Question

All work must be done in R programing. Consider this dataset provided to you as prob10.txt c1 t1 c2 t2 c3 t3 c4 t4 2650 3115 2619 2933 2331 2799 2750 3200 1200 1101 1200 1309 1888 1901 1315 980 1541 1358 1401 1499 1256 1238 1625 1421 1545 1910 1652 2028 1449 1901 1399 2002 1956 2999 2066 2880 1777 2898 1999 2798 1599 2710 1754 2765 1434 2689 1702 2402 2430 2589 2789 2899 2332 2300 2250 2741 1902 1910 2028 2100 1888 1901 2000 1899 1530 2329 1660 2332 1501 2298 1478 2287 2008 2485 2104 2871 1987 2650 2100 2520 (2) Read it in and set the row names to “Gene 1” through “Gene 10” It should look like this in R > prob10 c1 t1 c2 t2 c3 t3 c4 t4 Gene 1 2650 3115 2619 2933 2331 2799 2750 3200 Gene 2 1200 1101 1200 1309 1888 1901 1315 980 Gene 3 1541 1358 1401 1499 1256 1238 1625 1421 Gene 4 1545 1910 1652 2028 1449 1901 1399 2002 Gene 5 1956 2999 2066 2880 1777 2898 1999 2798 Gene 6 1599 2710 1754 2765 1434 2689 1702 2402 Gene 7 2430 2589 2789 2899 2332 2300 2250 2741 Gene 8 1902 1910 2028 2100 1888 1901 2000 1899 Gene 9 1530 2329 1660 2332 1501 2298 1478 2287 Gene 10 2008 2485 2104 2871 1987 2650 2100 2520 (3 )Perform a one-sample t-test to compare the hypothesis that the mean of the control expression values is 2000.

Answer 1

R code for this problem

#part a

data=matrix(c(2650, 3115, 2619, 2933, 2331, 2799, 2750 ,3200, 1200,
1101 ,1200, 1309, 1888, 1901, 1315, 980 ,1541 ,1358,
1401, 1499, 1256, 1238, 1625, 1421, 1545, 1910 ,1652,
2028, 1449, 1901, 1399, 2002, 1956, 2999, 2066 ,2880,
1777, 2898, 1999, 2798, 1599, 2710, 1754, 2765, 1434,
2689, 1702, 2402, 2430, 2589, 2789, 2899, 2332, 2300,
2250, 2741, 1902, 1910, 2028, 2100, 1888, 1901, 2000 ,
1899, 1530, 2329, 1660, 2332, 1501, 2298, 1478 ,2287,
2008, 2485, 2104, 2871, 1987, 2650, 2100, 2520),10,8,byrow=F)

#part b
row.names(data)= c("gane1","gane2","gane3",
"gane4","gane5","gane6","gane7",
"gane8","gane9","gane10")
colnames(data)=c("c1","t1","c2","t2","c3","t3","c4","t4")

#part c
test.data=as.vector(data[,c("c1","c2","c3","c4")])
t.test(test.data,mu=2000)
#remark: since p-value of this tet 0.5633 > 0.05
#we can comment that the mean of control expression is 2000 at
#95% confidence.

[please comment below if you have any problem of understanding]