what is the t-value in t-test?
353290
You are close.\n\nIn the T-test, you are evaluating the difference of the means between two samples normalized to the pooled standard errors of the two samples.\n\nIf mc is your control mean and mt is your test mean, then the T-value is the number of pooled standard errors difference mt is away from mc.\n\nYou can then look up this value in a table of T-values to find the probability of difference as a p-value for assessing the significance.  The "rote" significance level is 95% or a p-value of 0.05 (100% - 95%) which is a absolute T-value of 1.96 or more.
849629
Alpha, as related to level of significance, is purely up to the statistician.  It simply denotes how sure you are of your findings.  The p-value is more descriptive.  The p-value is basically the probability that the statistician will reject the null hypothesis when it in fact should not be rejected.  If the p-value is less than the chosen alpha level, the null hypothesis is rejected.
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Tf=(9/5)*Tc+32\n\nTf is temperature in Fahrenheit,\nTc is temperature in Celsius.
586725
The answer would be C.\n\nIn a two tailed test, you're looking to see if the mean is either significantly greater or less than k which is why it is called two-tailed.  The null hypothesis would be that mu = k and the alternate, not equal.  Thus, c
898945
Considering the data you had shown,\nyou must use a one-tailed test.\nyour null hypothesis would be:\nHo: There is no significant difference between the mean amount of time devoted to patient care per week by doctors using conventional methods and alternative methods.\nHo: mean(conventional) = mean(alternative)\n\nyour alternative hypothesis must be:\nHa: The mean amount of time devoted to patient care per week by doctors using conventional methods exceeds that of doctors using alternative methods\nHa: mean(conventional) > mean(alternative)\n\nNow that you have the hypotheses, next is to set the level of significance but it's all up to you depending on how confident are that you will not commit any kind of error.\n\nAnd your asking about the appropriate test?\nFor me, the choice depends on the given.\nBut you can only choose from the second and fourth test since you have to use a directional test (one-tailed test).\n\nUse Z-test if the standard deviation or variance of the population is known.\nUse t-test if the standard deviation or variance of the population is unknown.\nThat's the general rule of choosing the appropriate test.\nBut there's another consideration in the selection of test.\nyou must also take into consideration the number of your sample, whether it exceeds 30 or not.\n\nThat's all. I hope I've been a great help to you. Please inform me if there is something wrong with my lecture here.\nthank you.
484697
Find the inter quartile range, which is IQR = Q3 - Q1, where Q3 is the third quartile and Q1 is the first quartile.  Then find these two numbers:\n\na) Q1 - 1.5*IQR\nb) Q3 + 1.5*IQR\n\nAny observation that is below a) or above b) can be considered an outlier.\n\nedit: Chadwick, quartiles are considered robust, meaning that they are not highly effected by outliers.  This is because it takes location into account, not the values.  Let's look at your data set (sorted).\n\n2 3 6 9 13 18 21 106\n\nposition of Q1 = (8+1)/4 = 2.25\nQ1 = 0.75(3)+0.25(6) = 3.75\nposition of Q2 = (8+1)/2 = 4.5\nQ2 = (9+13)/2 = 11\nposition of Q3 = 3(8+1)/4 = 6.75\nQ3 = 0.25(18)+0.75(21) = 20.25\n\nNotice that none of these actually use the value 106.  Let's continue.\n\nSo IQR = Q3-Q1 = 20.25-3.75 = 16.5\n\nQ1-1.5*IQR = 3.75-1.5*16.5 = -21\nQ3+1.5*IQR = 20.25+1.5*16.5 = 45\n\nNo numbers are below -21, but 106 is above 45, so it can be considered an outlier.
458114
The confidence interval is the gap you set to determine a level of certainty in your answer. For example, in public health, we usually set alpha=0.05, meaning we have a confidence interval arbitrarily set to 95%.\n\nIf the p-value we get for whatever statistic we just crunched is less than 0.05, then we say that we are at least 95% certain of the validity of our answer. The lower out p-value is, the more certain we are. If we get p=0.00002, then that's 99.99998% confidence, which is higher than the 95% threshold we set.\n\nIn short, it's a measure of validity for our results.
