Displaying 20 results from an estimated 538 matches for "3q".
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2017 Dec 27
3
Answered time on channel
...st" channel. Below is an untested example, but is
> inspired by dialplan code I use in production. Maybe it will help.
>
> [outbound] ; this is called on the incoming (caller) channel
> exten => _X.,1,Noop
> same => n,Set(MASTER_CHANNEL(start_timestamp)=${STRFTIME(,,%s.%3q)})
> same => n,Set(CHANNEL(hangup_handler_push)=hangup_handler,s,1)
> same => n,Set(MASTER_CHANNEL(callid_ingress)=${SIPCALLID})
> same => n, *** unrelated dialplan, AGIs, etc. ***
> same => n,Dial(SIP/${EXTEN}@1.1.1.1,,U(answer_handler)b(pre_dial_
> handler^s^1)g
>...
2011 Sep 19
1
regression summary results pvalues and coefficients into a excel
...nction(x){x["Pr(>|t|)"][1:2,]}))
6) write.table(t(regpvalues), file = "regression-resultsheadshape.txt",
quote = F, sep ='\t')
> regressionresults
$CUST_54_PI410671829
Call:
lm(formula = morphtrait ~ x, data = m)
Residuals:
Min 1Q Median 3Q Max
-0.23217 -0.08980 -0.04592 -0.00947 1.07688
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.5985797 0.0364510 16.421 <2e-16 ***
x 0.0005372 0.0011161 0.481 0.632
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Re...
2009 Mar 17
2
formula question
...> test.df
y x
1 -0.9261650 1
2 1.5702700 2
3 0.1673920 3
4 0.7893085 4
5 0.3576875 5
6 -1.4620915 6
7 -0.5506215 7
8 -0.3480292 8
9 -1.2344036 9
10 0.8502660 10
> summary(lm(exp(y)~x))
Call:
lm(formula = exp(y) ~ x)
Residuals:
Min 1Q Median 3Q Max
-1.6360 -0.6435 -0.4722 0.4215 2.9127
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 2.1689 0.9782 2.217 0.0574 .
x -0.1368 0.1577 -0.868 0.4108
---
Signif. codes: 0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1
Residual standar...
2017 Dec 26
4
Answered time on channel
Hi,
I have a dial plan where I need to notify an external system when a call
was answered and when the call hung up. In both requests the start time
needs to be the same. My Dialplan looks something like this:
[outbound]
Exten => _X.,1,Dial(SIP/${EXTEN}@1.1.1.1,,U(call-answer-from-carrier))
Exten => h,1,NoOp(ANSWERED_TIME: ${ANSWEREDTIME} >>> DIAL_TIME:
${DIALEDTIME}
2004 Feb 28
2
questions about anova
...Sq Df Mean Sq F value Pr(>F)
Regression 215.7182 9 23.968693976 2686.508 < 2.22e-16
Deviation 480.1218 53814 0.008921876
Total 695.8401 53823
Multiple R-Squared: 0.31, Adjusted R-squared: 0.3099
AIC: (df = 53814) -146390.9
Fitted:
Min 1Q Median 3Q Max
0.007852 0.075619 0.100498 0.139042 0.338186
Residuals:
Min 1Q Median 3Q Max
-0.29758 -0.04418 -0.01411 0.02536 0.51484
>
So, what's the meaning of the "Pr(>F)?
2 - I have six trend surfaces, and I like to make a anova for the significance
of in...
2011 Nov 20
3
logistic regression by glm
...egression. and treat both response and
predictor as factor
In my first try:
*******************************************************************************
Call:
glm(formula = as.factor(diagnostic) ~ as.factor(7161521) +
as.factor(2281517), family = binomial())
Deviance Residuals:
Min 1Q Median 3Q Max
-1.5370 -1.0431 -0.9416 1.3065 1.4331
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -0.58363 0.27948 -2.088 0.0368 *
as.factor(7161521)2 1.39811 0.66618 2.099 0.0358 *
as.factor(7161521)3 0.28192 0.83255 0.339 0.7349
as.factor(2281517)2 -1.11284 0.63692 -1.747 0.0806 ....
2012 Jan 09
1
Different lm() Residuals Output
All but one of the summaries of multiple linear regressions in this
analysis set present the residuals by min, 1Q, median, 3Q, and max. Example:
lm(formula = TDS ~ Cond + Ca + Cl + Mg + Na + SO4, data = snow.cast)
Residuals:
Min 1Q Median 3Q Max -277.351 -32.551 -2.621
40.812 245.272
The one that doesn't has only a small number of rows (23) and presents the
results as:
lm(formula = TDS...
