university outpatient primary care clinic in Switzerl a. The intercept of the model is 30.594. What is the age, sex, and years of training of a physician whom this model would predict to have a DDPRQ-10 score of 30.594. - b. Is there evidence of a significant association between DDPRQ-10 score and any of the physician features? + b. Is there evidence of a discernible association between DDPRQ-10 score and any of the physician features? \clearpage @@ -447,9 +447,9 @@ In this exercise we study the relationship between the weight of the baby and tw a. There are three variables described in the figure, and each is paired with each other to create three different scatterplots. Rate the pairwise relationships from most correlated to least correlated. - b. When using only one variable to model the baby's `weight`, is `weeks` a significant predictor variable? Is `visits` a significant predictor variable? Explain. + b. When using only one variable to model the baby's `weight`, is `weeks` a discernible predictor variable? Is `visits` a discernible predictor variable? Explain. - c. When using both `visits` and `weeks` to predict the baby's `weight`, are both predictor variables still significant? Explain. + c. When using both `visits` and `weeks` to predict the baby's `weight`, are both predictor variables still discernible? Explain. \clearpage diff --git a/exercises/_26-ex-inf-model-logistic.qmd b/exercises/_26-ex-inf-model-logistic.qmd index 51a95260..b03198c8 100644 --- a/exercises/_26-ex-inf-model-logistic.qmd +++ b/exercises/_26-ex-inf-model-logistic.qmd @@ -20,7 +20,7 @@ The following model was fit to predict student drug use from parent drug use.^[T column_spec(2:5, width = "5em") ``` - a. State the hypotheses for evaluating whether parents' marijuana usage is a significant predictor of their kids' marijuana usage. + a. State the hypotheses for evaluating whether parents' marijuana usage is a discernible predictor of their kids' marijuana usage. b. Based on the regression output, state the sample statistic and the p-value of the test. @@ -46,7 +46,7 @@ The following model was fit to predict experiment outcome (`died` or `lived`, wh column_spec(2:5, width = "5em") ``` - a. State the hypotheses for evaluating whether experimental group is a significant predictor of treatment outcome. + a. State the hypotheses for evaluating whether experimental group is a discernible predictor of treatment outcome. b. Based on the regression output, state the sample statistic and the p-value of the test. diff --git a/exercises/_26-sa-inf-model-logistic.qmd b/exercises/_26-sa-inf-model-logistic.qmd index 1178ad84..42f71103 100644 --- a/exercises/_26-sa-inf-model-logistic.qmd +++ b/exercises/_26-sa-inf-model-logistic.qmd @@ -1,4 +1,4 @@ -1. \(a) $H_0: \beta_1 = 0$, the slope of the model predicting kids' marijuana use in college from their parents' marijuana use in college is 0. $H_A: \beta_1 \neq 0$, the slope of the model predicting kids' marijuana use in college from their parents' marijuana use in college is different than 0. (b) The test statistic is $Z = 4.09$ and the associated p-value is less than 0.0001. (c) With a small p-value we reject $H_0$. The data provide convincing evidence that the slope of the model predicting kids' marijuana use in college from their parents' marijuana use in college is different than 0, i.e. that parents' marijuana use in college is a significant predictor of kids' marijuana use in college. +1. \(a) $H_0: \beta_1 = 0$, the slope of the model predicting kids' marijuana use in college from their parents' marijuana use in college is 0. $H_A: \beta_1 \neq 0$, the slope of the model predicting kids' marijuana use in college from their parents' marijuana use in college is different than 0. (b) The test statistic is $Z = 4.09$ and the associated p-value is less than 0.0001. (c) With a small p-value we reject $H_0$. The data provide convincing evidence that the slope of the model predicting kids' marijuana use in college from their parents' marijuana use in college is different than 0, i.e. that parents' marijuana use in college is a discernible predictor of kids' marijuana use in college. \addtocounter{enumi}{1} 1. \(a) 26 observations are in Fold2. 