10

Multiple comparisons corrections are intended to control the familywise error rate--or something like it--so they should be applied across a "family" of related hypothesis tests. In your first example, the overarching goal probably to determine whether Groups A and B differ. If you didn't control for multiple comparisons, you could trivially find an effect ...


6

You are allowed to do this analysis but you will have to accept that your estimates of the coefficients will be rather imprecise. You also need to revisit the introductory material from which you learned regression as you do not check the predictor variables for their properties but instead examine the residuals from your model. With your sample size this is ...


6

Correlation and independence, in their technical senses in statistics, are conceptually different and only indirectly related. Correlation is a combination of second central moments of a bivariate random variable while the independence concerns the joint probability distribution. Although the relationship, as expressed in the question, is so familiar and ...


5

The qq-plot does not show normal residuals. You can see that the points have a shallow, concave-up curve and the ends move outside of the 95% confidence band (especially at the high end). Your data are more (positively) skewed than we would expect to find in a sample from a true normal distribution. In addition, your data are not independent since you ...


5

It looks like you have some kind of seasonality in your data, you might want to try seasonal adjustment programs. X-13ARIMA-SEATS can be a really powerful tool in this respect.


5

To answer (a), I would suggest a $z$-test for proportions. Your null hypothesis should be $H_0: p_Y=0.5$ against the alternative $H_A:p_Y\neq 0.5$, where $p_Y$ is the proportion of people who answered "Yes." To answer (b), I would suggest using a two-way table to display your data, then executing a chi-squared test with null $H_0:$ no association exists ...


5

The default test would be a paired-samples t-test (aka related-samples t-test or dependent-samples t-test). This test assumes that you have a normal distribution of difference scores, though, so if that assumption is not met you might consider a Wilcoxon signed rank test. Just one observation. The most basic pretest/posttest design is confounded with time....


4

I am assuming that one of the statements is your outcome of interest; something like, "I am satisfied with my customer experience". You want to tie back both the responses to other questions, as well as demographic/transactional/profile information about your customers to this outcome statement. If so, your question sounds like what's often called "key ...


4

What you want is called Gini coefficient. While it was created to measure economic inequality, it was used as the evaluation metric of a machine learning competition in a Kaggle featuring a regression task. It shares properties with the AUC. For example, it's only concerned with the ordering of the outcomes, not their actual value. High Gini means high ...


4

The Cliff's delta statistic (or Success Rate difference): $\delta = Pr(y_{i} >x_{j}) - Pr(y_{i}<x_{j})$ is the difference between the probability of $y_i$ greater than $x_j$ - with $y_i$ and $x_j$ randomly chosen in their respective groups - and the probability of the inverse. It is in all respects a population parameter, but it is not a location ...


4

I'm not sure why your advisor thought it was a good idea to remove a response option from the data after collection, as @RNM noted. As well, mean substitution is seldom a really good option for missing data. If you have to work with the altered data, one option you can do with SPSS if you have access to the MVA procedure is to use the EM imputation option ...


4

There are four common ways of using propensity scores (PS) to reduce confounding and arrive at an unbiased estimate of a causal effect. These are PS matching, PS weighting, PS subclassification, and regression on the PS. There have been systematic studies on the relative performance of these methods, but new variations of them always come out and it's not ...


4

The options under (d) are wrong, as a change score is associated with the baseline value. See this page, for example. Otherwise, it depends on what you mean by "taking into account the baseline measurement." You already note that option (a) doesn't do that at all. Option (b) looks only at the change from baseline as a function of Group. Based on ...


3

Honestly, the best case scenario is to present both mean and median. Doing so provides an accurate picture of your data, even if you don't include a picture. If you can only provide one, remember that these are both measures of center. Which number better measures the center of your data? Generally, I would compute both and if there are many cases where the ...


3

The Stata article you link discusses modeling an outcome as a proportion. More specifically, it discusses binomial regression, which is the same as frequency weighted logistic regression. If you don't know the denominator, nonlinear least squares or logistic regression with nonintegral outcomes and a dispersion parameter can handle the issue. However, you ...


3

Estimating integer parameters in STAN is impossible. HMC depends on gradient computations of the posterior density. Integer parameters aren't suitable for those computations. Rounding or truncating a real number to an integer won't help because the posterior surface will be flat in a region along that axis, so there will be no information for the sampler ...


