Say I am using a t statistic, the (two-tailed-t test) p-value as i understand is given as pr(t$\geq$|t| |null is true), why cant it be that pr(t=t|null) or at least some small relative frequency in an infinitesimal distance around t, such as f(t)dt ?
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After taking a statistics course and then trying to help fellow students, I noticed one subject that inspires much head-desk banging is interpreting the results of statistical hypothesis tests.  It seems that students easily learn how to perform the calculations required by a given test but get hung up on interpreting the results.  Many computerized tools report test results in terms of "p values" or "t values".  How would you explain the following points to college students taking their first course in statistics:   What does a "p-value" mean in relation to the hypothesis being tested?  Are there cases when one should be looking for a high p-value or a low p-value? What is the relationship between a p-value and a t-value?
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Suppose we have i.i.d. $n$ observations $(X_1,X_2,...X_n)$ from a   population with density $$f_\theta(x)=\begin{cases}\theta x^{\theta-1}&amp;\text{ if }0\leq x\leq 1\\0&amp;\text{otherwise.}\end{cases}$$ Note that $\theta&gt;0$. If $T_n$ denotes the   maximum likelihood estimator of $\theta$ given that $n$ is the sample   size, then show that $$\text{var}(T_n)\stackrel{n\to\infty}{\longrightarrow} 0$$   This is the problem I am trying to solve: I computed $$T_n=\dfrac{-n}{\sum_{i=1}^{n}\ln(X_i)}$$ But I am stuck in finding $\text{var}(T_n)$ and showing that $\text{var}(T_n)\to0$ as $n\to\infty$.
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While studying for my stats course, I was trying to understand the difference between one-tailed and two-tailed hypothesis tests. Specifically, why does the one-tailed test reject the null while the two-tailed one does not?   An example:
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What is the meaning of t value and Pr(&gt;|t|) when using summary() function on linear regression model in R?  Coefficients:                               Estimate Std. Error t value Pr(&gt;|t|)     (Intercept)                    10.1595     1.3603   7.469 1.11e-13 *** log(var)                        0.3422     0.1597   2.143   0.0322 *
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I was reading on the computation of the unbiased estimation of standard deviation and the source I read stated      (...) except in some important situations, the task has little relevance to   applications of statistics since its need is avoided by standard   procedures, such as the use of significance tests and confidence   intervals, or by using Bayesian analysis.   I was wondering if anyone could elucidate the reasoning behind this statement, for example doesn't the confidence interval use the standard deviation as part of the calculation? Therefore, wouldn't the confidence intervals be affected by a biased standard deviation?  EDIT:  Thanks for the answers so far, but I'm not quite sure I follow some of the reasoning for them so I'll add a very simple example. The point is that if the source is correct, then then something is wrong from my conclusion to the example and I would like someone to point how how the p-value doesn't depend on the standard deviation.  Suppose a researcher wished to test whether the mean score of fifth graders on a test in his or her city differed from the national mean of 76 with a significance level of 0.05. The researcher randomly sampled the scores of 20 students. The sample mean was 80.85 with a sample standard deviation of 8.87. This means: t = (80.85-76)/(8.87/sqrt(20)) = 2.44. A t-table is then used to calculate that the two-tailed probability value of a t of 2.44 with 19 df is 0.025. This is below our significance level of 0.05 so we reject the null hypothesis.  So in this example, wouldn't the p-value (and maybe your conclusion) change depending on how you estimated your sample standard deviation?
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Why do statisticians discourage us from referring to results as "highly significant" when the $p$-value is well below the conventional $\alpha$-level of $0.05$?  Is it really wrong to trust a result that has 99.9% chance of not being a Type I error ($p=0.001$) more than a result that only gives you that chance at 99% ($p=0.01$)?
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F-tests can be two-tailed (to test that $s_1^2 \ne s_2^2$) or one-tailed (to test that $s_1^2 &gt; s_2^2$).  How can I modify Levene/Brown-Forsythe to be "one-tailed", that is, to test $s_1^2 &gt; s_2^2$ instead of $s_1^2 \ne s_2^2$?  Here is a demo:    The image shows normally distributed training data (n=1000) and a model. An F-test is used to compare the variance of one point's residuals (n=2) to the variance of all of the residuals (n=2000), so the point is an outlier if its residual variance is "too large." The points are colored by p-value, where light points fit the model and dark points are outliers, and you can see that the two-tailed Brown-Forsythe rejects points that are too close to the model as well as too far.  Note: A different non-parametric, one-sided variance test would be fine as well.    Glen_b gave the information I needed, but I thought I would leave some implementation details (using scipy).  #basic F-test F = var(a) / var(b) Fp = stats.f.sf(F, df1, df2)  #Brown Forsythe BF, BFp = stats.levene(a, b, center='median')  #two tailed t-test on transformed data za = abs(a-median(a)) zb = abs(b-median(b)) t, tp_two_tailed = stats.ttest_ind(za, zb)  #the two tailed t test recapitulates the BF test assert(t**2 == BF) assert(p_BF == p_two_tailed)  #one tailed t test p value tp = stats.t.sf(t, df)     Above shows scatter plots of the p values from the one-tailed $F$-test and two-tailed BF-test (left), and the one-tailed $t$ tests (right). Red points are "too close" ($s_1^2 &lt; s_2^2$).
