CODEWITHSUNDEEP
rt-testsignificance
adame
adame

Reputation: 137

statistical test in r to show significant difference between groups

I am working with small scale survey data in r.

I would be grateful for input on what would be best/most simple test to use to show any row-wise significance between group differences for a series of options (opt1-opt9). When my data is grouped/aggregated it looks like this (respondents can multi select options):

opt group1_count group1_percent group2_count group2_percent diff_%
opt1 14 0.081395349 17 0.042821159 0.038574
opt2 23 0.13372093 59 0.14861461 -0.01489
opt3 29 0.168604651 65 0.16372796 0.004877
opt4 6 0.034883721 6 0.01511335 0.01977
opt5 2 0.011627907 7 0.017632242 -0.006
opt6 38 0.220930233 88 0.221662469 -0.00073
opt7 37 0.215116279 98 0.246851385 -0.03174
opt8 11 0.063953488 25 0.062972292 0.000981
opt9 12 0.069767442 32 0.080604534 -0.01084

Would a t-test be valid here to show whether there are significant differences between group 1 and group 2? If yes, is there a simple way of generating this row wise in r? If not, do you have any suggestions?

Here is first 3 rows as dput:

structure(list(opt = c("opt1", "opt2", "opt3"), group1_count = c(14, 
23, 29), group1_percent = c(0.081395349, 0.13372093, 0.168604651
), group2_count = c(17, 59, 65), group2_percent = c(0.042821159, 
0.14861461, 0.16372796), percent_diff = c(0.03857419, -0.01489368, 
0.00487669099999999)), row.names = c(NA, -3L), class = c("tbl_df", 
"tbl", "data.frame"))

Many thanks

Upvotes: 0

Views: 908

Answers (1)

David M.
David M.

Reputation: 4588

If you only want to compare the two groups in the first row, you can carry out a two-proportion z-test. For instance in R:

result <- prop.test(x = c(14, 17), n = c(172, 397))

where 172 = sum(group1_count) and 397 = sum(group2_count)

Output:

2-sample test for equality of proportions with continuity correction

data:  c(14, 17) out of c(172, 397)
X-squared = 2.758, df = 1, p-value = 0.09677
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.01105128  0.08819966
sample estimates:
    prop 1     prop 2 
0.08139535 0.04282116

If you want to compare your proportions all at once, you can use a chi-square test:

data <- as.table(cbind(c(14, 23, 29, 6, 2, 38, 37, 11, 12),
                       c(17, 59, 65, 6, 7, 88, 98, 25, 32)))

chisq <- chisq.test(data, simulate.p.value = TRUE)

Output:

Pearson's Chi-squared test with simulated p-value (based on 2000 replicates)

data:  data
X-squared = 6.671, df = NA, p-value = 0.5787

Upvotes: 2

Related Questions