8 correctly and 2 incorrectly predicted to be from Victoria. (b) 78 observations are used to build the model. (c) 2 coefficients for tail length; 3 coefficients for total length and sex. From 09dfda0b1b59f4b8920c702171f950f1911af94c Mon Sep 17 00:00:00 2001 From: hardin47 Date: Wed, 11 Oct 2023 15:27:52 -0700 Subject: [PATCH 2/8] more discernible --- 11-foundations-randomization.qmd | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/11-foundations-randomization.qmd b/11-foundations-randomization.qmd index f689e3ab..a5d2ea1f 100644 --- a/11-foundations-randomization.qmd +++ b/11-foundations-randomization.qmd @@ -676,7 +676,7 @@ This is also the case with hypothesis testing: *even if we fail to reject the nu Failing to find evidence in favor of the alternative hypothesis is not equivalent to finding evidence that the null hypothesis is true. We will see this idea in greater detail in Section \@ref(decerr). -### p-value and statistical significance +### p-value and statistically discernible In Section \@ref(caseStudySexDiscrimination) we encountered a study from the 1970's that explored whether there was strong evidence that female candidates were less likely to be promoted than male candidates. The research question -- are female candidates discriminated against in promotion decisions? From c302311c2b85ce6ec1cfd1f315698769eab51ff5 Mon Sep 17 00:00:00 2001 From: Jo Hardin Date: Wed, 11 Oct 2023 16:16:31 -0700 Subject: [PATCH 3/8] Update 11-foundations-randomization.qmd Co-authored-by: Mine Cetinkaya-Rundel --- 11-foundations-randomization.qmd | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/11-foundations-randomization.qmd b/11-foundations-randomization.qmd index a5d2ea1f..878b3d39 100644 --- a/11-foundations-randomization.qmd +++ b/11-foundations-randomization.qmd @@ -676,7 +676,7 @@ This is also the case with hypothesis testing: *even if we fail to reject the nu Failing to find evidence in favor of the alternative hypothesis is not equivalent to finding evidence that the null hypothesis is true. We will see this idea in greater detail in Section \@ref(decerr). -### p-value and statistically discernible +### p-value and statistical discernibility In Section \@ref(caseStudySexDiscrimination) we encountered a study from the 1970's that explored whether there was strong evidence that female candidates were less likely to be promoted than male candidates. The research question -- are female candidates discriminated against in promotion decisions? From 2050f2e7c173343ac6c1ce0edac2cd69e4c88ab6 Mon Sep 17 00:00:00 2001 From: Jo Hardin Date: Wed, 11 Oct 2023 16:16:54 -0700 Subject: [PATCH 4/8] Update 11-foundations-randomization.qmd Co-authored-by: Mine Cetinkaya-Rundel --- 11-foundations-randomization.qmd | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/11-foundations-randomization.qmd b/11-foundations-randomization.qmd index 878b3d39..189c394a 100644 --- a/11-foundations-randomization.qmd +++ b/11-foundations-randomization.qmd @@ -717,7 +717,7 @@ In each of these examples, the **point estimate** of the difference in proportio ::: When the p-value is small, i.e., less than a previously set threshold, we say the results are **statistically discernible**\index{statistically significant}\index{statistically discernible}. -This means the data provide such strong evidence against $H_0$ that we reject the null hypothesis in favor of the alternative hypothesis.^[Many texts use the phrase "statistically significant" instead of "statistically discernible". We have chosen to use "discernible" to indicate that a precise statistical event has happened, as opposed to an interesting effect which may or may not fit the statistical definition of discernible or significant.] +This means the data provide such strong evidence against $H_0$ that we reject the null hypothesis in favor of the alternative hypothesis.^[Many texts use the phrase "statistically significant" instead of "statistically discernible". We have chosen to use "discernible" to indicate that a precise statistical event has happened, as opposed to a notable effect which may or may not fit the statistical definition of discernible or significant.] The threshold is called the **significance level**\index{hypothesis testing!significance level}\index{significance level} and often represented by $\alpha$ (the Greek letter *alpha*). The value of $\alpha$ represents how rare an event needs to be in order for the null hypothesis to be rejected. Historically, many fields have set $\alpha = 0.05,$ meaning that the results need to occur less than 5% of the time, if the null hypothesis is to be rejected. From 8a0f94bafdcdd3d3da7a52a271e6605512311c44 Mon Sep 17 00:00:00 2001 From: Jo Hardin Date: Wed, 11 Oct 2023 16:17:07 -0700 Subject: [PATCH 5/8] Update 25-inf-model-mlr.qmd Co-authored-by: Mine Cetinkaya-Rundel --- 