2003 Nov 03
2
Odd r-squared
...of the same model, and should yield the same r-squared.
Any insight much appreciated
Simon.
> set.seed(10,kind=NULL)
> x <- runif(10)
> g <- gl(2,5)
> y <- runif(10)
>
> summary(lm(y ~ g*x))
Call:
lm(formula = y ~ g * x)
Residuals:
Min 1Q Median 3Q Max
-0.35205 -0.14021 0.02486 0.13958 0.39671
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.3138 0.2749 1.141 0.297
g2 -0.1568 0.4339 -0.361 0.730
x 0.3556 0.6082 0.585 0.580
g2:x 0.3018 1.0522...
2011 Feb 18
3
lm without intercept
Hi,
I am not a statistics expert, so I have this question. A linear model
gives me the following summary:
Call:
lm(formula = N ~ N_alt)
Residuals:
Min 1Q Median 3Q Max
-110.30 -35.80 -22.77 38.07 122.76
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 13.5177 229.0764 0.059 0.9535
N_alt 0.2832 0.1501 1.886 0.0739 .
---
Signif. codes: 0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1
Residual...
2004 Aug 26
1
Why terms are dropping out of an lm() model
...2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
> summary(d$r)
Min. 1st Qu. Median Mean 3rd Qu. Max.
18.68 19.88 21.94 21.48 22.64 24.36
> summary(d.lm1 <- lm(r ~ p1 + p2, data=d))
Call:
lm(formula = r ~ p1 + p2, data = d)
Residuals:
Min 1Q Median 3Q Max
-0.58107 -0.09248 0.02492 0.26061 0.49617
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 18.66417 0.06591 283.17 <2e-16 ***
p1 1.96145 0.04036 48.60 <2e-16 ***
p2 0.85801 0.04036 21.26 <2e-16 ***
---
Signif...
2008 Oct 02
1
missing output in summary() and anova()
...140.04, 142.44, 145.47, 144.34, 146.30, 147.54, 147.80)
> x<-c(194.5, 194.3, 197.9, 198.4, 199.4, 199.9, 200.9, 201.1, 201.4, 201.3,
203.6, 204.6, 209.5,208.6, 210.7, 211.9, 212.2)
> fitted.results<-lm(y~x)
> summary(fitted.results)
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-0.32220 -0.14473 -0.06664 0.02184 1.35978
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -42.13778 3.34020 -12.62 2.18e-09 ***
x 0.89549 0.01645 54.43 < 2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Res...
2013 Nov 08
1
Different output from lm() and lmPerm lmp() if categorical variables are included in the analysis
...+ runif(100,-20,20)
test <- data.frame(x1=testx1,x2=
testx2,y=testy)
atest <- lm(y ~ x1*x2,data=test)
aptest <- lmp(y ~ x1*x2,data=test,perm = "", seqs = TRUE, center = FALSE)
summary(atest)
Call:
lm(formula = y ~ x1 * x2, data = test)
Residuals:
Min 1Q Median 3Q Max
-17.1777 -9.5306 -0.9733 7.6840 22.2728
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -2.0036 3.2488 -0.617 0.539
x1 5.3346 0.2861 18.646 <2e-16 ***
x2b 2.4952 5.2160 0.478 0.633
x1:x2b -0.3833 0.4...
2008 Jan 05
2
Behavior of ordered factors in glm
...alled a "test
of trend"), at least as a starting point.
> age.mdl<-glm(actual~issuecat,data=xxx,family="poisson")
> summary(age.mdl)
Call:
glm(formula = actual ~ issuecat, family = "poisson", data = xxx)
Deviance Residuals:
Min 1Q Median 3Q Max
-0.3190 -0.2262 -0.1649 -0.1221 5.4776
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -4.31321 0.04865 -88.665 <2e-16 ***
issuecat.L 2.12717 0.13328 15.960 <2e-16 ***
issuecat.Q -0.06568 0.11842 -0.555 0.579
issuec...
2007 Dec 07
1
Adding a subset to a glm messes up factors?
...If I dont use subsets then all the factors are shown. I have copied the output from summary for both cases.