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This question is related to a couple of other questions ( and ) relating to some material that I teach in sampling theory.  Consider the critical points of the  defined as follows.  The value $t_{n-1, \alpha/2}$ is a function of $n$ defined by the implicit equation: $$\frac{\alpha}{2} = \frac{1}{\sqrt{(n-1) \pi}} \cdot \frac{\Gamma(\tfrac{n}{2})}{\Gamma(\tfrac{n-1}{2})} \int \limits_{t_{n-1, \alpha/2}}^\infty \Big( 1+ \frac{r^2}{n-1} \Big)^{-n/2} dr.$$ Normally we treat the input $n&gt;1$ as an integer, but it is possible to extend the treatment to consider it a real value (since the T-distribution is well-defined for non-integer degrees-of-freedom).  I'm trying to come up with a (relatively simple) proof that this critical point value is decreasing in $n$, but I seem to be making quite a mess.  Differentiation using Leibniz rule gets part way, but it quickly becomes a bit of a mess.  Consequently, I would like to solicit proofs of this result --- this simpler the better. Question: How do you prove that this critical point is strictly decreasing in $n$?
109049
I'm being asked to find the p-values for descriptive statistics. However, it's my understanding that p-values are for test statistics.  If I'm not mistaken, a p-value is the probability of observing a value as extreme as the test statistic if the null hypothesis were true.
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I use "boot" package to compute an approximated 2-sided bootstrapped p-value but the result is too far away from p-value of using t.test. I can't figure out what I did wrong in my R code. Can someone please give me a hint for this  time = c(14,18,11,13,18,17,21,9,16,17,14,15,          12,12,14,13,6,18,14,16,10,7,15,10) group=c(rep(1:2, each=12)) sleep = data.frame(time, group)  require(boot) diff = function(d1,i){     d = d1[i,]     Mean= tapply(X=d$time, INDEX=d$group, mean)     Diff = Mean[1]-Mean[2]     Diff }  set.seed(1234) b3 = boot(data = sleep, statistic = diff, R = 5000, strata=sleep$group)  pvalue = mean(abs(b3$t) &gt; abs(b3$t0)) pvalue    The 2-sided bootstrapped p-value (pvalue) = 0.4804 but the 2-sided p-value of t.test is 0.04342. Both p-values are around 11 times difference. How can this happen?
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I am unable to understand what exactly is critical value.My question here is why do we fail to reject a null hypothesis when the test statistic is within the range of critical value.  What i am trying to ask is what is the property of the critical value that makes the hypothesized value to fall within the range implied by the critical value.  Thanks in advance
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On the , there is this line:      $p(\mathrm{height}|\mathrm{male}) = 1.5789$ (A probability distribution over 1 is OK. It is the area under the bell curve that is equal to 1.)    How can a value $&gt;1$ be OK? I thought all probability values were expressed in the range $0 \leq p \leq 1$. Furthermore, given that it is possible to have such a value, how is that value obtained in the example shown on the page?
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It is true that $P(X&gt;t+s|X&gt;t)=P(X&gt;s)$ for certain values $t$ and $s$.  However, how can I show that this still holds if:   $T$ is a continuous random variable. That is $P(X&gt;T+s|X&gt;T)=P(X&gt;s)$ Both $T$ and $S$ are continuous RVs: $P(X&gt;T+S|X&gt;T)=P(X&gt;S)$   All the random variables are independent.
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I have three replicates of an experiment which gives me ratios of the abundance of different peptides.  If I take the data for one peptide:  pep1 &lt;- c(4.3, 5.7, 6.3)   I can use a one-sample t-test to test the hypothesis that the ratio is different from 1:  t.test(pep1,         alternative=c("greater"),        mu=1,        paired=FALSE,        var.equal=FALSE,        conf.level=0.95)   output:  data:  pep1 t = 7.4818, df = 2, p-value = 0.0087 alternative hypothesis: true mean is greater than 1 95 percent confidence interval:  3.703107      Inf sample estimates: mean of x   5.433333    However if I try and apply this to the rows of a dataframe called peptides, where each row is a separate peptide, with three ratios:  t_test &lt;- function(x) p_val =  t.test(x,                                        alternative=c("greater"),                                       mu=1,                                       paired=FALSE,                                       var.equal=FALSE,                                       conf.level=0.95)$p.value  apply(peptides, 1, t_test)   I get the data are essentially constant error.  I don't understand! I think apply is doing what I want because if I substitute 't-test' for 'mean' then it does the correct calculations.   My aim is to get the p-values for each peptide and then adjust these for multiple comparisons.  I'm relatively new to R, and any advice would be very much appreciated!
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What is a good rule of thumb for how to choose the question for the null hypothesis. For instance, if I want to check if the hypothesis B is true, should I use B as the null, B as the alternative hypothesis, or NOT B as the null? I hope the question is clear. I know that it has something to do with the error I want to minimize (Type I?), but I keep forgetting how it goes, because I don't have a clear intuition built for it. Thanks.
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I've read somewhere that the t-test is designed for small samples, why is that the case? What if I use a big sample out of a population to conduct the t-test? Would that be an issue or have potential implications?
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I'm new to stats and I need some help. Can anybody tell me, in the most "beginner friendly" way, how to perform hypothesis testing with a geometric distribution for 2 samples? please take into account that, until a month ago, Statistics for me was mean, mode and std dev...