25-inf-model-mlr.qmd | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/25-inf-model-mlr.qmd b/25-inf-model-mlr.qmd index 4ffda6e1..85cdbb65 100644 --- a/25-inf-model-mlr.qmd +++ b/25-inf-model-mlr.qmd @@ -62,7 +62,7 @@ lm(interest_rate ~ debt_to_income + term + credit_checks, data = loans) %>% mutate(p.value = ifelse(p.value < 0.0001, "<0.0001", round(p.value, 4))) %>% kbl( linesep = "", booktabs = TRUE, - caption = caption_helper("Summary of a linear model for predicting interest rate based on `debt_to_income`, `term`, and `credit_checks`. Each of the variables has its own coefficient estimate as well as a resulting p-value."), + caption = caption_helper("Summary of a linear model for predicting interest rate based on `debt_to_income`, `term`, and `credit_checks`. Each of the variables has its own coefficient estimate as well as a p-value."), digits = 2, align = "lrrrr" ) %>% kable_styling( From 595d889d7d4096cc5185768dca8665b95175a0c3 Mon Sep 17 00:00:00 2001 From: Jo Hardin Date: Wed, 11 Oct 2023 16:17:25 -0700 Subject: [PATCH 6/8] Update exercises/_17-ex-inference-two-props.qmd Co-authored-by: Mine Cetinkaya-Rundel --- exercises/_17-ex-inference-two-props.qmd | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/exercises/_17-ex-inference-two-props.qmd b/exercises/_17-ex-inference-two-props.qmd index 9829b427..1f0fa42d 100644 --- a/exercises/_17-ex-inference-two-props.qmd +++ b/exercises/_17-ex-inference-two-props.qmd @@ -461,4 +461,4 @@ A Gallup poll surveyed Americans about their employment status and whether they b. State appropriate hypotheses to test for difference in proportions of diabetes between employed and unemployed Americans. - c. The sample difference is about 1%. If we completed the hypothesis test, we would find that the p-value is very small (about 0), meaning the difference is statistically discernible Use this result to explain the difference between statistically discernible and practically different findings. + c. The sample difference is about 1%. If we completed the hypothesis test, we would find that the p-value is very small (about 0), meaning the difference is statistically discernible. Use this result to explain the difference between statistically discernible and practically different findings. From 55009162b0e3efdcc71f3c001a8463aa7a827fa1 Mon Sep 17 00:00:00 2001 From: Jo Hardin Date: Wed, 11 Oct 2023 16:18:02 -0700 Subject: [PATCH 7/8] Update exercises/_25-ex-inf-model-mlr.qmd Co-authored-by: Mine Cetinkaya-Rundel --- exercises/_25-ex-inf-model-mlr.qmd | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/exercises/_25-ex-inf-model-mlr.qmd b/exercises/_25-ex-inf-model-mlr.qmd index bff0f136..4d188f18 100644 --- a/exercises/_25-ex-inf-model-mlr.qmd +++ b/exercises/_25-ex-inf-model-mlr.qmd @@ -92,7 +92,7 @@ In this exercise we work with data from a survey of 55 Duke University students a. There are three variables described in the figure, and each is paired with each other to create three different scatterplots. Rate the pairwise relationships from most correlated to least correlated. - b. When using only one variable to model `gpa`, is `out` a discernible predictor variable? Is `sleepnight` a discernible predictor variable? Explain. + b. When using only one variable to model `gpa`, is `out` a discernible predictor? Is `sleepnight` a discernible predictor? Explain. c. When using both `out` and `sleepnight` to predict `gpa` in a multiple regression model, are either of the variables discernible? Explain. From 555157107c817c80def846b4460de865ce8e5e80 Mon Sep 17 00:00:00 2001 From: Jo Hardin Date: Wed, 11 Oct 2023 16:18:14 -0700 Subject: [PATCH 8/8] Update exercises/_25-ex-inf-model-mlr.qmd Co-authored-by: Mine Cetinkaya-Rundel --- exercises/_25-ex-inf-model-mlr.qmd | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/exercises/_25-ex-inf-model-mlr.qmd b/exercises/_25-ex-inf-model-mlr.qmd index 4d188f18..3ff06cba 100644 --- a/exercises/_25-ex-inf-model-mlr.qmd +++ b/exercises/_25-ex-inf-model-mlr.qmd @@ -94,7 +94,7 @@ In this exercise we work with data from a survey of 55 Duke University students b. When using only one variable to model `gpa`, is `out` a discernible predictor? Is `sleepnight` a discernible predictor? Explain. - c. When using both `out` and `sleepnight` to predict `gpa` in a multiple regression model, are either of the variables discernible? Explain. + c. When using both `out` and `sleepnight` to predict `gpa` in a multiple regression model, are either of the predictors discernible? Explain. \clearpage