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I'm supposing that by R-value you mean resistance value.\n\nIf so, there are two separate formulas. \nFirst for R-values in series, meaning, materials which are layered: \n\nRtotal =R1 + R2 +...+ Rn\n\nFor R-values in parallel, meaning two or more materials side by side:\n\n(1/Rtotal) = 1/R1 + 1/R2 +...+ 1/Rn
282133
sqrt(t(sqrt(t(sqrt(t))))) = sqrt(t(sqrt(t(sqrt(t)))))\nsqrt(t(sqrt(t(sqrt(t))))) = sqrt(t(sqrt(t*t^(1/2))))\nsqrt(t(sqrt(t(sqrt(t))))) = sqrt(t(sqrt(t^(3/2)))) \nsqrt(t(sqrt(t(sqrt(t))))) = sqrt(t(t^(3/4)))\nsqrt(t(sqrt(t(sqrt(t))))) = sqrt(t^(7/4))\nsqrt(t(sqrt(t(sqrt(t))))) = t^(7/8)\n\nThe derivative of sqrt(t(sqrt(t(sqrt(t))))) = the derivative of t^(7/8), so:\n\n(t^(7/8))' = (7/8)*t^((7/8)-(8/8))\n(t^(7/8))' = (7/8)*t^(-1/8)
803733
You need to refine you question.\n\nThe chi square compares observed to expected values.  Generate a table:\n\nobs, expected\n-------------------\n\nfill this in, and we'll go from there....
731912
An Anova is closely related to the t test. But the major difference is that, where the t test measures the difference between the means of two groups, an ANOVA tests the difference between the means of two or more groups.\n\nYou can use an ANOVA for the three treatment conditions in order to compare the three conditions, with a t-test you can not.
967657
When the population standard deviation is unknown, thats when you use the t-distribution, in calculating the confidence interval. \n\nThe t-distribution (also known as Student's t-distribution) measures the significance of a difference of means of small samples when two distributions are though to have the same variance, but possibly different means.\n\nBut as the sample size increases, the t-distribution gets close to the normal distribution.\n\nHope this helps!
104149
T-tests, orginally designed by Gossett for Guiness Brewery is a test of significance.  It is used when there are two groups and one wants to see if chance could account for differenes between the two groups.
861521
you would need to look at table for t values.\nyou are looking for a 95% confidence level, so you would divide .5 (which is 1-.95) by 2 and get .025 that is the column you find the value for t.\nyour n number, the row for the t number, is 15-1 or 14 and 12-1 or 11.\nLook at the table IV in colum t=0,025 and row 14 write down the number. that number is the t value for the first equation.\nLook at table IV in column t=0.025 and row 11, that is your t value for the second equation.\nNow that you have the t values you can plug them into the formula for confidence interval.\n\nGood luck
674991
For one-sample t-test, \nt = (sample mean - population mean) / estimated standard error. \n\nIn math notation, t = (X bar - mu not) / (s/sqrt(n))\n\nAssumptions: (normal population or sample size, n, is at least 30) and population standard deviation is unknown.\n\nThere are other variations of t-tests like two-sample(pooled and unpooled) and paired ...
753296
|t - 78.3| ≤ 38.3\n\nSo either t -78.3 ≤ 38.3 whence t ≤ 116.6\nOR - (t - 78.3) ≤ 38.3\nwhence - t ≤ - 40 ie 40 ≤ t\n\nSo 40 ≤ t ≤ 116.6
171180
Sorry, georgie,\n\nI cannot understand your question. Would you please first write it out fully in English, then set up the math formular - if any.\n\nI read, "What does a Z-test with two samples mean?" Please enlighten me; is this what you meant to write?