3

The standard errors of the scale parameters are small relative to the value of the parameter (they're measured in the same units) - the coefficients of variation are small. If you used different units, both those values would change; it's only the relative size that says much. I'm not sure I see any problem here. But let's imagine those standard errors ...


3

There are two approaches to handling data of this nature: fixed effects and mixed effects. The T-test is basically a linear regression model with the size of the fish as an outcome variable and the lake from which the fish was drawn as the indicator of interest (a 0/1 variable). We have to assume the fish you sampled are representative of the population of ...


3

Actually, censoring does occur, but not in the typical way. Ben Ogorek is right in that these data require using analyses that permit time-dependent covariates. There are many ways to do that, but they all require breaking up the follow-up period into pieces, where the pieces are defined by the boundaries where the covariate values change. Consider the ...


3

There's two sorts of things to do here, depending on what is meant by "wrong". If there are values other than 1, 2, 3, 4 and 5 (eg "6", "0", or "fantastic") you know something is wrong. This should be very obvious from the SAS frequency distribution, so obvious it might go without saying, but should be the first thing you check. Similarly, the counts ...


3

The density of the difference $Z=X-Y$ is $$g(z)=\int f(x)f(z+x)\text{d}x$$Then \begin{align*} g(\alpha z+(1-\alpha) w) &= \int f(x)f(\alpha z+(1-\alpha) w+x)\text{d}x\\&= \int f(x)f(\alpha (z+x)+(1-\alpha) (w+x))\text{d}x\\&= \int f(x)\exp\{\ln f(\alpha z+(1-\alpha) w+x)\}\text{d}x\\&\le\int f(x)\exp\{\alpha\ln f(z+x)+(1-\alpha)\ln f(w+x)\}\...


3

No. This just makes the estimation easier because you have complete data. Complete data gives more information than a data set that has some right-censored observations.


3

Yes, there are densities with negatively infinite entropy. To construct one, find a family of densities with arbitrarily negative entropy. (Clearly this requires the density functions to have arbitrarily large values and therefore there are points where they become "singular.") By shifting and scaling them into disjoint intervals within $[0,1],$ if we are ...


3

You have reported that you have transaction data . Since you don't have observations every day, you will need to aggregate/bucket your data to a time series i.e. some level of accumulation ...where non-zero observations are present. An example might be weekly or monthly buckets or even quarterly buckets. Buying patterns are seldomly weekly-based with ...


3

Pay attention :) A. "The economic theory of human capital maintains that education has a causal effect on the subsequent labor market earnings of individuals. [...] The complication [...] is that economists also accept that mental ability enhances productivity as well. Thus, because those with relatively high ability are assumed to be more likely to ...


3

One approach would be to use a chi-square test. To do this, you would estimate the overall distribution of sales in various genres from all of the data. For example, say your data revealed this distribution: Non-fiction science/industry: 20% Cookbooks: 35% Speculative fiction (sci-fi, fantasy, etc.): 25% Crime fiction: 10% Other fiction: 10% You then get ...


3

Simply put: you can't predict for epistemological reasons. Why is that? To predict, you must have time-varying $X$, that is, $X(t)$. But, if you have $X(t)$, then you must be making measurements on a customer, so they are not dead! Therefore, the fact that you have or don't have $X(t)$ tells you if the customer is alive or not, making any "survival"...


3

The image of the model is a bit inscrutable so I reproduced it. Here is what you're trying to estimate: $$ P_{it} = \gamma_i + \lambda_t + \eta T_i + P^{C}_{t} + P^{f_1}_{t} + P^{f_2}_{t} + \delta_1 (T_i \times P^{C}_{t}) + \delta_2 (T_i \times P^{f_1}_{t}) + \delta_3 (T_i \times P^{f_2}_{t}) + \epsilon_{it}, $$ where stock performance (e.g., return ...


3

This is a great question. Univariate means you have a single outcome you are studying -- chances of getting a job. To describe the multiple factors involved in analyzing this outcome, you would describe the analysis as multivariable. This refers to the independent variables age, gender, etc. The study could be described as both univariate (one outcome) ...


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