Thanks for the help,
Muri
> model<-glm(log(cpue)~year,family=gaussian)
Call:
glm(formula = log(cpue) ~ year, family = gaussian)
Deviance Residuals:
Min 1Q Median 3Q Max
-2.0962 -0.5851 -0.1241 0.4805 3.9236
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.8899 0.1844 4.825 1.42e-06 ***
year1990 -0.6107 0.1925 -3.173 0.00152 **
year1991 -1.7466 0.1902 -9.184 < 2e-16 ***
year1992...
2006 Sep 29
2
scatter3d() model.summary coefficients?
...amples$y, samples$x2, fit="linear",
residuals=TRUE, bg="white", axis.scales=TRUE, grid=TRUE,
ellipsoid=FALSE, xlab="x1", ylab="y", zlab="x2", model.summary=TRUE)
$linear
Call:
lm(formula = y ~ x + z)
Residuals:
Min 1Q Median 3Q Max
-0.096984 -0.022303 0.004758 0.029354 0.091188
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.708945 0.007005 101.20 <2e-16 ***
x 0.278540 0.011262 24.73 <2e-16 ***
z -0.688175 0.011605 -59.30 <2e-1...
2007 Sep 21
1
Stats 101 : lm with/without intercept
...actors a Population variable into 3 levels. I want to
investigate whether that categorical variable has some relation with my
dependent variable, so I go :
lm(Cout.ton ~ ClassePop33000, data=ech2)
Call:
lm(formula = Cout.ton ~ ClassePop33000, data = ech2)
Residuals:
Min 1Q Median 3Q Max
-182.24 -62.91 -22.76 66.38 277.39
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 231.66 11.50 20.141 < 2e-16 ***
ClassePop33000[T.[3000,25000)] -72.91 16.70 -4.366 2.19e-05 ***...
2011 Nov 15
2
Models with ordered and unordered factors
...18 38 62 30 78 34 ...
# $ Day: Factor w/ 3 levels "Day 1","Day 2",..: 1 1 1 1 1 1 2 2 2 2 ...
summary(lm(y~Day,data=dataf)) #Day 2 is not significantly different from
Day 1, but Day 3 is.
#
#Call:
#lm(formula = y ~ Day, data = dataf)
#
#Residuals:
# Min 1Q Median 3Q Max
#-39.833 -14.458 -3.833 13.958 42.167
#
#Coefficients:
# Estimate Std. Error t value Pr(>|t|)
#(Intercept) 29.833 9.755 3.058 0.00797 **
#DayDay 2 18.833 13.796 1.365 0.19234
#DayDay 3 37.000 13.796 2.682 0.01707 *
#---
#Signif. codes: 0 ‘...
2011 Nov 14
3
What is the CADF test criterion="BIC" report?
...ADF test
t-test statistic: -1.389086
p-value: 0.855681
Max lag of the diff. dependent variable: 1.000000
Call:
dynlm(formula = formula(model), start = obs.1, end = obs.T)
Residuals:
Min 1Q Median 3Q Max
-0.79726 -0.20587 -0.03332 0.23840 0.70460
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 24.342321 17.435476 1.396 0.167
trnd 0.009959 0.006941 1.435 0.156
L(y, 1) -0.026068 0.018767 -1.389 0.856
L(d(y), 1) 0.615983 0....
2004 Feb 08
1
APE: compar.gee( )
...uot;
[1] "running glm to get initial regression estimate"
[1] 7.9500000 0.5155172
Call:
formula: alt ~ R
Number of observations: 37
Model:
Link: identity
Variance to Mean Relation: gaussian
Summary of Residuals:
Min 1Q Median 3Q Max
-12.1267954 -9.4267954 -7.4267954 -0.4267954 20.7903982
Coefficients:
Estimate S.E. t Pr(T > |t|)
(Intercept) 9.209602 4.760274 1.934679 0.08798892
R1 3.217194 2.548273 1.262500 0.24130121
Estimated Scale Parameter: 86.39367
"Phylogene...
2010 Dec 15
1
lmList and lapply(... lm) different std. errors
...ays, using the following code: (data
is given at the end of this message)
reg <- lapply(split(TRY, VARIABLE2), function(X){lm(X2 ~ X3, data=X)})
lapply(reg, summary)
Which produces the following:
$`1`
Call:
lm(formula = X2 ~ X3, data = X)
Residuals:
Min 1Q Median 3Q Max
-1.24233 -0.30028 0.03706 0.46170 1.12408
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 3.0705 0.2323 13.215 5.95e-15 ***
X3 0.4744 0.2640 1.797 0.0813 .
---
Signif. codes: 0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1...