sidee loo xisaabiya T-two test. use the following codes sept by step
knitr::opts_chunk$set(echo = TRUE)
df<-mtcars
df## mpg cyl disp hp drat wt qsec vs am gear carb
## Mazda RX4 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4
## Mazda RX4 Wag 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4
## Datsun 710 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1
## Hornet 4 Drive 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1
## Hornet Sportabout 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2
## Valiant 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1
## Duster 360 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4
## Merc 240D 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2
## Merc 230 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2
## Merc 280 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4
## Merc 280C 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4
## Merc 450SE 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3
## Merc 450SL 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3
## Merc 450SLC 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3
## Cadillac Fleetwood 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4
## Lincoln Continental 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4
## Chrysler Imperial 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4
## Fiat 128 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1
## Honda Civic 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2
## Toyota Corolla 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1
## Toyota Corona 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1
## Dodge Challenger 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2
## AMC Javelin 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2
## Camaro Z28 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4
## Pontiac Firebird 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2
## Fiat X1-9 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1
## Porsche 914-2 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2
## Lotus Europa 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2
## Ford Pantera L 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4
## Ferrari Dino 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6
## Maserati Bora 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8
## Volvo 142E 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2df$am## [1] 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1df$am<-factor(df$am,labels = c("Automatic","Manual"))
table(df$am)##
## Automatic Manual
## 19 13df$am## [1] Manual Manual Manual Automatic Automatic Automatic Automatic
## [8] Automatic Automatic Automatic Automatic Automatic Automatic Automatic
## [15] Automatic Automatic Automatic Manual Manual Manual Automatic
## [22] Automatic Automatic Automatic Automatic Manual Manual Manual
## [29] Manual Manual Manual Manual
## Levels: Automatic Manualsido kale u kaal qaybi data dada ku jirta column (MPG) into manual and automatic
aggregate(mpg~am, data = df, mean)## am mpg
## 1 Automatic 17.14737
## 2 Manual 24.39231cat("is there same or different these two SD'S.\nCalculate assuming that sigma1 =sigma2 is different (pooled method).\n so that take the hypotheses test\n 1. H0 : mean1 =mean1.\n and H1= mean1 is not eqaul to Mean2.\n 2. take the SL = 0.5.\n 3. decision; P-value < 0.05. reject the null hypoth.\n 4. interpretation: there is significant difference between manual and automatic on mpg")## is there same or different these two SD'S.
## Calculate assuming that sigma1 =sigma2 is different (pooled method).
## so that take the hypotheses test
## 1. H0 : mean1 =mean1.
## and H1= mean1 is not eqaul to Mean2.
## 2. take the SL = 0.5.
## 3. decision; P-value < 0.05. reject the null hypoth.
## 4. interpretation: there is significant difference between manual and automatic on mpgweld<-t.test(mpg~am, data = df, var.equal=FALSE, alternative="two.sided")
weld##
## Welch Two Sample t-test
##
## data: mpg by am
## t = -3.7671, df = 18.332, p-value = 0.001374
## alternative hypothesis: true difference in means between group Automatic and group Manual is not equal to 0
## 95 percent confidence interval:
## -11.280194 -3.209684
## sample estimates:
## mean in group Automatic mean in group Manual
## 17.14737 24.39231cat(" we need to caclutate the two standard devaitions of population SD.\n So take the data is known 'SLEEP'\n then i make a two columns caled drig1 fror name= 1, and drug2 name=2 from df1 data ingroup column\n finally calculate using a t-test for t-paired.\n 1. H0:there is no diffrence the mean of the sleep hours of drug1 and drug2.\n H1: there is a diference (mean1 != mean2\n) ")## we need to caclutate the two standard devaitions of population SD.
## So take the data is known 'SLEEP'
## then i make a two columns caled drig1 fror name= 1, and drug2 name=2 from df1 data ingroup column
## finally calculate using a t-test for t-paired.
## 1. H0:there is no diffrence the mean of the sleep hours of drug1 and drug2.
## H1: there is a diference (mean1 != mean2
## )library(tidyr)
df1<-sleep
df1## extra group ID
## 1 0.7 1 1
## 2 -1.6 1 2
## 3 -0.2 1 3
## 4 -1.2 1 4
## 5 -0.1 1 5
## 6 3.4 1 6
## 7 3.7 1 7
## 8 0.8 1 8
## 9 0.0 1 9
## 10 2.0 1 10
## 11 1.9 2 1
## 12 0.8 2 2
## 13 1.1 2 3
## 14 0.1 2 4
## 15 -0.1 2 5
## 16 4.4 2 6
## 17 5.5 2 7
## 18 1.6 2 8
## 19 4.6 2 9
## 20 3.4 2 10df1_wide<-pivot_wider(df1,names_from = group,values_from = extra)
df1_wide## # A tibble: 10 × 3
## ID `1` `2`
## <fct> <dbl> <dbl>
## 1 1 0.7 1.9
## 2 2 -1.6 0.8
## 3 3 -0.2 1.1
## 4 4 -1.2 0.1
## 5 5 -0.1 -0.1
## 6 6 3.4 4.4
## 7 7 3.7 5.5
## 8 8 0.8 1.6
## 9 9 0 4.6
## 10 10 2 3.4t_paired<-t.test(df1_wide$'1', df1_wide$'2', paired = TRUE, alternative= "two.sided")
t_paired##
## Paired t-test
##
## data: df1_wide$"1" and df1_wide$"2"
## t = -4.0621, df = 9, p-value = 0.002833
## alternative hypothesis: true mean difference is not equal to 0
## 95 percent confidence interval:
## -2.4598858 -0.7001142
## sample estimates:
## mean difference
## -1.58cat(" 2.so that of SL= 0.05.\n 3. decision: reject the HO. so that\n 4. interpretation: there is a significant difference. ")## 2.so that of SL= 0.05.
## 3. decision: reject the HO. so that
## 4. interpretation: there is a significant difference.