Can die influence our attractiveness to mosquitoes? Malaria causes an estimated
ID: 3151528 • Letter: C
Question
Can die influence our attractiveness to mosquitoes? Malaria causes an estimated 700,000 deaths a year so this could be an important question. Study in West Africa, working with a malaria-carrying species of mosquito investigated whether drinking local beer reduced mosquito bites. Researchers opened a container with 50 mosquitoes next to 5 participants who had consumed the local beer and measured the number of mosquitoes that flew from the cage towards the participants. This procedure was also carried out on 6 people who drank water instead of beer. Test the hypothesis that mosquitoes are attracted to individuals based on what type of liquid they consume. Remember to include confidence inter in your conclusion. Additionally, to ensure that the assumption of the you are using are met, test the hypothesis that the standard deviation of the two population are the same (you do not need to include confidence intervals for this hypothesis test). Name the test that you would use to determine if the sample are normally distributed (but you do not need to conduct the test, you can may assume that the populations are normally distributed).Explanation / Answer
Our null hypothesis is average number of mosquitoes that flew towards the participants of beer group is same that of water group.
2) As the two sample is independent we will do a independent sample t-test for equality of means. But indepedent sample sample t-test assumes that the standard deviation of the two population from which the sample is taken are same. To do that we will conduct a test of equality of variance for the two sample based on F-test. We will do it in R.
Codes to do in R:
beer <- c(4,10,11,14,12) # taking the beer group in one vector
water <- c(28,23,26,18,27,25) # taking the water group in another vector
eq_sd<-var.test(beer,water, ratio = 1,
alternative = c("two.sided", "less", "greater"),
conf.level = 0.95)
eq_sd
Result :
F test to compare two variances
data: beer and water
F = 1.084, num df = 4, denom df = 5, p-value = 0.9073
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.1467226 10.1508004
sample estimates:
ratio of variances
1.083969
From the result as the p-value is greater than 0.05 so we will accept the null hypothesis to conclude that the standard deviation of the two population is equal.
3) Saphiro-Wilk test can be done to check whether the data is normally distributed or not.
1) Codes to conduct independent sample t-test
ttest<-t.test(beer,water,var.equal=TRUE) # as in 2) we tested that the variance of the two sample is equal.
ttest
Result:
Two Sample t-test
data: beer and water
t = -6.4063, df = 9, p-value = 0.0001243
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-19.349524 -9.250476
sample estimates:
mean of x mean of y
10.2 24.5
test statistic : -6.4063
p-value : 0.0001243
Conclusion :
As the p-value of the test is much less than 0.05 we will fail to accept the null hypothesis to conclude that the average number of mosquitoes that flew towards the participants of beer group is different from that of water group.
Confidence Interval :
From the result the a 95% confodence interval is
(-19.349524 , -9.250476)